Frank Brothers Solutions for ICSE Class 9 Mathematics Chapter 9 Indices

Exercise 9.1

Question 1. Evaluate:
(i) \( 6^0 \)
(ii) \( \left(\frac{1}{2}\right)^{-3} \)
(iii) \( (2^3)^2 \)
(iv) \( (3^2)^3 \)
(v) \( (0.008)^{\frac{2}{3}} \)
(vi) \( (0.00243)^{-\frac{3}{5}} \)
(vii) \( \sqrt[6]{25^3} \)
(viii) \( \left(2\frac{10}{27}\right)^{\frac{2}{3}} \)
Answer:
(i) \( 6^0 = 1 \)
(ii) \( \left(\frac{1}{2}\right)^{-3} = (2)^3 = 8 \)
(iii) \( (2^3)^2 = 2^6 = 64 \)
(iv) \( (3^2)^3 = 3^6 = 729 \)
(v) \( (0.008)^{\frac{2}{3}} = \left( (0.2)^3 \right)^{\frac{2}{3}} = (0.2)^{3 \times \frac{2}{3}} = (0.2)^2 = 0.04 \)
(vi) \( (0.00243)^{-\frac{3}{5}} = \frac{1}{(0.00243)^{\frac{3}{5}}} = \frac{1}{\left( (0.3)^5 \right)^{\frac{3}{5}}} = \frac{1}{(0.3)^3} = \frac{1}{0.027} \)
(vii) \( \sqrt[6]{25^3} = \sqrt[6]{(5^2)^3} = \sqrt[6]{5^6} = 5^{6 \times \frac{1}{6}} = 5 \)
(viii) \( \left(2\frac{10}{27}\right)^{\frac{2}{3}} = \left(\frac{64}{27}\right)^{\frac{2}{3}} = \left( \left(\frac{4}{3}\right)^3 \right)^{\frac{2}{3}} = \left(\frac{4}{3}\right)^{3 \times \frac{2}{3}} = \left(\frac{4}{3}\right)^2 = \frac{16}{9} \)
In simple words: To solve these, change decimals and fractions into simple base numbers with powers. Then, multiply or subtract the powers using standard exponent rules to find the final value.

Exam Tip: Remember that any non-zero number raised to the power of zero is always 1, and a negative exponent means you should take the reciprocal of the base.

Question 2A. Simplify: \( 9^4 \div 27^{-\frac{2}{3}} \)
Answer:
\( 9^4 \div 27^{-\frac{2}{3}} = \left[(3)^2\right]^4 \div \left[(3)^3\right]^{-\frac{2}{3}} \)
\( = (3)^{2 \times 4} \div (3)^{3 \times \left(-\frac{2}{3}\right)} \) [Using \( (a^m)^n = a^{mn} \)]
\( = (3)^8 \div (3)^{-2} \)
\( = (3)^{8 - (-2)} \) [Using \( a^m \div a^n = a^{m-n} \)]
\( = (3)^{8+2} \)
\( = 3^{10} \)
\( = (3)^{2 \times 5} \)
\( = \left[(3)^2\right]^5 \)
\( = [9]^5 \)
\( = 59049 \)
In simple words: Write both numbers with the same base of 3. Then, use exponent subtraction for division to combine the powers and calculate the final answer.

Exam Tip: Pay close attention to signs when subtracting negative exponents in division, as subtracting a negative power turns into addition.

Question 2B. Simplify: \( 7^{-4} \times (343)^{\frac{2}{3}} \div (49)^{-\frac{1}{2}} \)
Answer:
\( 7^{-4} \times (343)^{\frac{2}{3}} \div (49)^{-\frac{1}{2}} \)
\( = 7^{-4} \times (7^3)^{\frac{2}{3}} \div (7^2)^{-\frac{1}{2}} \)
\( = 7^{-4} \times 7^{3 \times \frac{2}{3}} \div 7^{2 \times \left(-\frac{1}{2}\right)} \)
\( = 7^{-4} \times 7^2 \div 7^{-1} \)
\( = 7^{-4+2-(-1)} \) [Using \( a^m \times a^n = a^{m+n} \) and \( a^m \div a^n = a^{m-n} \)]
\( = 7^{-4+2+1} \)
\( = 7^{-1} \)
\( = \frac{1}{7} \) [Using \( a^{-m} = \frac{1}{a^m} \)]
In simple words: Change 343 and 49 into powers of 7. Add the powers when multiplying and subtract them when dividing to find the simplified result.

Exam Tip: Simplify the base of each term to a prime factor like 7 first, which makes applying the laws of indices much easier.

Question 2C. Simplify: \( \left(\frac{64}{216}\right)^{-\frac{2}{3}} \times \left(\frac{16}{36}\right)^{-\frac{3}{2}} \)
Answer:
\( \left(\frac{64}{216}\right)^{-\frac{2}{3}} \times \left(\frac{16}{36}\right)^{-\frac{3}{2}} \)
\( = \left(\frac{8}{27}\right)^{-\frac{2}{3}} \times \left(\frac{4}{9}\right)^{-\frac{3}{2}} \) [Simplifying the fractions inside the parentheses]
\( = \left( \left(\frac{2}{3}\right)^3 \right)^{-\frac{2}{3}} \times \left( \left(\frac{2}{3}\right)^2 \right)^{-\frac{3}{2}} \)
\( = \left(\frac{2}{3}\right)^{3 \times \left(-\frac{2}{3}\right)} \times \left(\frac{2}{3}\right)^{2 \times \left(-\frac{3}{2}\right)} \)
\( = \left(\frac{2}{3}\right)^{-2} \times \left(\frac{2}{3}\right)^{-3} \)
\( = \left(\frac{3}{2}\right)^2 \times \left(\frac{3}{2}\right)^3 \)
\( = \frac{9}{4} \times \frac{27}{8} \)
\( = \frac{243}{32} \)
In simple words: First, simplify the fractions inside the brackets. Then, apply the negative exponent rule by flipping the fractions and multiply them to get the final answer.

Exam Tip: Simplifying fractions to their lowest terms before applying fractional exponents saves time and prevents calculation errors.

Question 3A. Simplify: \( (a^3)^5 \times a^4 \)
Answer:
\( (a^3)^5 \times a^4 \)
\( = a^{3 \times 5} \times a^4 \) [Using \( (a^m)^n = a^{mn} \)]
\( = a^{15} \times a^4 \)
\( = a^{15+4} \) [Using \( a^m \times a^n = a^{m+n} \)]
\( = a^{19} \)
In simple words: Multiply the powers inside and outside the bracket first, then add that result to the other power to get the final answer.

Exam Tip: Remember the difference: multiply exponents when raising a power to another power, but add exponents when multiplying terms with the same base.

Question 3B. Simplify: \( a^2 \times a^3 \div a^4 \)
Answer:
\( a^2 \times a^3 \div a^4 \)
\( = a^{2+3-4} \) [Using \( a^m \times a^n = a^{m+n} \) and \( a^m \div a^n = a^{m-n} \)]
\( = a^1 \)
\( = a \)
In simple words: Add the exponents when you multiply, and subtract the exponent when you divide.

Exam Tip: Any variable raised to the power of 1 is just the variable itself; do not leave the exponent 1 in your final simplified answer.

Question 3C. Simplify: \( a^{\frac{1}{3}} \div a^{-\frac{2}{3}} \)
Answer:
\( a^{\frac{1}{3}} \div a^{-\frac{2}{3}} \)
\( = a^{\frac{1}{3} - \left(-\frac{2}{3}\right)} \) [Using \( a^m \div a^n = a^{m-n} \)]
\( = a^{\frac{1}{3} + \frac{2}{3}} \)
\( = a^{\frac{1+2}{3}} \)
\( = a^{\frac{3}{3}} \)
\( = a^1 \)
\( = a \)
In simple words: Subtract the negative exponent because of the division, which turns it into addition, and simplify the fraction.

Exam Tip: Be careful with fractional exponents; ensure you find a common denominator before adding or subtracting them.

Question 3D. Simplify: \( a^{-3} \times a^2 \times a^0 \)
Answer:
\( a^{-3} \times a^2 \times a^0 \)
\( = a^{-3+2+0} \) [Using \( a^m \times a^n = a^{m+n} \)]
\( = a^{-1} \)
\( = \frac{1}{a} \)
In simple words: Add all the exponents together. Since one of them is zero, it doesn't change the value, leaving us with a negative power that we write as a fraction.

Exam Tip: Always write your final answer with positive indices unless the question specifically asks otherwise.

Question 3E. Simplify: \( (b^{-2} - a^{-2}) \div (b^{-1} - a^{-1}) \)
Answer:
\( (b^{-2} - a^{-2}) \div (b^{-1} - a^{-1}) \)
\( = \left( \frac{1}{b^2} - \frac{1}{a^2} \right) \div \left( \frac{1}{b} - \frac{1}{a} \right) \)
\( = \left( \frac{a^2 - b^2}{a^2b^2} \right) \div \left( \frac{a - b}{ab} \right) \)
\( = \frac{(a-b)(a+b)}{a^2b^2} \times \frac{ab}{a-b} \)
\( = \frac{a+b}{ab} \)
\( = \frac{a}{ab} + \frac{b}{ab} \)
\( = \frac{1}{b} + \frac{1}{a} = b^{-1} + a^{-1} \)
In simple words: Write the negative powers as fractions, find a common denominator to subtract them, and then simplify by factoring and canceling common terms.

Exam Tip: Use the algebraic identity \( a^2 - b^2 = (a-b)(a+b) \) to easily cancel common binomial terms in the numerator and denominator.

