Selina Concise Solutions for ICSE Class 9 Mathematics Chapter 7 Indices Exponents

Exercise 7(A)

Question 1. Simplify the following expressions:
(i) \( 3^3 \times (243)^{-\frac{2}{3}} \times 9^{-\frac{1}{3}} \)
(ii) \( 5^{-4} \times (125)^{\frac{5}{3}} \div (25)^{-\frac{1}{2}} \)
(iii) \( \left(\frac{27}{125}\right)^{\frac{2}{3}} \times \left(\frac{9}{25}\right)^{-\frac{3}{2}} \)
(iv) \( 7^0 \times (25)^{-\frac{3}{2}} - 5^{-3} \)
(v) \( \left(\frac{16}{81}\right)^{-\frac{3}{4}} \times \left(\frac{49}{9}\right)^{\frac{3}{2}} \div \left(\frac{343}{216}\right)^{\frac{2}{3}} \)
Answer:
(i) \( 3^3 \times (243)^{-\frac{2}{3}} \times 9^{-\frac{1}{3}} \)
\( \implies 3^3 \times (3 \times 3 \times 3 \times 3 \times 3)^{-\frac{2}{3}} \times (3 \times 3)^{-\frac{1}{3}} \)
\( \implies 3^3 \times \left(3^5\right)^{-\frac{2}{3}} \times \left(3^2\right)^{-\frac{1}{3}} \)
\( \implies 3^3 \times 3^{5 \times \left(-\frac{2}{3}\right)} \times 3^{2 \times \left(-\frac{1}{3}\right)} \)
\( \implies 3^3 \times 3^{-\frac{10}{3}} \times 3^{-\frac{2}{3}} \)
\( \implies 3^{3 - \frac{10}{3} - \frac{2}{3}} \)
\( \implies 3^{\frac{9 - 10 - 2}{3}} \)
\( \implies 3^{\frac{-3}{3}} \)
\( \implies 3^{-1} \)
\( \implies \frac{1}{3} \)

(ii) \( 5^{-4} \times (125)^{\frac{5}{3}} \div (25)^{-\frac{1}{2}} \)
\( \implies 5^{-4} \times (5 \times 5 \times 5)^{\frac{5}{3}} \div (5 \times 5)^{-\frac{1}{2}} \)
\( \implies 5^{-4} \times \left(5^3\right)^{\frac{5}{3}} \div \left(5^2\right)^{-\frac{1}{2}} \)
\( \implies 5^{-4} \times 5^{3 \times \frac{5}{3}} \div 5^{2 \times \left(-\frac{1}{2}\right)} \)
\( \implies \frac{5^{-4} \times 5^5}{5^{-1}} \)
\( \implies \frac{5^{5-4}}{5^{-1}} \)
\( \implies \frac{5^1}{5^{-1}} \)
\( \implies 5^{1 - (-1)} \)
\( \implies 5^2 \)
\( \implies 25 \)

(iii) \( \left(\frac{27}{125}\right)^{\frac{2}{3}} \times \left(\frac{9}{25}\right)^{-\frac{3}{2}} \)
\( \implies \left(\frac{3 \times 3 \times 3}{5 \times 5 \times 5}\right)^{\frac{2}{3}} \times \left(\frac{3 \times 3}{5 \times 5}\right)^{-\frac{3}{2}} \)
\( \implies \left[\left(\frac{3}{5}\right)^3\right]^{\frac{2}{3}} \times \left[\left(\frac{3}{5}\right)^2\right]^{-\frac{3}{2}} \)
\( \implies \left(\frac{3}{5}\right)^{3 \times \frac{2}{3}} \times \left(\frac{3}{5}\right)^{2 \times \left(-\frac{3}{2}\right)} \)
\( \implies \left(\frac{3}{5}\right)^2 \times \left(\frac{3}{5}\right)^{-3} \)
\( \implies \left(\frac{3}{5}\right)^{2-3} \)
\( \implies \left(\frac{3}{5}\right)^{-1} \)
\( \implies \frac{1}{\frac{3}{5}} \)
\( \implies \frac{5}{3} \)

(iv) \( 7^0 \times (25)^{-\frac{3}{2}} - 5^{-3} \)
\( \implies 7^0 \times (5 \times 5)^{-\frac{3}{2}} - 5^{-3} \)
\( \implies 7^0 \times \left(5^2\right)^{-\frac{3}{2}} - \frac{1}{5^3} \)
\( \implies 7^0 \times 5^{2 \times \left(-\frac{3}{2}\right)} - \frac{1}{5^3} \)
\( \implies 7^0 \times 5^{-3} - \frac{1}{5^3} \)
\( \implies 1 \times 5^{-3} - \frac{1}{5^3} \)
\( \implies \frac{1}{5^3} - \frac{1}{5^3} \)
\( \implies \frac{1-1}{125} \)
\( \implies 0 \)

(v) \( \left(\frac{16}{81}\right)^{-\frac{3}{4}} \times \left(\frac{49}{9}\right)^{\frac{3}{2}} \div \left(\frac{343}{216}\right)^{\frac{2}{3}} \)
\( \implies \left(\frac{2 \times 2 \times 2 \times 2}{3 \times 3 \times 3 \times 3}\right)^{-\frac{3}{4}} \times \left(\frac{7 \times 7}{3 \times 3}\right)^{\frac{3}{2}} \div \left(\frac{7 \times 7 \times 7}{6 \times 6 \times 6}\right)^{\frac{2}{3}} \)
\( \implies \left[\left(\frac{2}{3}\right)^4\right]^{-\frac{3}{4}} \times \left[\left(\frac{7}{3}\right)^2\right]^{\frac{3}{2}} \div \left[\left(\frac{7}{6}\right)^3\right]^{\frac{2}{3}} \)
\( \implies \left(\frac{2}{3}\right)^{-3} \times \left(\frac{7}{3}\right)^3 \div \left(\frac{7}{6}\right)^2 \)
\( \implies \frac{1}{\left(\frac{2}{3}\right)^3} \times \left(\frac{7}{3}\right)^3 \times \frac{1}{\left(\frac{7}{6}\right)^2} \)
\( \implies \frac{1}{\frac{2}{3} \times \frac{2}{3} \times \frac{2}{3}} \times \frac{7}{3} \times \frac{7}{3} \times \frac{7}{3} \times \frac{1}{\frac{7}{6} \times \frac{7}{6}} \)
\( \implies \frac{1 \times 3 \times 3 \times 3}{2 \times 2 \times 2} \times \frac{7}{3} \times \frac{7}{3} \times \frac{7}{3} \times \frac{1 \times 6 \times 6}{7 \times 7} \)
\( \implies \frac{7 \times 3 \times 3}{2} \)
\( \implies \frac{63}{2} \)
\( \implies 31.5 \)
In simple words: To simplify expressions with exponents, express the bases as powers of prime numbers first. Then, use the laws of exponents to combine the powers and calculate the final value.

Exam Tip: Be extra careful with negative fractional exponents. Always convert them to positive exponents by taking the reciprocal of the base, i.e., \( x^{-n} = \frac{1}{x^n} \), to avoid sign errors during calculations.

Question 2. Simplify the following expressions:
(i) \( \left(8x^3 \div 125y^3\right)^{\frac{2}{3}} \)
(ii) \( (a+b)^{-1} \cdot \left(a^{-1} + b^{-1}\right) \)
(iii) \( \frac{5^{n+3} - 6 \times 5^{n+1}}{9 \times 5^n - 5^n \times 2^2} \)
(iv) \( \left(3x^2\right)^{-3} \times \left(x^9\right)^{\frac{2}{3}} \)
Answer:
(i) \( \left(8x^3 \div 125y^3\right)^{\frac{2}{3}} \)
\( \implies \left(\frac{8x^3}{125y^3}\right)^{\frac{2}{3}} \)
\( \implies \left(\frac{2x \times 2x \times 2x}{5y \times 5y \times 5y}\right)^{\frac{2}{3}} \)
\( \implies \left[\left(\frac{2x}{5y}\right)^3\right]^{\frac{2}{3}} \)
\( \implies \left(\frac{2x}{5y}\right)^{3 \times \frac{2}{3}} \)
\( \implies \left(\frac{2x}{5y}\right)^2 \)
\( \implies \frac{2x}{5y} \times \frac{2x}{5y} \)
\( \implies \frac{4x^2}{25y^2} \)

(ii) \( (a+b)^{-1} \cdot \left(a^{-1} + b^{-1}\right) \)
\( \implies \frac{1}{a+b} \times \left(\frac{1}{a} + \frac{1}{b}\right) \)
\( \implies \frac{1}{a+b} \times \left(\frac{b+a}{ab}\right) \)
\( \implies \frac{1}{a+b} \times \frac{a+b}{ab} \)
\( \implies \frac{1}{ab} \)

(iii) \( \frac{5^{n+3} - 6 \times 5^{n+1}}{9 \times 5^n - 5^n \times 2^2} \)
\( \implies \frac{5^{n+1} \times 5^2 - 6 \times 5^{n+1}}{9 \times 5^n - 5^n \times 2^2} \)
\( \implies \frac{5^{n+1} \times \left(5^2 - 6\right)}{5^n \times (9 - 4)} \)
\( \implies \frac{5^n \times 5^1 \times (25 - 6)}{5^n \times (9 - 4)} \)
\( \implies \frac{5^1 \times 19}{5} \)
\( \implies 19 \)

(iv) \( \left(3x^2\right)^{-3} \times \left(x^9\right)^{\frac{2}{3}} \)
\( \implies \frac{1}{\left(3x^2\right)^3} \times x^{9 \times \frac{2}{3}} \)
\( \implies \frac{1}{3^3 \times x^{2 \times 3}} \times x^6 \)
\( \implies \frac{1}{27x^6} \times x^6 \)
\( \implies \frac{1}{27} \)
In simple words: When simplifying algebra with exponents, first write division as fractions and negative exponents as reciprocals. Group like terms together so you can cancel them out to get the simplest form.

Exam Tip: In problems involving variables in exponents (like \(5^n\)), always factor out the common terms from both the numerator and the denominator. This allows the variable part to cancel out cleanly, leaving behind simple numerical calculations.

Question 3. Evaluate the following:
(i) \( \sqrt{\frac{1}{4}} + (0.01)^{-\frac{1}{2}} - (27)^{\frac{2}{3}} \)
(ii) \( \left(\frac{27}{8}\right)^{\frac{2}{3}} - \left(\frac{1}{4}\right)^{-2} + 5^0 \)
Answer:
(i) \( \sqrt{\frac{1}{4}} + (0.01)^{-\frac{1}{2}} - (27)^{\frac{2}{3}} \)
\( \implies \sqrt{\frac{1}{2} \times \frac{1}{2}} + (0.1 \times 0.1)^{-\frac{1}{2}} - (3 \times 3 \times 3)^{\frac{2}{3}} \)
\( \implies \frac{1}{2} + \left[(0.1)^2\right]^{-\frac{1}{2}} - \left(3^3\right)^{\frac{2}{3}} \)
\( \implies \frac{1}{2} + (0.1)^{2 \times \left(-\frac{1}{2}\right)} - 3^{3 \times \frac{2}{3}} \)
\( \implies \frac{1}{2} + (0.1)^{-1} - 3^2 \)
\( \implies \frac{1}{2} + \frac{1}{0.1} - 9 \)
\( \implies \frac{1}{2} + \frac{10}{1} - 9 \)
\( \implies \frac{1 + 20 - 18}{2} \)
\( \implies \frac{3}{2} \)

(ii) \( \left(\frac{27}{8}\right)^{\frac{2}{3}} - \left(\frac{1}{4}\right)^{-2} + 5^0 \)
\( \implies \left(\frac{3 \times 3 \times 3}{2 \times 2 \times 2}\right)^{\frac{2}{3}} - \left(\frac{1 \times 1}{2 \times 2}\right)^{-2} + 5^0 \)
\( \implies \left[\left(\frac{3}{2}\right)^3\right]^{\frac{2}{3}} - \left[\left(\frac{1}{2}\right)^2\right]^{-2} + 1 \)
\( \implies \left(\frac{3}{2}\right)^{3 \times \frac{2}{3}} - \left(\frac{1}{2}\right)^{2 \times (-2)} + 1 \)
\( \implies \left(\frac{3}{2}\right)^2 - \left(\frac{1}{2}\right)^{-4} + 1 \)
\( \implies \frac{3}{2} \times \frac{3}{2} - \frac{1}{\left(\frac{1}{2}\right)^4} + 1 \)
\( \implies \frac{9}{4} - \frac{1}{\frac{1}{16}} + 1 \)
\( \implies \frac{9}{4} - 16 + 1 \)
\( \implies \frac{9 - 64 + 4}{4} \)
\( \implies -\frac{51}{4} \)
In simple words: Remember that any non-zero number to the power of 0 is always 1. Convert decimals like 0.01 into powers of 0.1 or fractions, and turn negative exponents into positive ones before combining the terms.