Question 4. Simplify the following expressions:
(i) \( \frac{2^3 \times 3^5 \times 24^2}{12^2 \times 18^3 \times 27} \)
(ii) \( \frac{4^3 \times 3^7 \times 5^6}{5^8 \times 2^7 \times 3^3} \)
(iii) \( \frac{12^2 \times 75^{-2} \times 35 \times 400}{48^2 \times 15^{-3} \times 525} \)
(iv) \( \frac{2^6 \times 5^{-4} \times 3^{-3} \times 4^2}{8^3 \times 15^{-3} \times 25^{-1}} \)
Answer:
(i)
\( \frac{2^3 \times 3^5 \times 24^2}{12^2 \times 18^3 \times 27} \)
\( = \frac{2^3 \times 3^5 \times (2^3 \times 3)^2}{(2^2 \times 3)^2 \times (2 \times 3^2)^3 \times (3^3)} \)
\( = \frac{2^3 \times 3^5 \times 2^6 \times 3^2}{2^4 \times 3^2 \times 2^3 \times 3^6 \times 3^3} \)
\( = \frac{2^{3+6} \times 3^{5+2}}{2^{4+3} \times 3^{2+6+3}} \)
\( = \frac{2^9 \times 3^7}{2^7 \times 3^{11}} \)
\( = 2^{9-7} \times 3^{7-11} \)
\( = 2^2 \times 3^{-4} \)
\( = \frac{2^2}{3^4} = \frac{4}{81} \)

(ii)
\( \frac{4^3 \times 3^7 \times 5^6}{5^8 \times 2^7 \times 3^3} \)
\( = \frac{(2^2)^3 \times 3^{7-3}}{5^{8-6} \times 2^7} \)
\( = \frac{2^6 \times 3^4}{5^2 \times 2^7} \)
\( = \frac{3^4}{5^2 \times 2^{7-6}} \)
\( = \frac{81}{25 \times 2^1} = \frac{81}{50} \)

(iii)
\( \frac{12^2 \times 75^{-2} \times 35 \times 400}{48^2 \times 15^{-3} \times 525} \)
\( = \frac{(2^2 \times 3)^2 \times (7 \times 5) \times (2^4 \times 5^2) \times (3 \times 5)^3}{(2^4 \times 3)^2 \times (3 \times 5^2 \times 7) \times (3 \times 5^2)^2} \)
\( = \frac{2^4 \times 3^2 \times 7 \times 5 \times 2^4 \times 5^2 \times 3^3 \times 5^3}{2^8 \times 3^2 \times 3 \times 5^2 \times 7 \times 3^2 \times 5^4} \)
\( = \frac{2^{4+4} \times 3^{2+3} \times 5^{1+2+3} \times 7}{2^8 \times 3^{2+1+2} \times 5^{2+4} \times 7} \)
\( = \frac{2^8 \times 3^5 \times 5^6 \times 7}{2^8 \times 3^5 \times 5^6 \times 7} \)
\( = 1 \)

(iv)
\( \frac{2^6 \times 5^{-4} \times 3^{-3} \times 4^2}{8^3 \times 15^{-3} \times 25^{-1}} \)
\( = \frac{2^6 \times (2^2)^2 \times (3 \times 5)^3 \times (5^2)^1}{(2^3)^3 \times 5^4 \times 3^3} \)
\( = \frac{2^6 \times 2^4 \times 3^3 \times 5^3 \times 5^2}{2^9 \times 5^4 \times 3^3} \)
\( = \frac{2^{6+4} \times 3^3 \times 5^{3+2}}{2^9 \times 3^3 \times 5^4} \)
\( = \frac{2^{10} \times 3^3 \times 5^5}{2^9 \times 3^3 \times 5^4} \)
\( = 2^{10-9} \times 3^{3-3} \times 5^{5-4} \)
\( = 2^1 \times 1 \times 5^1 = 2 \times 5 = 10 \)
In simple words: Break down large numbers like 24, 12, 15, and 75 into prime bases like 2, 3, 5, and 7. Move negative powers across the fraction bar to make them positive, then add or subtract exponents for each base to simplify.

Exam Tip: Splitting composite bases into their prime factors (e.g., \(24 = 2^3 \times 3\)) at the beginning avoids arithmetic mistakes and ensures all powers are easily simplified.

Question 5A. Simplify: \( 3p^{-2}q^3 \div 2p^3q^{-2} \)
Answer:
\( 3p^{-2}q^3 \div 2p^3q^{-2} = \frac{3p^{-2}q^3}{2p^3q^{-2}} \)
\( = \frac{3}{2} \left[ \frac{p^{-2}}{p^3} \times \frac{q^3}{q^{-2}} \right] \)
\( = \frac{3}{2} \left[ (p^{-2-3}) \times (q^{3-(-2)}) \right] \) [Using \( a^m \div a^n = a^{m-n} \)]
\( = \frac{3}{2} \left[ (p^{-5}) \times (q^5) \right] \)
\( = \frac{3}{2} \left[ \left(\frac{1}{p^5}\right) \times (q^5) \right] \)
\( = \frac{3q^5}{2p^5} \)
In simple words: Write the division as a fraction. Subtract the exponents of like variables and rewrite any negative power as a positive power in the denominator.

Exam Tip: Always keep the numerical coefficients (like 3 and 2) separate from the variable exponents to prevent accidental mixing during simplification.

Question 5B. Simplify: \( \left[ (p^{-3})^{\frac{2}{3}} \right]^{\frac{1}{2}} \)
Answer:
\( \left[ (p^{-3})^{\frac{2}{3}} \right]^{\frac{1}{2}} = p^{-3 \times \frac{2}{3} \times \frac{1}{2}} \) [Using \( (a^m)^n = a^{mn} \)]
\( = p^{-1} \)
\( = \frac{1}{p} \)
In simple words: Multiply all the nested powers together. Since they cancel out to leave negative one, write the variable as a fraction.

Exam Tip: When dealing with nested parentheses, you can multiply all the powers together in a single step to save time.

Question 6. Evaluate the following:
(i) \( \left(1 - \frac{15}{64}\right)^{-\frac{1}{2}} \)
(ii) \( \left(\frac{8}{27}\right)^{-\frac{2}{3}} - \left(\frac{1}{3}\right)^{-2} - 7^0 \)
(iii) \( 9^{\frac{5}{2}} - 3 \times 5^0 - \left(\frac{1}{81}\right)^{-\frac{1}{2}} \)
(iv) \( (27)^{\frac{2}{3}} \times 8^{-\frac{1}{6}} \div 18^{-\frac{1}{2}} \)
(v) \( 16^{\frac{3}{4}} + 2 \left(\frac{1}{2}\right)^{-1} \times 3^0 \)
(vi) \( \sqrt{\frac{1}{4}} + (0.01)^{-\frac{1}{2}} - (27)^{\frac{2}{3}} \)
Answer:
(i)
\( \left(1 - \frac{15}{64}\right)^{-\frac{1}{2}} = \left(\frac{64 - 15}{64}\right)^{-\frac{1}{2}} \)
\( = \left(\frac{49}{64}\right)^{-\frac{1}{2}} \)
\( = \left(\frac{64}{49}\right)^{\frac{1}{2}} \)
\( = \frac{8}{7} \)

(ii)
\( \left(\frac{8}{27}\right)^{-\frac{2}{3}} - \left(\frac{1}{3}\right)^{-2} - 7^0 \)
\( = \left(\frac{27}{8}\right)^{\frac{2}{3}} - (3)^2 - 1 \)
\( = \left( \left(\frac{3}{2}\right)^3 \right)^{\frac{2}{3}} - 9 - 1 \)
\( = \left(\frac{3}{2}\right)^2 - 10 \)
\( = \frac{9}{4} - 10 = \frac{9 - 40}{4} = \frac{-31}{4} \)

(iii)
\( 9^{\frac{5}{2}} - 3 \times 5^0 - \left(\frac{1}{81}\right)^{-\frac{1}{2}} \)
\( = (3^2)^{\frac{5}{2}} - 3 \times 1 - (81)^{\frac{1}{2}} \)
\( = 3^5 - 3 - 9 \)
\( = 243 - 3 - 9 \)
\( = 231 \)

(iv)
\( (27)^{\frac{2}{3}} \times 8^{-\frac{1}{6}} \div 18^{-\frac{1}{2}} \)
\( = (3^3)^{\frac{2}{3}} \times \frac{1}{8^{\frac{1}{6}}} \div \frac{1}{18^{\frac{1}{2}}} \)
\( = 3^2 \times \frac{1}{(2^3)^{\frac{1}{6}}} \div \frac{1}{(2 \times 3^2)^{\frac{1}{2}}} \)
\( = \frac{3^2}{2^{\frac{1}{2}}} \times (2 \times 3^2)^{\frac{1}{2}} \)
\( = \frac{3^2}{2^{\frac{1}{2}}} \times 2^{\frac{1}{2}} \times 3 \)
\( = 3^2 \times 3 = 3^3 = 27 \)

(v)
\( 16^{\frac{3}{4}} + 2 \left(\frac{1}{2}\right)^{-1} \times 3^0 \)
\( = (2^4)^{\frac{3}{4}} + 2 \times 2 \times 1 \)
\( = 2^3 + 4 \)
\( = 8 + 4 = 12 \)

(vi)
\( \sqrt{\frac{1}{4}} + (0.01)^{-\frac{1}{2}} - (27)^{\frac{2}{3}} \)
\( = \left(\left(\frac{1}{2}\right)^2\right)^{\frac{1}{2}} + (0.1)^{-1} - (3^3)^{\frac{2}{3}} \)
\( = \frac{1}{2} + \frac{1}{0.1} - 3^2 \)
\( = \frac{1}{2} + 10 - 9 \)
\( = \frac{1}{2} + 1 \)
\( = \frac{3}{2} \)
In simple words: Solve each part step-by-step by turning decimals and fractions into their prime base powers. Simplify negative powers by flipping the numbers, then calculate the basic arithmetic to get the final total.

Exam Tip: Be mindful of square roots and cube roots; converting decimals like \(0.01\) to fractional exponents makes them much simpler to handle.

Question 7A. Simplify: \( (27x^9)^{\frac{2}{3}} \)
Answer:
\( (27x^9)^{\frac{2}{3}} = (3^3 \cdot x^9)^{\frac{2}{3}} \)
\( = (3^3)^{\frac{2}{3}} \cdot (x^9)^{\frac{2}{3}} \) [Using \( (a \times b)^n = a^n \times b^n \)]
\( = (3)^{3 \times \frac{2}{3}} \cdot (x)^{9 \times \frac{2}{3}} \) [Using \( (a^m)^n = a^{mn} \)]
\( = (3)^2 \cdot x^6 \)
\( = 9x^6 \)
In simple words: Apply the exponent outside the bracket to both the number and the variable inside. Multiply the powers and simplify.

Exam Tip: Do not forget to apply the outer exponent to the numeric coefficient as well as the variables inside the parenthesis.