Exam Tip: Never forget that \(a^0 = 1\) (for \(a \neq 0\)). Students often mistakenly write \(5^0 = 0\) or \(5^0 = 5\), which is a very common error that can cost you marks on simple evaluations.

Question 4. Simplify the following expressions:
(i) \( \left(\frac{3^{-4}}{2^{-8}}\right)^{\frac{1}{4}} \)
(ii) \( \left(\frac{27^{-3}}{9^{-3}}\right)^{\frac{1}{5}} \)
(iii) \( (32)^{-\frac{2}{5}} \div (125)^{-\frac{2}{3}} \)
(iv) \( \left[1 - \left\{1 - (1-n)^{-1}\right\}^{-1}\right]^{-1} \)
Answer:
(i) \( \left(\frac{3^{-4}}{2^{-8}}\right)^{\frac{1}{4}} \)
\( \implies \left(\frac{2^8}{3^4}\right)^{\frac{1}{4}} \)
\( \implies \frac{\left(2^8\right)^{\frac{1}{4}}}{\left(3^4\right)^{\frac{1}{4}}} \)
\( \implies \frac{2^{8 \times \frac{1}{4}}}{3^{4 \times \frac{1}{4}}} \)
\( \implies \frac{2^2}{3} \)
\( \implies \frac{4}{3} \)

(ii) \( \left(\frac{27^{-3}}{9^{-3}}\right)^{\frac{1}{5}} \)
\( \implies \left(\frac{9^3}{27^3}\right)^{\frac{1}{5}} \)
\( \implies \left[\frac{\left(3^2\right)^3}{\left(3^3\right)^3}\right]^{\frac{1}{5}} \)
\( \implies \left[\frac{3^6}{3^9}\right]^{\frac{1}{5}} \)
\( \implies \left[\left(\frac{3^2}{3^3}\right)^3\right]^{\frac{1}{5}} \)
\( \implies \left[\left(\frac{1}{3}\right)^3\right]^{\frac{1}{5}} \)
\( \implies \left(\frac{1}{3}\right)^{3 \times \frac{1}{5}} \)
\( \implies \frac{1}{3^{\frac{3}{5}}} \)

(iii) \( (32)^{-\frac{2}{5}} \div (125)^{-\frac{2}{3}} \)
\( \implies \frac{(32)^{-\frac{2}{5}}}{(125)^{-\frac{2}{3}}} \)
\( \implies \frac{(125)^{\frac{2}{3}}}{(32)^{\frac{2}{5}}} \)
\( \implies \frac{(5 \times 5 \times 5)^{\frac{2}{3}}}{(2 \times 2 \times 2 \times 2 \times 2)^{\frac{2}{5}}} \)
\( \implies \frac{\left(5^3\right)^{\frac{2}{3}}}{\left(2^5\right)^{\frac{2}{5}}} \)
\( \implies \frac{5^2}{2^2} \)
\( \implies \frac{25}{4} \)
\( \implies 6\frac{1}{4} \)

(iv) \( \left[1 - \left\{1 - (1-n)^{-1}\right\}^{-1}\right]^{-1} \)
\( \implies \frac{1}{\left[1 - \left\{1 - (1-n)^{-1}\right\}^{-1}\right]} \)
\( \implies \frac{1}{1 - \frac{1}{1 - (1-n)^{-1}}} \)
\( \implies \frac{1}{1 - \frac{1}{1 - \frac{1}{1-n}}} \)
\( \implies \frac{1}{1 - \frac{1}{\frac{(1-n)-1}{1-n}}} \)
\( \implies \frac{1}{1 - \frac{1}{\frac{-n}{1-n}}} \)
\( \implies \frac{1}{1 - \frac{1-n}{-n}} \)
\( \implies \frac{1}{1 + \frac{1-n}{n}} \)
\( \implies \frac{1}{\frac{n + (1-n)}{n}} \)
\( \implies \frac{1}{\frac{n+1-n}{n}} \)
\( \implies \frac{n}{1} \)
\( \implies n \)
In simple words: When dealing with nested negative exponents like \(x^{-1}\), solve from the innermost bracket first. Work step-by-step outwards, converting negative powers to fractions and simplifying them at each level.

Exam Tip: For complex algebraic expressions with multiple nested levels of negative powers (such as part iv), solve the innermost expression first and move outward systematically. Rushing this calculation often leads to sign or reciprocal errors.

Question 5. If \( 2160 = 2^a \times 3^b \times 5^c \), find the value of \( a \), \( b \), and \( c \). Hence, calculate the value of \( 3^a \times 2^{-b} \times 5^{-c} \).
Answer:
\( 2160 = 2^a \times 3^b \times 5^c \)
\( \implies 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5 = 2^a \times 3^b \times 5^c \)
\( \implies 2^4 \times 3^3 \times 5^1 = 2^a \times 3^b \times 5^c \)
Equating the exponents of 2, 3, and 5 from both sides, we get:
\( a=4 \), \( b=3 \), and \( c=1 \)
Now, substitute these values into \( 3^a \times 2^{-b} \times 5^{-c} \):
\( \implies 3^4 \times 2^{-3} \times 5^{-1} \)
\( \implies 3 \times 3 \times 3 \times 3 \times \frac{1}{2^3} \times \frac{1}{5} \)
\( \implies 81 \times \frac{1}{8} \times \frac{1}{5} \)
\( \implies \frac{81}{40} \)
\( \implies 2\frac{1}{40} \)
In simple words: First, find the prime factors of 2160 to find the values of a, b, and c by matching powers. Then put these numbers into the second formula to get the final fraction.

Exam Tip: Prime factorization must be done very carefully. Always check that the sum of the prime-factored numbers multiplies back to the original number before comparing exponents.

Question 6. If \( 1960 = 2^a \times 5^b \times 7^c \), find the values of \( a \), \( b \), and \( c \). Hence, find the value of \( 2^{-a} \times 7^b \times 5^{-c} \).
Answer:
\( 1960 = 2^a \times 5^b \times 7^c \)
\( \implies 2 \times 2 \times 2 \times 5 \times 7 \times 7 = 2^a \times 5^b \times 7^c \)
\( \implies 2^3 \times 5^1 \times 7^2 = 2^a \times 5^b \times 7^c \)
Comparing the exponents of the prime bases 2, 5, and 7 on both sides, we get:
\( a=3 \), \( b=1 \), and \( c=2 \)
Substituting these values into \( 2^{-a} \times 7^b \times 5^{-c} \):
\( \implies 2^{-3} \times 7^1 \times 5^{-2} \)
\( \implies \frac{1}{2^3} \times 7 \times \frac{1}{5^2} \)
\( \implies \frac{1}{8} \times 7 \times \frac{1}{25} \)
\( \implies \frac{7}{200} \)
In simple words: Find the prime factorization of 1960 first, then compare the powers of 2, 5, and 7 to find a, b, and c. Plug these into the expression to compute the final fraction.

Exam Tip: Keep your work tidy. Clearly label the prime bases and their respective exponents on both sides so you do not accidentally swap the values of \(a\), \(b\), or \(c\).

Question 7. Simplify the following expressions:
(i) \( \frac{8^{3a} \times 2^5 \times 2^{2a}}{4 \times 2^{11a} \times 2^{-2a}} \)
(ii) \( \frac{3 \times 27^{n+1} + 9 \times 3^{3n-1}}{8 \times 3^{3n} - 5 \times 27^n} \)
Answer:
(i) \( \frac{8^{3a} \times 2^5 \times 2^{2a}}{4 \times 2^{11a} \times 2^{-2a}} \)
\( \implies \frac{\left(2^3\right)^{3a} \times 2^5 \times 2^{2a}}{2^2 \times 2^{11a} \times 2^{-2a}} \)
\( \implies \frac{2^{9a} \times 2^5 \times 2^{2a}}{2^2 \times 2^{11a} \times 2^{-2a}} \)
\( \implies 2^{9a + 5 + 2a - (2 + 11a - 2a)} \)
\( \implies 2^{9a + 5 + 2a - 2 - 11a + 2a} \)
\( \implies 2^{2a+3} \)

(ii) \( \frac{3 \times 27^{n+1} + 9 \times 3^{3n-1}}{8 \times 3^{3n} - 5 \times 27^n} \)
\( \implies \frac{3 \times (3^3)^{n+1} + 3^2 \times 3^{3n-1}}{2^3 \times 3^{3n} - 5 \times (3^3)^n} \)
\( \implies \frac{3 \times 3^{3n+3} + 3^{3n+1}}{2^3 \times (3^3)^n - 5 \times (3^3)^n} \)
\( \implies \frac{3^{3n+4} + 3^{3n+1}}{2^3 \times (3^3)^n - 5 \times (3^3)^n} \)
\( \implies \frac{3^{3n} \times 3^4 + 3^{3n} \times 3^1}{2^3 \times \left(3^3\right)^n - 5 \times \left(3^3\right)^n} \)
\( \implies \frac{3^{3n} \left(3^4 + 3^1\right)}{\left(3^3\right)^n (8 - 5)} \)
\( \implies \frac{3^{3n} \left(3^4 + 3^1\right)}{3^{3n} \times 3} \)
\( \implies \frac{3^4 + 3^1}{3} \)
\( \implies \frac{3 \times 3 \times 3 \times 3 + 3}{3} \)
\( \implies \frac{81+3}{3} \)
\( \implies \frac{84}{3} \)
\( \implies 28 \)
In simple words: To simplify fractional expressions with powers, rewrite all bases using their prime factors (like changing 8 to \(2^3\), 27 to \(3^3\), and 9 to \(3^2\)). Then group common exponential terms in the numerator and denominator so they can be simplified or canceled out.

Exam Tip: Be careful when simplifying terms like \(3^{3n+4}\). You can write it as \(3^{3n} \times 3^4\), which allows you to factor out the \(3^{3n}\) term and easily cancel it from the fraction.

Question 8. Prove that: \( \left(\frac{a^m}{a^{-n}}\right)^{m-n} \times \left(\frac{a^n}{a^{-l}}\right)^{n-l} \times \left(\frac{a^l}{a^{-m}}\right)^{l-m} = 1 \)
Answer:
LHS:
\( \left(\frac{a^m}{a^{-n}}\right)^{m-n} \times \left(\frac{a^n}{a^{-l}}\right)^{n-l} \times \left(\frac{a^l}{a^{-m}}\right)^{l-m} \)
\( \implies \left(a^m \times a^n\right)^{m-n} \times \left(a^n \times a^l\right)^{n-l} \times \left(a^l \times a^m\right)^{l-m} \)
\( \implies \left(a^{m+n}\right)^{m-n} \times \left(a^{n+l}\right)^{n-l} \times \left(a^{l+m}\right)^{l-m} \)
\( \implies a^{(m+n)(m-n)} \times a^{(n+l)(n-l)} \times a^{(l+m)(l-m)} \)
\( \implies a^{m^2 - n^2} \times a^{n^2 - l^2} \times a^{l^2 - m^2} \)
\( \implies a^{m^2 - n^2 + n^2 - l^2 + l^2 - m^2} \)
\( \implies a^0 \)
\( \implies 1 = \text{RHS} \)
In simple words: First, simplify the fractions inside each bracket by moving the negative exponents from the bottom to the top. Then, multiply the powers together, which results in the terms canceling out to give a final exponent of 0. Since any number raised to the power of 0 is 1, the result is 1.

Exam Tip: Remember the algebraic identity \((x-y)(x+y) = x^2 - y^2\). Applying this identity directly to the exponents saves time and prevents algebraic mistakes.

Question 9. If \( a = x^{m+n} \cdot x^l \), \( b = x^{n+l} \cdot x^m \), and \( c = x^{l+m} \cdot x^n \), prove that \( a^{m-n} \cdot b^{n-l} \cdot c^{l-m} = 1 \).
Answer:
LHS:
\( a^{m-n} \cdot b^{n-l} \cdot c^{l-m} \)
Substituting the values of \( a \), \( b \), and \( c \) into the expression:
\( \implies \left(x^{m+n} \cdot x^l\right)^{m-n} \cdot \left(x^{n+l} \cdot x^m\right)^{n-l} \cdot \left(x^{l+m} \cdot x^n\right)^{l-m} \)
\( \implies x^{(m+n)(m-n)} \cdot x^{l(m-n)} \cdot x^{(n+l)(n-l)} \cdot x^{m(n-l)} \cdot x^{(l+m)(l-m)} \cdot x^{n(l-m)} \)
\( \implies x^{m^2 - n^2} \cdot x^{ml - nl} \cdot x^{n^2 - l^2} \cdot x^{mn - ml} \cdot x^{l^2 - m^2} \cdot x^{nl - mn} \)
\( \implies x^{(m^2 - n^2 + ml - nl + n^2 - l^2 + mn - ml + l^2 - m^2 + nl - mn)} \)
Grouping and canceling the exponents of \( x \):
\( \implies x^0 \)
\( \implies 1 = \text{RHS} \)
In simple words: Substitute the given formulas for a, b, and c into the expression. Expand the powers using multiplication, and then add all the exponents of x. Every single term cancels out, leaving \(x^0\), which equals 1.