Question 7B. Simplify: \( (8x^6y^3)^{\frac{2}{3}} \)
Answer:
\( (8x^6y^3)^{\frac{2}{3}} = (2^3 \cdot x^6 \cdot y^3)^{\frac{2}{3}} \)
\( = (2^3)^{\frac{2}{3}} \cdot (x^6)^{\frac{2}{3}} \cdot (y^3)^{\frac{2}{3}} \) [Using \( (a \times b)^n = a^n \times b^n \)]
\( = (2)^{3 \times \frac{2}{3}} \cdot (x)^{6 \times \frac{2}{3}} \cdot (y)^{3 \times \frac{2}{3}} \) [Using \( (a^m)^n = a^{mn} \)]
\( = (2)^2 \cdot (x)^4 \cdot (y)^2 \)
\( = 4x^4y^2 \)
In simple words: Distribute the power of two-thirds to all three factors inside the bracket, multiply their exponents, and simplify the base numbers.

Exam Tip: Clearly write out the step where you distribute the power to each individual term to show full working out.

Question 7C. Simplify: \( \left(\frac{64a^{12}}{27b^6}\right)^{-\frac{2}{3}} \)
Answer:
\( \left(\frac{64a^{12}}{27b^6}\right)^{-\frac{2}{3}} = \left(\frac{2^6 a^{12}}{3^3 b^6}\right)^{-\frac{2}{3}} \)
\( = \frac{2^{6 \times \left(-\frac{2}{3}\right)} \cdot a^{12 \times \left(-\frac{2}{3}\right)}}{3^{3 \times \left(-\frac{2}{3}\right)} \cdot b^{6 \times \left(-\frac{2}{3}\right)}} \) [Using \( (ab)^n = a^b \times b^n \) and \( \left(\frac{a}{b}\right)^n = \frac{a^n}{b^n} \)]
\( = \frac{2^{-4} a^{-8}}{3^{-2} b^{-4}} \)
\( = \frac{3^2 b^4}{2^4 a^8} \) [Using \( a^{-m} = \frac{1}{a^m} \)]
\( = \frac{9b^4}{16a^8} \)
In simple words: Write numbers as powers of 2 and 3. Multiply all powers inside by the negative fraction power, then flip negative powers to make them positive.

Exam Tip: Alternatively, you can first flip the entire fraction to change the negative outer exponent to a positive one, which makes the subsequent steps less prone to sign errors.

Question 7D. Simplify: \( \left(\frac{36m^{-4}}{49n^{-2}}\right)^{-\frac{3}{2}} \)
Answer:
\( \left(\frac{36m^{-4}}{49n^{-2}}\right)^{-\frac{3}{2}} = \left(\frac{6^2 m^{-4}}{7^2 n^{-2}}\right)^{-\frac{3}{2}} \)
\( = \frac{6^{2 \times \left(-\frac{3}{2}\right)} \cdot m^{-4 \times \left(-\frac{3}{2}\right)}}{7^{2 \times \left(-\frac{3}{2}\right)} \cdot n^{-2 \times \left(-\frac{3}{2}\right)}} \) [Using \( (a \times b)^n = a^n \times b^n \) and \( \left(\frac{a}{b}\right)^n = \frac{a^n}{b^n} \)]
\( = \frac{6^{-3} m^6}{7^{-3} n^3} \)
\( = \frac{7^3 m^6}{6^3 n^3} \) [Using \( a^{-m} = \frac{1}{a^m} \)]
\( = \frac{343m^6}{216n^3} \)
In simple words: Represent 36 and 49 as squares. Distribute the outer fraction power, multiply the powers together carefully, and flip any negative powers to simplify.

Exam Tip: Be extremely careful with signs; multiplying two negative exponents (like -4 and -3/2) results in a positive power.

Question 7E. Simplify: \( \left( a^{\frac{1}{3}} + a^{-\frac{1}{3}} \right) \left( a^{\frac{2}{3}} - 1 + a^{-\frac{2}{3}} \right) \)
Answer:
\( \left( a^{\frac{1}{3}} + a^{-\frac{1}{3}} \right) \left( a^{\frac{2}{3}} - 1 + a^{-\frac{2}{3}} \right) \)
\( = a^{\frac{1}{3}} \left( a^{\frac{2}{3}} - 1 + a^{-\frac{2}{3}} \right) + a^{-\frac{1}{3}} \left( a^{\frac{2}{3}} - 1 + a^{-\frac{2}{3}} \right) \)
\( = \left( a^{\frac{1}{3}} \cdot a^{\frac{2}{3}} - a^{\frac{1}{3}} \cdot 1 + a^{\frac{1}{3}} \cdot a^{-\frac{2}{3}} \right) + \left( a^{-\frac{1}{3}} \cdot a^{\frac{2}{3}} - a^{-\frac{1}{3}} \cdot 1 + a^{-\frac{1}{3}} \cdot a^{-\frac{2}{3}} \right) \)
\( = \left( a^{\frac{1}{3}+\frac{2}{3}} - a^{\frac{1}{3}} + a^{\frac{1}{3}-\frac{2}{3}} \right) + \left( a^{-\frac{1}{3}+\frac{2}{3}} - a^{-\frac{1}{3}} + a^{-\frac{1}{3}-\frac{2}{3}} \right) \) [Using \( a^m \times a^n = a^{m+n} \)]
\( = \left( a^1 - a^{\frac{1}{3}} + a^{-\frac{1}{3}} \right) + \left( a^{\frac{1}{3}} - a^{-\frac{1}{3}} + a^{-1} \right) \)
\( = a - a^{\frac{1}{3}} + a^{-\frac{1}{3}} + a^{\frac{1}{3}} - a^{-\frac{1}{3}} + \frac{1}{a} \)
\( = a + \frac{1}{a} \)
In simple words: Multiply each term of the first bracket by each term of the second. Add the exponents of like terms, and cancel out the opposite terms to get the final answer.

Exam Tip: This expression follows the identity \( (x+y)(x^2 - xy + y^2) = x^3 + y^3 \) where \( x = a^{1/3} \) and \( y = a^{-1/3} \). Recognizing this allows you to find the answer directly as \( a + a^{-1} \).

Question 7F. Simplify: \( \sqrt[3]{x^4y^2} \div \sqrt[6]{x^5y^{-5}} \)
Answer:
\( \sqrt[3]{x^4y^2} \div \sqrt[6]{x^5y^{-5}} \)
\( = (x^4y^2)^{\frac{1}{3}} \div (x^5y^{-5})^{\frac{1}{6}} \)
\( = \left( x^{4 \times \frac{1}{3}} y^{2 \times \frac{1}{3}} \right) \div \left( x^{5 \times \frac{1}{6}} y^{-5 \times \frac{1}{6}} \right) \) [Using \( (a^m)^n = a^{mn} \)]
\( = \left( x^{\frac{4}{3}} y^{\frac{2}{3}} \right) \div \left( x^{\frac{5}{6}} y^{-\frac{5}{6}} \right) \)
\( = \frac{x^{\frac{4}{3}} y^{\frac{2}{3}}}{x^{\frac{5}{6}} y^{-\frac{5}{6}}} \)
\( = x^{\frac{4}{3} - \frac{5}{6}} y^{\frac{2}{3} - \left(-\frac{5}{6}\right)} \) [Using \( a^m \div a^n = a^{m-n} \)]
\( = x^{\frac{1}{2}} y^{\frac{3}{2}} \)
\( = x^{\frac{1}{2}} (y^3)^{\frac{1}{2}} \) [Using \( (a^m)^n = a^{mn} \)]
\( = \sqrt{x} \sqrt{y^3} \)
\( = \sqrt{xy^3} \)
In simple words: Turn the roots into fractional powers. Subtract the powers for division, and combine them back under a single square root at the end.

Exam Tip: When subtracting fractions, always find a common denominator (for example, \( \frac{4}{3} - \frac{5}{6} = \frac{8-5}{6} = \frac{3}{6} = \frac{1}{2} \)) before simplifying.

Question 7G. Simplify: \( \left\{ (a^m)^{m - \frac{1}{m}} \right\}^{\frac{1}{m+1}} \)
Answer:
\( \left\{ (a^m)^{m - \frac{1}{m}} \right\}^{\frac{1}{m+1}} = (a)^{m \times \left( m - \frac{1}{m} \right) \times \left( \frac{1}{m+1} \right)} \) [Using \( (a^m)^n = a^{mn} \)]
Consider the product of the exponents: \( m \times \left( m - \frac{1}{m} \right) \times \left( \frac{1}{m+1} \right) \)
\( = \left(m^2 - 1\right) \times \left( \frac{1}{m+1} \right) \)
\( = m^2 \times \left( \frac{1}{m+1} \right) - 1 \times \left( \frac{1}{m+1} \right) \)
\( = \frac{m^2}{m+1} - \frac{1}{m+1} \)
\( = \frac{m^2 - 1}{m+1} \)
\( = \frac{(m-1)(m+1)}{m+1} \)
\( = m - 1 \)
Substituting this back:
\( (a)^{m \times \left( m - \frac{1}{m} \right) \times \left( \frac{1}{m+1} \right)} = a^{m-1} \)
In simple words: Multiply all the powers of \(a\) together. Use algebraic factoring on the exponents to simplify the term down to a single power of \(m-1\).

Exam Tip: Factorize the expression \( m^2 - 1 \) as \( (m-1)(m+1) \) to easily cancel out the factor of \( m+1 \) in the denominator.

Question 7H. Simplify: \( x^{m+2n} \cdot x^{3m-8n} \div x^{5m-60} \)
Answer:
\( x^{m+2n} \cdot x^{3m-8n} \div x^{5m-60} \)
\( = x^{m+2n+3m-8n-(5m-60)} \) [Using \( a^m \times a^n = a^{m+n} \) and \( a^m \div a^n = a^{m-n} \)]
\( = x^{m+2n+3m-8n-5m+60} \)
\( = x^{-m-6n+60} \)
In simple words: Add the powers when multiplying terms and subtract the power when dividing. Combine the like terms of \(m\) and \(n\) to get the final power.

Exam Tip: Be careful to apply the negative sign to all terms inside the parentheses when subtracting the dividing exponent.

Question 7I. Simplify: \( (81)^{\frac{3}{4}} - \left(\frac{1}{32}\right)^{-\frac{2}{5}} + 8^{\frac{1}{3}} \left(\frac{1}{2}\right)^{-1} \cdot 2^0 \)
Answer:
\( (81)^{\frac{3}{4}} - \left(\frac{1}{32}\right)^{-\frac{2}{5}} + 8^{\frac{1}{3}} \left(\frac{1}{2}\right)^{-1} \cdot 2^0 \)
\( = (3^4)^{\frac{3}{4}} - \left(\frac{1}{2^5}\right)^{-\frac{2}{5}} + (2^3)^{\frac{1}{3}} \cdot (2)^1 \times 1 \) [Using \( a^0 = 1 \)]
\( = 3^{4 \times \frac{3}{4}} - \frac{1}{2^{5 \times \left(-\frac{2}{5}\right)}} + 2^{3 \times \frac{1}{3}} \cdot (2)^1 \) [Using \( (a^m)^n = a^{mn} \)]
\( = 3^3 - \frac{1}{2^{-2}} + 2^1 \cdot (2)^1 \)
\( = 3^3 - 2^2 + 2^{1+1} \) [Using \( a^m \times a^n = a^{m+n} \), \( a^m \div a^n = a^{m-n} \), and \( \frac{1}{a^{-1}} = a^1 \)]
\( = 3^3 - 2^2 + 2^2 \)
\( = 27 \)
In simple words: Convert each base into its prime power. Simplify the fractional powers, resolve negative exponents, and then add or subtract the results.

Exam Tip: Notice that \( -2^2 + 2^2 \) cancels out, leaving only \( 3^3 \) as the final answer, which saves calculation time.

Question 7J. Simplify: \( \left(\frac{27}{343}\right)^{\frac{2}{3}} \div \frac{1}{\left(\frac{625}{1296}\right)^{\frac{1}{4}}} \times \frac{536}{\sqrt[3]{27}} \)
Answer:
\( \left(\frac{27}{343}\right)^{\frac{2}{3}} \div \frac{1}{\left(\frac{625}{1296}\right)^{\frac{1}{4}}} \times \frac{536}{\sqrt[3]{27}} \)
\( = \left(\frac{3^3}{7^3}\right)^{\frac{2}{3}} \div \frac{1}{\left(\frac{5^4}{2^4 \times 3^4}\right)^{\frac{1}{4}}} \times \frac{2^3 \times 67}{\sqrt[3]{3^3}} \)
\( = \frac{3^2}{7^2} \div \frac{1}{\left(\frac{5}{2 \times 3}\right)} \times \frac{2^3 \times 67}{3} \)
\( = \frac{3^2}{7^2} \div \left(\frac{2 \times 3}{5}\right) \times \frac{2^3 \times 67}{3} \)
\( = \frac{3^2}{7^2} \times \frac{5}{2 \times 3} \times \frac{2^3 \times 67}{3} \)
\( = \frac{3^2 \times 5 \times 2^3 \times 67}{7^2 \times 2 \times 3 \times 3} \)
\( = \frac{3^2 \times 5 \times 2^3 \times 67}{7^2 \times 2 \times 3^2} \)
\( = \frac{5 \times 2^2 \times 67}{7^2} \)
\( = \frac{5 \times 4 \times 67}{49} \)
\( = \frac{1340}{49} \)
In simple words: Convert the large numbers into prime bases, simplify the fractional powers, turn division into multiplication by flipping fractions, and then simplify the remaining terms.

Exam Tip: Always make sure to write powers in the denominator with a negative exponent if you bring them to the numerator, or keep them in the denominator during simplification.

Question 8. Simplify the following indices expressions:
(i) \( \frac{5^x \times 7 - 5^x}{5^{x+2} - 5^{x+1}} \)
(ii) \( \frac{3^{x+1} + 3^x}{3^{x+3} - 3^{x+1}} \)
(iii) \( \frac{2^m \times 3 - 2^m}{2^{m+4} - 2^{m+1}} \)
(iv) \( \frac{5^{n+2} - 6 \cdot 5^{n+1}}{13 \cdot 5^n - 2 \cdot 5^{n+1}} \)
Answer:
(i)
\( \frac{5^x \times 7 - 5^x}{5^{x+2} - 5^{x+1}} = \frac{5^x (7 - 1)}{5^{x+1} (5 - 1)} \)
\( = \frac{5^x \times 6}{5^x \times 5^1 \times 4} \)
\( = \frac{6}{5 \times 4} = \frac{6}{20} = \frac{3}{10} \)

(ii)
\( \frac{3^{x+1} + 3^x}{3^{x+3} - 3^{x+1}} = \frac{3^x (3 + 1)}{3^{x+1} (3^2 - 1)} \)
\( = \frac{3^x \times 4}{3^x \times 3^1 \times (9 - 1)} \)
\( = \frac{4}{3 \times 8} = \frac{4}{24} = \frac{1}{6} \)

(iii)
\( \frac{2^m \times 3 - 2^m}{2^{m+4} - 2^{m+1}} = \frac{2^m (3-1)}{2^m (2^4 - 2^1)} \)
\( = \frac{2}{16-2} = \frac{2}{14} = \frac{1}{7} \)

(iv)
\( \frac{5^{n+2} - 6 \cdot 5^{n+1}}{13 \cdot 5^n - 2 \cdot 5^{n+1}} = \frac{5^n (5^2 - 6 \times 5)}{5^n (13 - 2 \times 5)} \)
\( = \frac{25 - 30}{13 - 10} = \frac{-5}{3} \)
In simple words: Factor out the common base raised to the variable power from both the numerator and the denominator. Once they cancel out, simplify the remaining numbers to get the answer.

Exam Tip: Factoring out the variable power (like \(5^x\) or \(2^m\)) first is the easiest way to solve these kinds of problems, as it eliminates the algebraic variable completely.

Question 9A. Solve for \( x \): \( 2^{2x+1} = 8 \)
Answer:
\( 2^{2x+1} = 8 \)
\implies \( 2^{2x+1} = 2^3 \)
\implies \( 2x + 1 = 3 \)
\implies \( 2x = 2 \)
\implies \( x = 1 \)
In simple words: Write 8 as a power of 2 so that both sides have the same base. Then, set the exponents equal to each other and solve for \( x \).

Exam Tip: When bases are equal on both sides of an equation, you can directly equate their exponents to solve for the variable.

Question 9B. Solve for \( x \): \( 3 \times 7^x = 7 \times 3^x \)
Answer:
\( 3 \times 7^x = 7 \times 3^x \)
\implies \( \frac{7^x}{7} = \frac{3^x}{3} \)
\implies \( 7^{x-1} = 3^{x-1} \) [Using \( a^m \div a^n = a^{m-n} \)]
\implies \( 7^{x-1} = 3^{x-1} \times 1 \)
\implies \( \frac{7^{x-1}}{3^{x-1}} = 1 \)
\implies \( \left(\frac{7}{3}\right)^{x-1} = \left(\frac{7}{3}\right)^0 \) [Using \( a^0 = 1 \)]
\implies \( x - 1 = 0 \)
\implies \( x = 1 \)
In simple words: Group the terms with the same bases together. Since two different bases with the same exponent can only be equal when the exponent is zero, solve for \( x \) by setting the exponent to zero.

Exam Tip: Remember that if \( a^y = b^y \) where \( a \neq b \), then \( y \) must equal 0.

Question 9C. Solve for \( x \): \( 2^{x+3} + 2^{x+1} = 320 \)
Answer:
\( 2^{x+3} + 2^{x+1} = 320 \)
\implies \( 2^{x+3} + 2^{x+1} = 2^6 \times 5 \)
\implies \( 2^x \cdot 2^3 + 2^x \cdot 2^1 = 2^6 \times 5 \)
\implies \( 2^x (2^3 + 2^1) = 2^6 \times 5 \)
\implies \( 2^x (8 + 2) = 2^6 \times 5 \)
\implies \( 2^x (10) = 2^6 \times 5 \)
\implies \( 2^x \left(\frac{10}{5}\right) = 2^6 \)
\implies \( 2^x \cdot 2 = 2^6 \)
\implies \( \frac{2^{x+1}}{2^6} = 1 \)
\implies \( 2^{x+1-6} = 1 \times 2^0 \)
\implies \( 2^{x-5} = 2^0 \)
\implies \( x - 5 = 0 \)
\implies \( x = 5 \)
In simple words: Factor out \( 2^x \) on the left side and write 320 as prime factors on the right. Simplify both sides to find that \( 2^x = 32 \), which means \( x = 5 \).

Exam Tip: Factoring out \( 2^x \) turns a addition of indices into a multiplication, which can then be easily simplified with the other side of the equation.

Question 9D. Solve for \( x \): \( 9 \times 3^x = (27)^{2x-5} \)
Answer:
\( 9 \times 3^x = (27)^{2x-5} \)
\implies \( 3^2 \times 3^x = (3^3)^{2x-5} \)
\implies \( 3^2 \times 3^x = 3^{3(2x-5)} \)
\implies \( 3^{2+x} = 3^{6x-15} \)
\implies \( 1 = \frac{3^{6x-15}}{3^{2+x}} \)
\implies \( 1 = 3^{6x-15-2-x} \)
\implies \( 3^0 = 3^{5x-17} \)
\implies \( 5x - 17 = 0 \)
\implies \( x = \frac{17}{5} \)
In simple words: Convert 9 and 27 to powers of 3. Combine the powers on both sides, equate the exponents, and solve the simple equation for \( x \).

Exam Tip: Be careful when distributing the exponent on the right side: \( 3(2x-5) \) becomes \( 6x - 15 \), not \( 6x - 5 \).

Question 9E. Solve for \( x \): \( 2^{2x+3} - 9 \times 2^x + 1 = 0 \)
Answer:
\( 2^{2x+3} - 9 \times 2^x + 1 = 0 \)
\( \implies 2^{2x} \cdot 2^3 - 9 \times 2^x + 1 = 0 \)
Let \( 2^x = t \), which means \( 2^{2x} = t^2 \).
Substituting these into the equation, we get:
\( 8t^2 - 9t + 1 = 0 \)
\( \implies 8t^2 - 8t - t + 1 = 0 \)
\( \implies 8t(t-1) - (t-1) = 0 \)
\( \implies (t-1)(8t-1) = 0 \)
\( \implies t-1 = 0 \) or \( 8t-1 = 0 \)
\( \implies t = 1 \) or \( t = \frac{1}{8} \)
Now, replacing \( t \) with \( 2^x \):
\( \implies 2^x = 1 \) or \( 2^x = \frac{1}{2^3} \)
\( \implies 2^x = 2^0 \) or \( 2^x = 2^{-3} \)
\( \implies x = 0 \) or \( x = -3 \)
In simple words: Solve the given quadratic-like exponential equation by substituting \( 2^x = t \), finding the values of \( t \), and then solving back for \( x \).