Exam Tip: Be methodical when expanding the exponents. Crossing out matching positive and negative terms (like \(+ml\) and \(-ml\)) with a pencil during your rough work ensures you do not miss any terms.

Question 10. Simplify the following expressions:
(i) \( \left(\frac{x^a}{x^b}\right)^{a^2 + ab + b^2} \times \left(\frac{x^b}{x^c}\right)^{b^2 + bc + c^2} \times \left(\frac{x^c}{x^a}\right)^{c^2 + ca + a^2} \)
(ii) \( \left(\frac{x^a}{x^{-b}}\right)^{a^2 - ab + b^2} \times \left(\frac{x^b}{x^{-c}}\right)^{b^2 - bc + c^2} \times \left(\frac{x^c}{x^{-a}}\right)^{c^2 - ca + a^2} \)
Answer:
(i) \( \left(\frac{x^a}{x^b}\right)^{a^2 + ab + b^2} \times \left(\frac{x^b}{x^c}\right)^{b^2 + bc + c^2} \times \left(\frac{x^c}{x^a}\right)^{c^2 + ca + a^2} \)
\( \implies \left(x^{a-b}\right)^{a^2 + ab + b^2} \times \left(x^{b-c}\right)^{b^2 + bc + c^2} \times \left(x^{c-a}\right)^{c^2 + ca + a^2} \)
Using the identity \( (y-z)\left(y^2 + yz + z^2\right) = y^3 - z^3 \):
\( \implies x^{a^3 - b^3} \times x^{b^3 - c^3} \times x^{c^3 - a^3} \)
\( \implies x^{a^3 - b^3 + b^3 - c^3 + c^3 - a^3} \)
\( \implies x^0 \)
\( \implies 1 \)

(ii) \( \left(\frac{x^a}{x^{-b}}\right)^{a^2 - ab + b^2} \times \left(\frac{x^b}{x^{-c}}\right)^{b^2 - bc + c^2} \times \left(\frac{x^c}{x^{-a}}\right)^{c^2 - ca + a^2} \)
\( \implies \left(x^{a+b}\right)^{a^2 - ab + b^2} \times \left(x^{b+c}\right)^{b^2 - bc + c^2} \times \left(x^{c+a}\right)^{c^2 - ca + a^2} \)
Using the identity \( (y+z)\left(y^2 - yz + z^2\right) = y^3 + z^3 \):
\( \implies x^{a^3 + b^3} \times x^{b^3 + c^3} \times x^{c^3 + a^3} \)
\( \implies x^{a^3 + b^3 + b^3 + c^3 + c^3 + a^3} \)
\( \implies x^{2a^3 + 2b^3 + 2c^3} \)
\( \implies x^{2\left(a^3 + b^3 + c^3\right)} \)
In simple words: Use algebraic identity rules for the sum and difference of cubes, i.e., \((y-z)(y^2+yz+z^2) = y^3-z^3\) and \((y+z)(y^2-yz+z^2) = y^3+z^3\). Once you write the exponents in this form, simplify them by addition.

Exam Tip: Memorizing algebraic identities for cubes is highly recommended. Recognizing these patterns instantly tells you how the exponent terms will simplify, which prevents long-winded expansions.

Exercise 7(B)

Question 1. Solve the following equations for \( x \):
(i) \( 2^{2x+1} = 8 \)
(ii) \( 2^{5x-1} = 4 \times 2^{3x+1} \)
(iii) \( 3^{4x+1} = 27^{x+1} \)
(iv) \( 49^{x+4} = 7^2(343)^{x+1} \)
Answer:
(i) \( 2^{2x+1} = 8 \)
\( \implies 2^{2x+1} = 2^3 \)
Since the bases are identical on both sides, their exponents must be equal:
\( \implies 2x + 1 = 3 \)
\( \implies 2x = 3 - 1 \)
\( \implies 2x = 2 \)
\( \implies x = \frac{2}{2} \)
\( \implies x = 1 \)

(ii) \( 2^{5x-1} = 4 \times 2^{3x+1} \)
\( \implies 2^{5x-1} = 2^2 \times 2^{3x+1} \)
\( \implies 2^{5x-1} = 2^{3x+1+2} \)
\( \implies 2^{5x-1} = 2^{3x+3} \)
Comparing the exponents from both sides since the bases are equal:
\( \implies 5x - 1 = 3x + 3 \)
\( \implies 5x - 3x = 3 + 1 \)
\( \implies 2x = 4 \)
\( \implies x = \frac{4}{2} \)
\( \implies x = 2 \)

(iii) \( 3^{4x+1} = 27^{x+1} \)
\( \implies 3^{4x+1} = \left(3^3\right)^{x+1} \)
\( \implies 3^{4x+1} = 3^{3x+3} \)
Since the bases are equal, their powers must be equal:
\( \implies 4x + 1 = 3x + 3 \)
\( \implies 4x - 3x = 3 - 1 \)
\( \implies x = 2 \)

(iv) \( 49^{x+4} = 7^2(343)^{x+1} \)
\( \implies (7 \times 7)^{x+4} = 7^2(7 \times 7 \times 7)^{x+1} \)
\( \implies \left(7^2\right)^{x+4} = 7^2\left(7^3\right)^{x+1} \)
\( \implies 7^{2x+8} = 7^2 \times 7^{3x+3} \)
\( \implies 7^{2x+8} = 7^{3x+3+2} \)
\( \implies 7^{2x+8} = 7^{3x+5} \)
Equating the exponents as the bases are identical on both sides:
\( \implies 2x + 8 = 3x + 5 \)
\( \implies 3x - 2x = 8 - 5 \)
\( \implies x = 3 \)
In simple words: To solve exponential equations, rewrite both sides so they have the exact same base. Once the bases match, you can drop them and set the exponents equal to each other to solve for x.