Exam Tip: Substitution is extremely helpful in simplifying exponential equations that look like quadratics. Remember to solve for the original variable \( x \) at the end, not just \( t \).

Question 9F. Solve for \( x \): \( 1 = p^x \)
Answer:
\( 1 = p^x \)
\( \implies p^0 = p^x \) (Applying the exponent property where \( a^0 = 1 \))
\( \implies x = 0 \)
In simple words: Any non-zero number raised to the power of 0 is always equal to 1. Comparing the exponents gives the solution.

Exam Tip: Always remember the fundamental rule of indices where \( a^0 = 1 \) for any non-zero base \( a \).

Question 9G. Solve for \( x \): \( p^3 \times p^{-2} = p^x \)
Answer:
\( p^3 \times p^{-2} = p^x \)
\( \implies p^{3+(-2)} = p^x \) (Using the property \( a^m \times a^n = a^{m+n} \))
\( \implies p^1 = p^x \)
\( \implies x = 1 \)
In simple words: Multiply bases by adding their exponents. Then, equate the power of the left side to the right side.

Exam Tip: Ensure that you keep track of negative signs when adding exponents of the same base.

Question 9H. Solve for \( x \): \( p^{-5} = \frac{1}{p^{x+1}} \)
Answer:
\( p^{-5} = \frac{1}{p^{x+1}} \)
\( \implies p^{-5} \times p^{x+1} = 1 \)
\( \implies p^{-5+x+1} = 1 \)
\( \implies p^{x-4} = p^0 \)
\( \implies x-4 = 0 \)
\( \implies x = 4 \)
In simple words: Write the reciprocal term with a negative exponent, combine the exponents on the left, and compare them with the right side.

Exam Tip: Be careful with signs when transferring terms between the numerator and denominator using negative powers.

Question 9I. Solve for \( x \): \( 2^{2x} + 2^{x+2} - 4 \times 2^3 = 0 \)
Answer:
\( 2^{2x} + 2^{x+2} - 4 \times 2^3 = 0 \)
\( \implies 2^{2x} + 2^{x+2} - 2^2 \times 2^3 = 0 \)
\( \implies 2^{2x} + 2^{x+2} - 2^{2+3} = 0 \) (Applying \( a^m \times a^n = a^{m+n} \))
\( \implies 2^{2x} + 2^x \cdot 2^2 - 2^5 = 0 \)
\( \implies 2^{2x} + 4 \cdot 2^x - 32 = 0 \)
Let \( 2^x = t \), which gives \( 2^{2x} = t^2 \). This simplifies the equation to:
\( t^2 + 4t - 32 = 0 \)
\( \implies (t+8)(t-4) = 0 \)
\( \implies t+8 = 0 \) or \( t-4 = 0 \)
\( \implies t = -8 \) or \( t = 4 \)
\( \implies 2^x = -8 \) or \( 2^x = 4 \)
Since a real positive base raised to a power cannot yield a negative result, we discard \( 2^x = -8 \).
By comparing the exponents in the second equation \( 2^x = 2^2 \), we obtain:
\( \implies x = 2 \)
In simple words: Substitute \( 2^x = t \) to turn the exponential equation into a standard quadratic equation, solve it, and discard any impossible negative base solutions.

Exam Tip: Real bases raised to a variable exponent cannot yield negative results, so always discard negative roots when solving back.

Question 9J. Solve for \( x \): \( 9 \times 81 = \frac{1}{27^{x-3}} \)
Answer:
\( 9 \times 81 = \frac{1}{27^{x-3}} \)
\( \implies 3^2 \times 3^4 = \frac{1}{3^{3(x-3)}} \)
\( \implies 3^6 = \frac{1}{3^{3x-9}} \) (Applying the rule \( (a^m)^n = a^{mn} \))
\( \implies 3^6 \times 3^{3x-9} = 1 \)
\( \implies 3^{6 + 3x - 9} = 3^0 \)
\( \implies 3x - 3 = 0 \)
\( \implies 3x = 3 \)
\( \implies x = 1 \)
In simple words: Express both sides using a common base of 3, simplify the exponents using indices rules, and equate the powers.

Exam Tip: Finding a common prime base is the most standard and efficient way to solve exponent-based equations.

Question 9K. Solve for \( x \): \( 2^{2x-1} - 9 \times 2^{x-2} + 1 = 0 \)
Answer:
\( 2^{2x-1} - 9 \times 2^{x-2} + 1 = 0 \)
\( \implies 2^{2x} \cdot 2^{-1} - 9 \times 2^x \cdot 2^{-2} + 1 = 0 \)
Let \( 2^x = t \), so \( 2^{2x} = t^2 \).
The expression can now be written as:
\( \frac{t^2}{2} - 9 \times \frac{t}{2^2} + 1 = 0 \)
\( \implies \frac{t^2}{2} - \frac{9t}{4} + 1 = 0 \)
Multiplying the entire equation by 4 to clear denominators:
\( \implies 2t^2 - 9t + 4 = 0 \)
\( \implies 2t^2 - 8t - t + 4 = 0 \)
\( \implies 2t(t-4) - 1(t-4) = 0 \)
\( \implies (t-4)(2t-1) = 0 \)
\( \implies t-4 = 0 \) or \( 2t-1 = 0 \)
\( \implies t = 4 \) or \( t = \frac{1}{2} \)
Substituting back \( t = 2^x \):
\( \implies 2^x = 4 \) or \( 2^x = \frac{1}{2} \)
\( \implies 2^x = 2^2 \) or \( 2^x = 2^{-1} \)
\( \implies x = 2 \) or \( x = -1 \)
In simple words: Break down the powers of 2, use substitution \( 2^x = t \) to simplify the fraction equation, solve the quadratic equation, and convert back to find \( x \).

Exam Tip: When dealing with fractional coefficients in quadratic equations, multiply the whole equation by the common denominator first to avoid mistakes.

Question 9L. Solve for \( x \): \( 5^{x^2} : 5^x = 25 : 1 \)
Answer:
\( 5^{x^2} : 5^x = 25 : 1 \)
\( \implies \frac{5^{x^2}}{5^x} = \frac{25}{1} \)
\( \implies \frac{5^{x^2}}{5^x} = \frac{5^2}{1} \)
\( \implies 5^{x^2} = 5^2 \times 5^x \)
\( \implies 5^{x^2} = 5^{2+x} \)
\( \implies x^2 = 2+x \)
\( \implies x^2 - x - 2 = 0 \)
\( \implies (x-2)(x+1) = 0 \)
\( \implies x-2 = 0 \) or \( x+1 = 0 \)
\( \implies x = 2 \) or \( x = -1 \)
In simple words: Write the division as a fraction, multiply to cross-multiply, add exponents on the right, and solve the resulting quadratic equation for \( x \).

Exam Tip: Always test both roots of the final quadratic equation back in the original expression to ensure they are valid.

Question 9M. Solve for \( x \): \( \sqrt{8^0 + \frac{2}{3}} = (0.6)^{2-3x} \)
Answer:
\( \sqrt{8^0 + \frac{2}{3}} = (0.6)^{2-3x} \)
\( \implies \left(1 + \frac{2}{3}\right)^{\frac{1}{2}} = \left(\frac{6}{10}\right)^{2-3x} \)
\( \implies \left(\frac{5}{3}\right)^{\frac{1}{2}} = \left(\frac{3}{5}\right)^{2-3x} \)
\( \implies \left(\frac{3}{5}\right)^{-\frac{1}{2}} = \left(\frac{3}{5}\right)^{2-3x} \)
\( \implies -\frac{1}{2} = 2-3x \)
\( \implies -1 = 4-6x \)
\( \implies -5 = -6x \)
\( \implies x = \frac{5}{6} \)
In simple words: Convert the decimal to a fraction and simplify. Rewrite the square root as a fractional power of 1/2, then equate exponents using reciprocal properties.

Exam Tip: Recognize that \( 0.6 = \frac{3}{5} \), which is the exact reciprocal of the left side's base \( \frac{5}{3} \).

Question 9N. Solve for \( x \): \( \sqrt{\left(\frac{3}{5}\right)^{x+3}} = \frac{27^{-1}}{125^{-1}} \)
Answer:
\( \sqrt{\left(\frac{3}{5}\right)^{x+3}} = \frac{27^{-1}}{125^{-1}} \)
\( \implies \left(\left(\frac{3}{5}\right)^{x+3}\right)^{\frac{1}{2}} = \frac{\left(3^3\right)^{-1}}{\left(5^3\right)^{-1}} \)
\( \implies \left(\frac{3}{5}\right)^{\frac{x+3}{2}} = \left(\frac{3}{5}\right)^{-3} \)
\( \implies \frac{x+3}{2} = -3 \)
\( \implies x+3 = -6 \)
\( \implies x = -9 \)
In simple words: Express the square root as a power of 1/2. Write the right-hand numbers as powers of 3 and 5 to match the base on the left, then solve for \( x \).

Exam Tip: Take extra care when dealing with negative fractional exponents to ensure bases are correctly inverted.

Question 9O. Solve for \( x \): \( 9^{x+4} = 3^2 \times (27)^{x+1} \)
Answer:
\( 9^{x+4} = 3^2 \times (27)^{x+1} \)
\( \implies 9^{x+4} = 3^2 \times \left(3^3\right)^{x+1} \)
\( \implies 3^{2(x+4)} = 3^2 \times 3^{3x+3} \)
\( \implies 3^{2x+8} = 3^{2+3x+3} \)
\( \implies 2x+8 = 2+3x+3 \)
\( \implies 2x+8 = 3x+5 \)
\( \implies x = 3 \)
In simple words: Convert all bases to the base of 3, simplify using multiplication and power-to-power rules of exponents, and equate the powers.

Exam Tip: Clearly write down each simplification step for the exponents to avoid simple addition errors.

Question 10. Find the value of \( k \) in each of the following:
(i) \( \left(\sqrt[3]{8}\right)^{-\frac{1}{2}} = 2^k \)
(ii) \( \sqrt[4]{\sqrt[3]{x^2}} = x^k \)
(iii) \( (\sqrt{9})^{-7} \times (\sqrt{3})^{-5} = 3^k \)
(iv) \( \left(\frac{1}{3}\right)^{-4} \div 9^{-\frac{1}{3}} = 3^k \)
Answer:
(i)
\( \left(\sqrt[3]{8}\right)^{-\frac{1}{2}} = 2^k \)
\( \implies 8^{\frac{1}{3} \times -\frac{1}{2}} = 2^k \)
\( \implies \left(2^3\right)^{-\frac{1}{6}} = 2^k \)
\( \implies 2^{3 \times -\frac{1}{6}} = 2^k \)
\( \implies 2^{-\frac{1}{2}} = 2^k \)
\( \implies k = -\frac{1}{2} \)

(ii)
\( \sqrt[4]{\sqrt[3]{x^2}} = x^k \)
\( \implies \left\{\left(x^2\right)^{\frac{1}{3}}\right\}^{\frac{1}{4}} = x^k \)
\( \implies \left(x^2\right)^{\frac{1}{12}} = x^k \)
\( \implies x^{\frac{2}{12}} = x^k \)
\( \implies x^{\frac{1}{6}} = x^k \)
\( \implies k = \frac{1}{6} \)

(iii)
\( (\sqrt{9})^{-7} \times (\sqrt{3})^{-5} = 3^k \)
\( \implies \left\{\left(3^2\right)^{\frac{1}{2}}\right\}^{-7} \times \left(3^{\frac{1}{2}}\right)^{-5} = 3^k \)
\( \implies 3^{-7} \times 3^{-\frac{5}{2}} = 3^k \)
\( \implies 3^{-7 - \frac{5}{2}} = 3^k \)
\( \implies 3^{\frac{-14-5}{2}} = 3^k \)
\( \implies 3^{-\frac{19}{2}} = 3^k \)
\( \implies k = -\frac{19}{2} \)

(iv)
\( \left(\frac{1}{3}\right)^{-4} \div 9^{-\frac{1}{3}} = 3^k \)
\( \implies \left(3^{-1}\right)^{-4} \div \left(3^2\right)^{-\frac{1}{3}} = 3^k \)
\( \implies 3^4 \div 3^{-\frac{2}{3}} = 3^k \)
\( \implies 3^{4 + \frac{2}{3}} = 3^k \)
\( \implies 3^{\frac{14}{3}} = 3^k \)
\( \implies k = \frac{14}{3} \)
In simple words: Use the rules of indices to express both sides of each equation with a single base, then compare the exponents to solve for \( k \).

Exam Tip: When simplifying multiple nested roots, multiply the fractional exponents together to find the single equivalent power.

Question 11. If \( a = 2^{\frac{1}{3}} - 2^{-\frac{1}{3}} \), prove that \( 2a^3 + 6a = 3 \).
Answer:
\( a = 2^{\frac{1}{3}} - 2^{-\frac{1}{3}} \)
\( \implies a = 2^{\frac{1}{3}} - \frac{1}{2^{\frac{1}{3}}} \)
Taking the cube of both sides of the equation:
\( \implies a^3 = \left(2^{\frac{1}{3}} - \frac{1}{2^{\frac{1}{3}}}\right)^3 \)
\( = 2 - \frac{1}{2} - 3 \left(2^{\frac{1}{3}} \cdot \frac{1}{2^{\frac{1}{3}}}\right) \left(2^{\frac{1}{3}} - \frac{1}{2^{\frac{1}{3}}}\right) \)
\( \implies a^3 = \frac{4-1}{2} - 3a \) (since the expression in brackets is \( a \))
\( \implies a^3 = \frac{3}{2} - 3a \)
\( \implies 2a^3 + 6a = 3 \)
Thus, the relation is proved.
In simple words: Cube both sides of the given equation and use the identity \( (x-y)^3 = x^3 - y^3 - 3xy(x-y) \) to substitute the original value of \( a \) back into the result.

Exam Tip: Remember that cubing both sides of a rational exponent equation helps eliminate the fractional powers.

Question 12. If \( x = 3^{\frac{2}{3}} + 3^{\frac{1}{3}} \), show that \( x^3 - 9x - 12 = 0 \).
Answer:
\( x = 3^{\frac{2}{3}} + 3^{\frac{1}{3}} \)
Taking the cube of both sides:
\( \implies x^3 = \left(3^{\frac{2}{3}} + 3^{\frac{1}{3}}\right)^3 \)
Using the identity \( (a + b)^3 = a^3 + b^3 + 3ab(a+b) \):
\( \implies x^3 = \left(3^{\frac{2}{3}}\right)^3 + \left(3^{\frac{1}{3}}\right)^3 + 3 \times 3^{\frac{2}{3}} \times 3^{\frac{1}{3}} \left(3^{\frac{2}{3}} + 3^{\frac{1}{3}}\right) \)
\( \implies x^3 = 9 + 3 + 3 \times 3^{\frac{2}{3} + \frac{1}{3}} (x) \)
\( \implies x^3 = 12 + 9x \)
\( \implies x^3 - 9x - 12 = 0 \)
Hence proved.
In simple words: Take the cube of both sides of the equation and apply the algebraic expansion formula to simplify the expression into the desired quadratic or cubic form.

Exam Tip: Group terms carefully during cubing to substitute the initial equation and simplify the resulting expression.

Question 13. If \( \sqrt[x]{a} = \sqrt[y]{b} = \sqrt[z]{c} \) and \( abc = 1 \), prove that \( x+y+z = 0 \).
Answer:
Let \( \sqrt[x]{a} = \sqrt[y]{b} = \sqrt[z]{c} = k \)
\( \implies a^{\frac{1}{x}} = k \), \( b^{\frac{1}{y}} = k \), \( c^{\frac{1}{z}} = k \)
\( \implies a = k^x \), \( b = k^y \), \( c = k^z \)
We are also given the condition \( abc = 1 \).
Substituting these values:
\( \implies k^x \times k^y \times k^z = 1 \)
\( \implies k^{x+y+z} = k^0 \)
\( \implies x+y+z = 0 \)
Which proves the statement.
In simple words: Set the equal roots to a constant \( k \), find expressions for \( a \), \( b \), and \( c \) in terms of \( k \), and use the product \( abc = 1 \) to find the sum of powers.

Exam Tip: Introducing a temporary constant \( k \) is a highly effective algebraic tool when dealing with three equal ratios or powers.

Question 14. If \( a^x = b^y = c^z \) and \( b^2 = ac \), prove that \( y = \frac{2zx}{x+z} \).
Answer:
Let \( a^x = b^y = c^z = k \)
\( \implies a = k^{\frac{1}{x}} \), \( b = k^{\frac{1}{y}} \), \( c = k^{\frac{1}{z}} \)
We are also provided with the relation \( b^2 = ac \).
Substituting these expressions into the relation:
\( \implies \left(k^{\frac{1}{y}}\right)^2 = k^{\frac{1}{x}} \times k^{\frac{1}{z}} \)
\( \implies k^{\frac{2}{y}} = k^{\frac{1}{x} + \frac{1}{z}} \)
\( \implies \frac{2}{y} = \frac{1}{x} + \frac{1}{z} \)
\( \implies \frac{2}{y} = \frac{z+x}{zx} \)
\( \implies y = \frac{2zx}{z+x} \)
Hence proved.
In simple words: Let the equal terms be a constant \( k \), express \( a \), \( b \), and \( c \) in terms of \( k \), substitute these in the given relation, and solve for \( y \).

Exam Tip: Always write down the base equations clearly before substituting variables to avoid arithmetic confusion.

Question 15. Prove that \( \frac{1}{1+a^{p-q}} + \frac{1}{1+a^{q-p}} = 1 \).
Answer:
LHS = \( \frac{1}{1+a^{p-q}} + \frac{1}{1+a^{q-p}} \)
\( = \frac{1 + a^{q-p} + 1 + a^{p-q}}{\left(1+a^{p-q}\right)\left(1+a^{q-p}\right)} \)
\( = \frac{2 + a^{-(p-q)} + a^{p-q}}{\left(1+a^{p-q}\right)\left(1+a^{-(p-q)}\right)} \)
\( = \frac{2 + a^{-(p-q)} + a^{p-q}}{1 + a^{-(p-q)} + a^{p-q} + a^{p-q} \cdot a^{-(p-q)}} \)
\( = \frac{2 + a^{-(p-q)} + a^{p-q}}{1 + a^{-(p-q)} + a^{p-q} + a^{p-q-p+q}} \)
\( = \frac{2 + a^{-(p-q)} + a^{p-q}}{1 + a^{-(p-q)} + a^{p-q} + a^0} \)
\( = \frac{2 + a^{-(p-q)} + a^{p-q}}{1 + a^{-(p-q)} + a^{p-q} + 1} \)
\( = \frac{2 + a^{-(p-q)} + a^{p-q}}{2 + a^{-(p-q)} + a^{p-q}} \)
\( = 1 \) = RHS
Thus, the relation is proved.
In simple words: Find a common denominator to add the fractions, simplify the resulting expression using index laws, and verify that the numerator and denominator are identical.

Exam Tip: Remember that \( a^{p-q} \cdot a^{q-p} = a^0 = 1 \). This simplification is the key to solving this proof quickly.

Question 16. If \( 9^{p+2} - 9^p = 240 \), find the value of \( (8p)^p \).
Answer:
\( 9^{p+2} - 9^p = 240 \)
\( \implies 9^p\left(9^2 - 1\right) = 240 \)
\( \implies 9^p(80) = 240 \)
\( \implies 9^p = 3 \)
\( \implies 3^{2p} = 3^1 \)
\( \implies 2p = 1 \)
\( \implies p = \frac{1}{2} \)
Now, calculating the value of \( (8p)^p \):
\( (8p)^p = \left(2^3 p\right)^p \)
\( = \left(2^3 \cdot \frac{1}{2}\right)^{\frac{1}{2}} \)
\( = \left(2^{3-1}\right)^{\frac{1}{2}} \)
\( = \left(2^2\right)^{\frac{1}{2}} \)
\( = 2 \)
In simple words: Factor out the common term \( 9^p \) to solve for \( p \), then substitute this value back into the target expression and simplify the power.