Exam Tip: Always make sure to simplify the powers on one side into a single term before equating exponents. For example, in part (iv), combine \(7^2 \times 7^{3x+3}\) into \(7^{3x+5}\) before setting exponents equal.

Question 2. Solve the following equations for \( x \):
(i) \( 4^{2x} = \frac{1}{32} \)
(ii) \( \sqrt{2}^{x+3} = 16 \)
(iii) \( \left(\sqrt{\frac{3}{5}}\right)^{x+1} = \frac{125}{27} \)
(iv) \( \left(\sqrt[3]{\frac{2}{3}}\right)^{x-1} = \frac{27}{8} \)
Answer:
(i) \( 4^{2x} = \frac{1}{32} \)
\( \implies \left(2^2\right)^{2x} = \frac{1}{2^5} \)
\( \implies 2^{4x} = 2^{-5} \)
Equating the exponents since the bases are equal:
\( \implies 4x = -5 \)
\( \implies x = -\frac{5}{4} \)

(ii) \( \sqrt{2}^{x+3} = 16 \)
\( \implies \left(2^{\frac{1}{2}}\right)^{x+3} = 2^4 \)
\( \implies 2^{\frac{x+3}{2}} = 2^4 \)
Comparing the exponents from both sides:
\( \implies \frac{x+3}{2} = 4 \)
\( \implies x + 3 = 8 \)
\( \implies x = 8 - 3 \)
\( \implies x = 5 \)

(iii) \( \left(\sqrt{\frac{3}{5}}\right)^{x+1} = \frac{125}{27} \)
\( \implies \left[\left(\frac{3}{5}\right)^{\frac{1}{2}}\right]^{x+1} = \frac{5 \times 5 \times 5}{3 \times 3 \times 3} \)
\( \implies \left(\frac{3}{5}\right)^{\frac{x+1}{2}} = \left(\frac{5}{3}\right)^3 \)
\( \implies \left(\frac{3}{5}\right)^{\frac{x+1}{2}} = \left(\frac{3}{5}\right)^{-3} \)
Equating the exponents as the bases are equal:
\( \implies \frac{x+1}{2} = -3 \)
\( \implies x + 1 = -6 \)
\( \implies x = -6 - 1 \)
\( \implies x = -7 \)

(iv) \( \left(\sqrt[3]{\frac{2}{3}}\right)^{x-1} = \frac{27}{8} \)
\( \implies \left[\left(\frac{2}{3}\right)^{\frac{1}{3}}\right]^{x-1} = \frac{3^3}{2^3} \)
\( \implies \left(\frac{2}{3}\right)^{\frac{x-1}{3}} = \left(\frac{3}{2}\right)^3 \)
\( \implies \left(\frac{2}{3}\right)^{\frac{x-1}{3}} = \left(\frac{2}{3}\right)^{-3} \)
Equating the exponents since the bases match:
\( \implies \frac{x-1}{3} = -3 \)
\( \implies x - 1 = -9 \)
\( \implies x = -9 + 1 \)
\( \implies x = -8 \)
In simple words: Remember that roots can be written as fractional exponents (square root is power of 1/2, cube root is power of 1/3). If you need to flip a fraction to make the bases match, just change the sign of its exponent.

Exam Tip: When dealing with fractional bases, remember that \( \left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n \). Using this reciprocal rule helps you easily match bases on both sides of the equation.

Question 3. Solve the following equations for \( x \):
(i) \( 4^{x-2} - 2^{x+1} = 0 \)
(ii) \( 3^{x^2} : 3^x = 9 : 1 \)
Answer:
(i) \( 4^{x-2} - 2^{x+1} = 0 \)
\( \implies 4^{x-2} = 2^{x+1} \)
\( \implies \left(2^2\right)^{x-2} = 2^{x+1} \)
\( \implies 2^{2x-4} = 2^{x+1} \)
Equating the exponents as the bases are identical:
\( \implies 2x - 4 = x + 1 \)
\( \implies 2x - x = 4 + 1 \)
\( \implies x = 5 \)

(ii) \( 3^{x^2} : 3^x = 9 : 1 \)
\( \implies \frac{3^{x^2}}{3^x} = \frac{9}{1} \)
\( \implies 3^{x^2} = 9 \times 3^x \)
\( \implies 3^{x^2} = 3^2 \times 3^x \)
\( \implies 3^{x^2} = 3^{x+2} \)
Since the bases are equal, we equate their powers:
\( \implies x^2 = x + 2 \)
\( \implies x^2 - x - 2 = 0 \)
Factoring the quadratic equation:
\( \implies x^2 - 2x + x - 2 = 0 \)
\( \implies x(x-2) + 1(x-2) = 0 \)
\( \implies (x+1)(x-2) = 0 \)
This gives:
\( \implies x + 1 = 0 \text{ or } x - 2 = 0 \)
\( \implies x = -1 \text{ or } x = 2 \)
In simple words: For equations involving quadratic exponents, equate the powers to set up a quadratic equation in terms of x. Then solve the quadratic equation using splitting-the-middle-term to find the possible values of x.

Exam Tip: Be sure to write down both solutions of the quadratic equation. Students often forget to list the negative value, but both values are valid answers unless there are specific constraints given in the problem.

Question 4(i). Solve for \( x \): \( 8 \times 2^{2x} + 4 \times 2^{x+1} = 1 + 2^x \)
Answer:
Given equation:
\( 8 \times 2^{2x} + 4 \times 2^{x+1} = 1 + 2^x \)
\( \implies 8(2^x)^2 + 4 \times 2^x \times 2^1 = 1 + 2^x \)
\( \implies 8(2^x)^2 + 8(2^x) - 2^x - 1 = 0 \)
\( \implies 8(2^x)^2 + 2^x(8 - 1) - 1 = 0 \)
\( \implies 8(2^x)^2 + 7(2^x) - 1 = 0 \)

Let \( 2^x = y \). Substituting this variable into our quadratic equation gives:
\( 8y^2 + 7y - 1 = 0 \)
\( \implies 8y^2 + 8y - y - 1 = 0 \)
\( \implies 8y(y + 1) - 1(y + 1) = 0 \)
\( \implies (8y - 1)(y + 1) = 0 \)
\( \implies 8y - 1 = 0 \text{ or } y + 1 = 0 \)
\( \implies y = \frac{1}{8} \text{ or } y = -1 \)

Substituting \( y = 2^x \) back into our equations:
\( 2^x = \frac{1}{8} \text{ or } 2^x = -1 \)

Since an exponential function with a positive base can never produce a negative output, \( 2^x = -1 \) has no real solution.
Solving the other part:
\( 2^x = 2^{-3} \)
\( \implies x = -3 \)

In simple words: Change the expression by using a substitute letter \( y \) for \( 2^x \) to make a quadratic equation. Once solved, find \( x \) using the positive value since exponential terms cannot yield negative numbers.

Exam Tip: Be sure to explicitly mention that negative outputs like \( 2^x = -1 \) are impossible, as examiners look for this justification when awarding full marks.

Question 4(ii). Solve for \( x \): \( 2^{2x} + 2^{x+2} - 4 \times 2^3 = 0 \)
Answer:
Given equation:
\( 2^{2x} + 2^{x+2} - 4 \times 2^3 = 0 \)
\( \implies (2^x)^2 + 2^x \times 2^2 - 4 \times 8 = 0 \)
\( \implies (2^x)^2 + 4(2^x) - 32 = 0 \)

Let \( 2^x = y \). Substituting this variable gives:
\( y^2 + 4y - 32 = 0 \)
\( \implies y^2 + 8y - 4y - 32 = 0 \)
\( \implies y(y + 8) - 4(y + 8) = 0 \)
\( \implies (y + 8)(y - 4) = 0 \)
\( \implies y = -8 \text{ or } y = 4 \)

Substituting \( y = 2^x \) back into our equations:
\( 2^x = -8 \text{ or } 2^x = 4 \)

Since \( 2^x \) must be greater than zero, we ignore the negative equation.
Solving the other part:
\( 2^x = 2^2 \)
\( \implies x = 2 \)

In simple words: Convert the equation into a quadratic form using a temporary variable. Factor the quadratic equation to get two values, and discard the negative one to find the final value of \( x \).

Exam Tip: Pay close attention to factorization signs. A small sign error while breaking up the middle term can lead to incorrect roots.

Question 4(iii). Solve for \( x \): \( (\sqrt{3})^{x-3} = (\sqrt[4]{3})^{x+1} \)
Answer:
Given equation:
\( (\sqrt{3})^{x-3} = (\sqrt[4]{3})^{x+1} \)
\( \implies \left( 3^{\frac{1}{2}} \right)^{x-3} = \left( 3^{\frac{1}{4}} \right)^{x+1} \)
\( \implies 3^{\frac{x-3}{2}} = 3^{\frac{x+1}{4}} \)

Since the bases on both sides are identical, their exponents must be equal:
\( \frac{x-3}{2} = \frac{x+1}{4} \)
\( \implies 4(x - 3) = 2(x + 1) \)
\( \implies 4x - 12 = 2x + 2 \)
\( \implies 4x - 2x = 12 + 2 \)
\( \implies 2x = 14 \)
\( \implies x = \frac{14}{2} \)
\( \implies x = 7 \br />
In simple words: Rewrite the square roots as fractional exponents. Once the bases on both sides match, set the power expressions equal and solve the linear equation for \( x \).

Exam Tip: Simplify the fraction values before cross-multiplying to keep the numbers small and reduce any chances of basic calculation errors.

Question 5. If \( 4^{2m} = (\sqrt[3]{16})^{-\frac{6}{n}} = (\sqrt{8})^2 \), find the values of \( m \) and \( n \).
Answer:
We can split the given chain of equations into two individual equations:
\( 4^{2m} = (\sqrt{8})^2 \) - (Equation 1)
\( (\sqrt[3]{16})^{-\frac{6}{n}} = (\sqrt{8})^2 \) - (Equation 2)

From Equation (1):
\( 4^{2m} = (\sqrt{8})^2 \)
\( \implies (2^2)^{2m} = \left(\sqrt{2^3}\right)^2 \)
\( \implies 2^{4m} = \left( (2^3)^{\frac{1}{2}} \right)^2 \)
\( \implies 2^{4m} = 2^{3 \times \frac{1}{2} \times 2} \)
\( \implies 2^{4m} = 2^3 \)

Since the bases are identical, we equate the exponents:
\( 4m = 3 \)
\( \implies m = \frac{3}{4} \)

From Equation (2):
\( (\sqrt[3]{16})^{-\frac{6}{n}} = (\sqrt{8})^2 \)
\( \implies \left( \sqrt[3]{2 \times 2 \times 2 \times 2} \right)^{-\frac{6}{n}} = \left(\sqrt{2 \times 2 \times 2}\right)^2 \)
\( \implies \left( \sqrt[3]{2^4} \right)^{-\frac{6}{n}} = \left(\sqrt{2^3}\right)^2 \)
\( \implies \left[ (2^4)^{\frac{1}{3}} \right]^{-\frac{6}{n}} = \left[ (2^3)^{\frac{1}{2}} \right]^2 \)
\( \implies \left[ 2^{\frac{4}{3}} \right]^{-\frac{6}{n}} = \left[ 2^{\frac{3}{2}} \right]^2 \)
\( \implies 2^{\frac{4}{3} \times \left(-\frac{6}{n}\right)} = 2^{\frac{3}{2} \times 2} \)
\( \implies 2^{-\frac{8}{n}} = 2^3 \)

Equating the exponents:
\( -\frac{8}{n} = 3 \)
\( \implies n = -\frac{8}{3} \)

Hence, \( m = \frac{3}{4} \) and \( n = -\frac{8}{3} \).

In simple words: Equate the first and second terms separately to the final term. Convert all terms to base 2 to solve the simple power equations for \( m \) and \( n \).

Exam Tip: Be careful with fractional exponent signs. Negative fractions in denominators often lead to simple sign mistakes.