Exam Tip: When factoring exponential expressions, always pull out the term with the smallest power to simplify the brackets.

Question 17. If \( a^x = b^y = c^z \) and \( abc = 1 \), prove that \( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0 \).
Answer:
\( a^x = b^y = c^z \)
So, \( a^x = b^y \implies a = b^{\frac{y}{x}} \) (Using the rule \( u^n = v \implies u = v^{\frac{1}{n}} \))
And, \( b^y = c^z \implies c = b^{\frac{y}{z}} \) (Using the rule \( u^n = v \implies u = v^{\frac{1}{n}} \))
We are also given the condition \( abc = 1 \).
Substituting these expressions into the product equation:
\( \implies b^{\frac{y}{x}} \cdot b \cdot b^{\frac{y}{z}} = 1 \)
\( \implies b^{\frac{y}{x}} \cdot b^1 \cdot b^{\frac{y}{z}} = 1 \)
\( \implies b^{\frac{y}{x} + 1 + \frac{y}{z}} = b^0 \) (Using the property \( a^0 = 1 \))
\( \implies \frac{y}{x} + 1 + \frac{y}{z} = 0 \)
Dividing the entire equation by \( y \):
\( \implies \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0 \)
Hence proved.
In simple words: Express \( a \) and \( c \) in terms of \( b \) using index rules, substitute them into the product \( abc = 1 \), simplify the base \( b \), and divide by \( y \) to complete the proof.

Exam Tip: Ensure you carefully carry the fractional powers when converting the equations before substituting them into the product equation.

Question 18. If \( x^{\frac{1}{3}} + y^{\frac{1}{3}} + z^{\frac{1}{3}} = 0 \), prove that \( (x+y+z)^3 = 27xyz \).
Answer:
\( x^{\frac{1}{3}} + y^{\frac{1}{3}} + z^{\frac{1}{3}} = 0 \)
\( \implies x^{\frac{1}{3}} + y^{\frac{1}{3}} = -z^{\frac{1}{3}} \)
Taking the cube of both sides gives:
\( \implies \left(x^{\frac{1}{3}} + y^{\frac{1}{3}}\right)^3 = \left(-z^{\frac{1}{3}}\right)^3 \)
\( \implies x + y + 3 x^{\frac{1}{3}}y^{\frac{1}{3}}\left(x^{\frac{1}{3}} + y^{\frac{1}{3}}\right) = -z \)
Substituting the initial relationship once more:
\( \implies x + y + 3 x^{\frac{1}{3}}y^{\frac{1}{3}}\left(-z^{\frac{1}{3}}\right) + z = 0 \)
\( \implies x + y + z = 3 x^{\frac{1}{3}}y^{\frac{1}{3}}z^{\frac{1}{3}} \)
Cubing both sides of this equation again:
\( \implies (x+y+z)^3 = 27xyz \)
In simple words: Isolate one term on one side, cube both sides to eliminate the fractional exponent, replace the remaining sum with the initial condition, and cube once more.

Exam Tip: Do not hesitate to apply algebraic operations twice if fractional powers still remain after the first round of cubing.

Question 19. Given \( 2250 = 2^a \cdot 3^b \cdot 5^c \), find the values of \( a \), \( b \), and \( c \). Hence, evaluate \( 3^a \times 2^{-b} \times 5^{-c} \).
Answer:
Given \( 2250 = 2^a \cdot 3^b \cdot 5^c \)
\( \implies 3^2 \times 5^3 \times 2 = 2^a \cdot 3^b \cdot 5^c \)
By equating the bases, we find:
\( \implies a = 1, b = 2, c = 3 \)
Now, evaluate \( 3^a \times 2^{-b} \times 5^{-c} \):
\( = 3^1 \times 2^{-2} \times 5^{-3} \)
\( = \frac{3}{2^2 \times 5^3} \)
\( = \frac{3}{500} \)
In simple words: Find the prime factorization of 2250 to determine the values of \( a \), \( b \), and \( c \), then substitute these into the second expression to compute the final fraction.

Exam Tip: Double-check your prime factorization of the constant to ensure the exponents of 2, 3, and 5 are completely accurate.

Question 20. Given \( 2400 = 2^x \times 3^y \times 5^z \), find the values of \( x \), \( y \), and \( z \). Hence, evaluate \( 2^{-x} \times 3^y \times 5^z \).
Answer:
\( 2400 = 2^x \times 3^y \times 5^z \)
\( 2400 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 5 \times 5 \)
\( \therefore 2^x \times 3^y \times 5^z = 2^5 \times 3^1 \times 5^2 \)
Comparing exponents on both sides:
\( \implies x = 5, y = 1, z = 2 \)
Substituting these into the target expression:
\( \implies 2^{-x} \times 3^y \times 5^z = 2^{-5} \times 3^1 \times 5^2 \)
\( = \frac{1}{32} \times 3 \times 25 = \frac{75}{32} \)
In simple words: Factorize 2400 into its prime components to find \( x \), \( y \), and \( z \). Then substitute these values into the given equation to evaluate the final fraction.

Exam Tip: Ensure that you handle the negative sign in the exponent \( 2^{-x} \) properly by writing it as a reciprocal in the final fraction calculation.

Question 21. If \( 2^x = 3^y = 12^z \), show that \( \frac{1}{z} = \frac{2}{x} + \frac{1}{y} \).
Answer:
Let \( 2^x = 3^y = 12^z = k \)
\( \implies 2 = k^{\frac{1}{x}} \), \( 3 = k^{\frac{1}{y}} \), \( 12 = k^{\frac{1}{z}} \)
Now, \( 12 = 2 \times 2 \times 3 \)
Substituting our values of \( 2, 3 \) and \( 12 \):
\( \implies k^{\frac{1}{z}} = k^{\frac{1}{x}} \times k^{\frac{1}{x}} \times k^{\frac{1}{y}} \)
\( \implies k^{\frac{1}{z}} = k^{\frac{1}{x} + \frac{1}{x} + \frac{1}{y}} \)
Comparing the power indices:
\( \implies \frac{1}{z} = \frac{1}{x} + \frac{1}{x} + \frac{1}{y} \)
\( \implies \frac{1}{z} = \frac{2}{x} + \frac{1}{y} \)
In simple words: Define the terms as equal to \( k \), express each base in terms of \( k \), write 12 as a product of its prime factors, substitute, and compare powers.

Exam Tip: Decomposing a composite number like 12 into prime factors \( 2^2 \times 3 \) is crucial for solving these kinds of algebraic comparisons.

Question 22A. If \( 9^{2a} = (\sqrt[3]{81})^{-\frac{6}{b}} = (\sqrt{27})^2 \), find the values of \( a \) and \( b \).
Answer:
\( 9^{2a} = \left(\sqrt[3]{81}\right)^{-\frac{6}{b}} = (\sqrt{27})^2 \)
\( \implies 9^{2a} = \left(\sqrt[3]{3^4}\right)^{-\frac{6}{b}} = \left(\sqrt{3^3}\right)^2 \)
\( \implies \left(3^2\right)^{2a} = \left(3^{\frac{4}{3}}\right)^{-\frac{6}{b}} = \left(3^{\frac{3}{2}}\right)^2 \)
\( \implies 3^{4a} = 3^{-\frac{8}{b}} = 3^3 \)
Comparing the first and third parts:
\( \implies 4a = 3 \implies a = \frac{3}{4} \)
Comparing the second and third parts:
\( \implies -\frac{8}{b} = 3 \implies b = -\frac{8}{3} \)
In simple words: Convert all expressions to the base of 3, equate the powers across the equal terms to get linear equations, and solve for \( a \) and \( b \).

Exam Tip: Always verify that the fractional indices of the roots are correctly calculated before setting up equations to solve for variables.

Question 22B. Solve the system of equations for \( a \) and \( b \): \( (\sqrt{243})^a \div 3^{b+1} = 1 \) and \( 27^b - 81^{4-\frac{a}{2}} = 0 \).
Answer:
\( (\sqrt{243})^a \div 3^{b+1} = 1 \) and \( 27^b - 81^{4-\frac{a}{2}} = 0 \)
\( \implies \left(\sqrt{3^5}\right)^a \div 3^{b+1} = 1 \) and \( \left(3^3\right)^b - \left(3^4\right)^{4-\frac{a}{2}} = 0 \)
\( \implies \left(3^5\right)^{\frac{a}{2}} \div 3^{b+1} = 1 \) and \( 3^{3b} - \left(3^4\right)^{4-\frac{a}{2}} = 0 \)
\( \implies 3^{\frac{5a}{2}} \div 3^{b+1} = 1 \) and \( 3^{3b} - 3^{4\left(4-\frac{a}{2}\right)} = 0 \)
\( \implies 3^{\frac{5a}{2}-b-1} = 1 \) and \( 3^{3b} - 3^{16-2a} = 0 \)
\( \implies 3^{\frac{5a}{2}-b-1} = 3^0 \) and \( 3^{3b} = 3^{16-2a} \)
Equating the exponents:
\( \implies \frac{5a}{2} - b - 1 = 0 \) and \( 3b = 16 - 2a \)
\( \implies \frac{5a}{2} - b = 1 \) and \( 2a + 3b = 16 \)
\( \implies 5a - 2b = 2 \) and \( 2a + 3b = 16 \)
Next, multiply the first equation by 3 and the second equation by 2:
\( \implies 15a - 6b = 6 \) and \( 4a + 6b = 32 \)
Adding these two equations together gives:
\( 19a = 38 \)
\( \implies a = 2 \)
Now, substitute this value of \( a \) back into \( 5a - 2b = 2 \) to calculate \( b \):
\( 5a - 2b = 2 \)
\( \implies 5(2) - 2b = 2 \)
\( \implies 10 - 2b = 2 \)
\( \implies b = 4 \)
Therefore, we find that \( a = 2 \) and \( b = 4 \).
In simple words: Simplify both given equations using base 3, establish two linear equations in terms of \( a \) and \( b \), and solve them using the elimination method.

Exam Tip: When solving simultaneous equations, look for coefficients that can be easily matched by simple multiplication of the whole equation.

Question 23A. Prove that \( \sqrt{x^{-1}y} \cdot \sqrt{y^{-1}z} \cdot \sqrt{z^{-1}x} = 1 \).
Answer:
LHS = \( \sqrt{x^{-1}y} \cdot \sqrt{y^{-1}z} \cdot \sqrt{z^{-1}x} \)
\( = \sqrt{\frac{y}{x}} \cdot \sqrt{\frac{z}{y}} \cdot \sqrt{\frac{x}{z}} \) (Using the rule \( (a^m)^n = a^{mn} \))
\( = \sqrt{\left(\frac{y}{x}\right)\left(\frac{z}{y}\right)\left(\frac{x}{z}\right)} \)
\( = \sqrt{x^{1-1} \cdot y^{1-1} \cdot z^{1-1}} \)
\( = \sqrt{x^0 \cdot y^0 \cdot z^0} \)
\( = \sqrt{1 \cdot 1 \cdot 1} \)
\( = 1 \) (Applying the property \( a^0 = 1 \))
= RHS
Thus, the relation is verified.
In simple words: Express the negative powers as fractions, combine all terms under a single square root, cancel common terms, and use the property that any base to power 0 equals 1.

Exam Tip: Grouping all terms under a single radical sign makes the cancellation process much clearer and reduces the chance of errors.

Question 23B. Prove that \( \left(\frac{a^m}{a^n}\right)^{m+n-l} \cdot \left(\frac{a^n}{a^l}\right)^{n+l-m} \cdot \left(\frac{a^l}{a^m}\right)^{l+m-n} = 1 \).
Answer:
LHS = \( \left(\frac{a^m}{a^n}\right)^{m+n-l} \cdot \left(\frac{a^n}{a^l}\right)^{n+l-m} \cdot \left(\frac{a^l}{a^m}\right)^{l+m-n} \)
\( = \frac{a^{m(m+n-l)}}{a^{n(m+n-l)}} \cdot \frac{a^{n(n+l-m)}}{a^{l(n+l-m)}} \cdot \frac{a^{l(l+m-n)}}{a^{m(l+m-n)}} \) (Using the rule \( (a^m)^n = a^{mn} \))
\( = \frac{a^{m^2+mn-ml}}{a^{mn+n^2-nl}} \cdot \frac{a^{n^2+nl-mn}}{a^{nl+l^2-ml}} \cdot \frac{a^{l^2+lm-nl}}{a^{lm+m^2-mn}} \)
\( = a^{m^2+mn-ml-(mn+n^2-nl)} \cdot a^{n^2+nl-mn-(nl+l^2-ml)} \cdot a^{l^2+lm-nl-(lm+m^2-mn)} \) (Using \( a^u \div a^v = a^{u-v} \))
\( = a^{m^2+mn-ml-mn-n^2+nl} \cdot a^{n^2+nl-mn-nl-l^2+ml} \cdot a^{l^2+lm-nl-lm-m^2+mn} \)
\( = a^{m^2-n^2-ml+nl+n^2-l^2-mn+ml+l^2-m^2-nl+mn} \) (Using \( a^u \times a^v = a^{u+v} \))
\( = a^0 \)
\( = 1 \) (Applying the property \( a^0 = 1 \))
= RHS
Hence proved.
In simple words: Apply the quotient rule of indices for terms inside the brackets, expand the powers, multiply the simplified terms together, and sum up the exponents to get zero.

Exam Tip: When expanding the product of algebraic terms in the exponents, organize them cyclically to easily identify which positive and negative terms cancel out.

Question 23C. Prove that \( \left(\frac{a^m}{a^n}\right)^{m+n-l} \cdot \left(\frac{a^n}{a^l}\right)^{n+l-m} \cdot \left(\frac{a^l}{a^m}\right)^{l+m-n} = 1 \).
Answer:
LHS = \( \left(\frac{a^m}{a^n}\right)^{m+n-l} \cdot \left(\frac{a^n}{a^l}\right)^{n+l-m} \cdot \left(\frac{a^l}{a^m}\right)^{l+m-n} \)
\( = \frac{a^{m(m+n-l)}}{a^{n(m+n-l)}} \cdot \frac{a^{n(n+l-m)}}{a^{l(n+l-m)}} \cdot \frac{a^{l(l+m-n)}}{a^{m(l+m-n)}} \) (Using the rule \( (a^m)^n = a^{mn} \))
\( = \frac{a^{m^2+mn-ml}}{a^{mn+n^2-nl}} \cdot \frac{a^{n^2+nl-mn}}{a^{nl+l^2-ml}} \cdot \frac{a^{l^2+lm-nl}}{a^{lm+m^2-mn}} \)
\( = a^{m^2+mn-ml-(mn+n^2-nl)} \cdot a^{n^2+nl-mn-(nl+l^2-ml)} \cdot a^{l^2+lm-nl-(lm+m^2-mn)} \) (Using \( a^u \div a^v = a^{u-v} \))
\( = a^{m^2+mn-ml-mn-n^2+nl} \cdot a^{n^2+nl-mn-nl-l^2+ml} \cdot a^{l^2+lm-nl-lm-m^2+mn} \)
\( = a^{m^2-n^2-ml+nl+n^2-l^2-mn+ml+l^2-m^2-nl+mn} \) (Using \( a^u \times a^v = a^{u+v} \))
\( = a^0 \)
\( = 1 \) (Applying the property \( a^0 = 1 \))
= RHS
Hence proved.
In simple words: Subtract exponents inside the brackets first, multiply by the outer exponents, then combine the terms using exponent addition to show that the result is 1.

Exam Tip: Systematic expansion of the exponents prevents algebraic sign errors when combining several fractions together.

Question 23D. Prove that \( \sqrt[ab]{\frac{x^a}{x^b}} \cdot \sqrt[bc]{\frac{x^b}{x^c}} \cdot \sqrt[ca]{\frac{x^c}{x^a}} = 1 \).
Answer:
LHS = \( \sqrt[ab]{\frac{x^a}{x^b}} \cdot \sqrt[bc]{\frac{x^b}{x^c}} \cdot \sqrt[ca]{\frac{x^c}{x^a}} \)
\( = \left(\frac{x^a}{x^b}\right)^{\frac{1}{ab}} \cdot \left(\frac{x^b}{x^c}\right)^{\frac{1}{bc}} \cdot \left(\frac{x^c}{x^a}\right)^{\frac{1}{ca}} \)
\( = \frac{x^{\frac{a}{ab}}}{x^{\frac{b}{ab}}} \cdot \frac{x^{\frac{b}{bc}}}{x^{\frac{c}{bc}}} \cdot \frac{x^{\frac{c}{ca}}}{x^{\frac{a}{ca}}} \) (Using the rule \( (a^m)^n = a^{mn} \))
\( = \frac{x^{\frac{1}{b}}}{x^{\frac{1}{a}}} \cdot \frac{x^{\frac{1}{c}}}{x^{\frac{1}{b}}} \cdot \frac{x^{\frac{1}{a}}}{x^{\frac{1}{c}}} \)
\( = x^{\frac{1}{b} - \frac{1}{a}} \cdot x^{\frac{1}{c} - \frac{1}{b}} \cdot x^{\frac{1}{a} - \frac{1}{c}} \) (Using \( a^u \div a^v = a^{u-v} \))
\( = x^{\frac{a-b}{ab}} \cdot x^{\frac{b-c}{bc}} \cdot x^{\frac{c-a}{ca}} \)
\( = x^{\frac{a-b}{ab} + \frac{b-c}{bc} + \frac{c-a}{ac}} \) (Using \( a^u \times a^v = a^{u+v} \))
\( = x^{\frac{ac-bc+ab-ac+bc-ab}{abc}} \)
\( = x^0 \)
\( = 1 \) (Applying the property \( a^0 = 1 \))
= RHS
Hence proved.
In simple words: Convert the roots into fractional exponents, expand the terms, subtract the denominator powers from the numerator powers, and add all exponents together to simplify to 1.

Exam Tip: Remember that an n-th root is represented as a fractional exponent \( \frac{1}{n} \), which can be distributed to both numerator and denominator.

Question 23E. Prove that \( (x^a)^{b-c} \cdot (x^b)^{c-a} \cdot (x^c)^{a-b} = 1 \).
Answer:
LHS = \( (x^a)^{b-c} \cdot (x^b)^{c-a} \cdot (x^c)^{a-b} \)
\( = x^{a(b-c)} \cdot x^{b(c-a)} \cdot x^{c(a-b)} \) (Using the rule \( (a^m)^n = a^{mn} \))
\( = x^{ab-ac} \cdot x^{bc-ab} \cdot x^{ac-bc} \)
\( = x^{ab-ac+bc-ab+ac-bc} \) (Using \( a^u \times a^v = a^{u+v} \))
\( = x^0 \)
\( = 1 \) = RHS
Hence proved.
In simple words: Multiply the inner exponents with the outer exponents, multiply the resulting terms by adding their powers together, and simplify the exponent sum to 0.

Exam Tip: Be thorough when multiplying the terms in the exponents to ensure you do not drop any negative signs.

Question 23F. Prove that \( \frac{x^{p(q-r)}}{x^{q(p-r)}} \div \left(\frac{x^q}{x^p}\right)^r = 1 \).
Answer:
LHS = \( \frac{x^{p(q-r)}}{x^{q(p-r)}} \div \left(\frac{x^q}{x^p}\right)^r \)
\( = \frac{x^{p(q-r)}}{x^{q(p-r)}} \div \frac{x^{qr}}{x^{pr}} \) (Using the rule \( (a^m)^n = a^{mn} \))
\( = \frac{x^{p(q-r)}}{x^{q(p-r)}} \times \frac{x^{pr}}{x^{qr}} \)
\( = \frac{x^{pq-pr}}{x^{pq-qr}} \times \frac{x^{pr}}{x^{qr}} \)
\( = \frac{x^{pq-pr+pr}}{x^{pq-qr+qr}} \) (Using \( a^u \times a^v = a^{u+v} \))
\( = \frac{x^{pq}}{x^{pq}} \)
\( = 1 \) = RHS
Hence proved.
In simple words: Simplify the fractions on both sides of the division sign by applying power rules, convert the division into multiplication of the reciprocal, and add the powers to get 1.

Exam Tip: When dividing by a fractional exponent, remember that it is equivalent to multiplying by its reciprocal, which simply swaps the positions of the terms.