Question 6. Solve for \( x \) and \( y \): \( (\sqrt{32})^x \div 2^{y+1} = 1 \) and \( 8^y - 16^{4-\frac{x}{2}} = 0 \).
Answer:
Let us consider the first equation:
\( (\sqrt{32})^x \div 2^{y+1} = 1 \)
\( \implies \left(\sqrt{2 \times 2 \times 2 \times 2 \times 2}\right)^x \div 2^{y+1} = 1 \)
\( \implies \left(\sqrt{2^5}\right)^x \div 2^{y+1} = 1 \)
\( \implies \left[ (2^5)^{\frac{1}{2}} \right]^x \div 2^{y+1} = 2^0 \)
\( \implies 2^{\frac{5x}{2}} \div 2^{y+1} = 2^0 \)
\( \implies \frac{5x}{2} - (y + 1) = 0 \)
\( \implies 5x - 2(y + 1) = 0 \)
\( \implies 5x - 2y - 2 = 0 \)
\( \implies 5x - 2y = 2 \) - (Equation i)

Now, let us consider the second equation:
\( 8^y - 16^{4-\frac{x}{2}} = 0 \)
\( \implies (2^3)^y - (2^4)^{4-\frac{x}{2}} = 0 \)
\( \implies 2^{3y} - 2^{4\left(4-\frac{x}{2}\right)} = 0 \)
\( \implies 2^{3y} = 2^{4\left(4-\frac{x}{2}\right)} \)
\( \implies 3y = 4\left(4 - \frac{x}{2}\right) \)
\( \implies 3y = 16 - 2x \)
\( \implies 2x + 3y = 16 \) - (Equation ii)

We can solve equations (i) and (ii) simultaneously. Multiply Equation (i) by 3, and Equation (ii) by 2:
\( 15x - 6y = 6 \) - (Equation iii)
\( 4x + 6y = 32 \) - (Equation iv)

Adding Equations (iii) and (iv):
\( 19x = 38 \)
\( \implies x = 2 \)

Substitute this value of \( x \) into Equation (i):
\( 5(2) - 2y = 2 \)
\( \implies 10 - 2y = 2 \)
\( \implies 2y = 10 - 2 \)
\( \implies 2y = 8 \)
\( \implies y = 4 \)

Hence, the final solutions are \( x = 2 \) and \( y = 4 \).

In simple words: Convert both exponential expressions into standard linear equations by bringing both sides to base 2. Once done, solve the two simultaneous equations to calculate \( x \) and \( y \).

Exam Tip: Clearly show the steps of simultaneous equation solving, including multiplication factors and additions, to earn step-wise marking credit.

Question 7. Prove that:
(i) \( \left( \frac{x^a}{x^b} \right)^{a+b-c} \times \left( \frac{x^b}{x^c} \right)^{b+c-a} \times \left( \frac{x^c}{x^a} \right)^{c+a-b} = 1 \)
(ii) \( \frac{x^{a(b-c)}}{x^{b(a-c)}} \div \left( \frac{x^b}{x^a} \right)^c = 1 \)
Answer:
(i)
Let's simplify the Left Hand Side (L.H.S.):
\( \text{L.H.S.} = \left( x^{a-b} \right)^{a+b-c} \times \left( x^{b-c} \right)^{b+c-a} \times \left( x^{c-a} \right)^{c+a-b} \)
\( = x^{(a-b)(a+b-c)} \times x^{(b-c)(b+c-a)} \times x^{(c-a)(c+a-b)} \)
\( = x^{a^2+ab-ac-ab-b^2+bc} \times x^{b^2+bc-ab-cb-c^2+ac} \times x^{c^2+ac-bc-ac-a^2+ab} \)
\( = x^{a^2-b^2-ac+bc} \times x^{b^2-c^2-ab+ac} \times x^{c^2-a^2-bc+ab} \)
\( = x^{a^2-b^2-ac+bc+b^2-c^2-ab+ac+c^2-a^2-bc+ab} \)
\( = x^0 \)
\( = 1 = \text{R.H.S.} \)
Hence, proved.

(ii)
Let's simplify the Left Hand Side (L.H.S.):
\( \text{L.H.S.} = \frac{x^{a(b-c)}}{x^{b(a-c)}} \div \left(\frac{x^b}{x^a}\right)^c \)
\( = x^{a(b-c)-b(a-c)} \div \frac{x^{bc}}{x^{ac}} \)
\( = x^{ab-ac-ab+bc} \div x^{bc-ac} \)
\( = x^{bc-ac} \div x^{bc-ac} \)
\( = x^{bc-ac-(bc-ac)} \)
\( = x^0 \)
\( = 1 = \text{R.H.S.} \)
Hence, proved.

In simple words: Subtract division exponents under the same base. When multiplying variables with identical bases, add their overall powers together, which simplifies the entire index to 0, giving a value of 1.

Exam Tip: Be very methodical while multiplying binomials and trinomials like \( (a-b)(a+b-c) \). A single wrong sign will prevent the exponents from cancelling out to 0.

Question 8. If \( a^x = b \), \( b^y = c \), and \( c^z = a \), prove that \( xyz = 1 \).
Answer:
Given equations:
\( a^x = b \), \( b^y = c \), and \( c^z = a \)

Consider the first equation:
\( a^x = b \)

Raise both sides to the power of \( yz \):
\( \implies a^{xyz} = b^{yz} \)
\( \implies a^{xyz} = (b^y)^z \)

Substitute \( b^y = c \) into this relation:
\( \implies a^{xyz} = (c)^z \)
\( \implies a^{xyz} = c^z \)

Now substitute \( c^z = a \) into our relation:
\( \implies a^{xyz} = a \)
\( \implies a^{xyz} = a^1 \)

Comparing the exponents on both sides:
\( \implies xyz = 1 \)
Hence, proved.

In simple words: Take the first equation and raise both sides to the product power of the other exponents. Use substitution with the remaining equations to show the relation holds.

Exam Tip: Raising to the product power is a rapid and clean approach for cyclic exponent proofs. Practice this method to save time during tests.

Question 9. If \( a^x = b^y = c^z \) and \( b^2 = ac \), prove that \( y = \frac{2xz}{z+x} \).
Answer:
Let:
\( a^x = b^y = c^z = k \)

We can express the bases \( a, b, \text{ and } c \) in terms of the constant \( k \):
\( a = k^{\frac{1}{x}} \)
\( b = k^{\frac{1}{y}} \)
\( c = k^{\frac{1}{z}} \)

We are given the relation:
\( b^2 = ac \)

Substitute the values of \( a, b, \text{ and } c \) into this equation:
\( \left( k^{\frac{1}{y}} \right)^2 = \left( k^{\frac{1}{x}} \right) \times \left( k^{\frac{1}{z}} \right) \)
\( \implies k^{\frac{2}{y}} = k^{\frac{1}{x} + \frac{1}{z}} \)
\( \implies k^{\frac{2}{y}} = k^{\frac{z+x}{xz}} \)

Since the bases on both sides are matching, we equate their powers:
\( \frac{2}{y} = \frac{z+x}{xz} \)
\( \implies y = \frac{2xz}{z+x} \)
Hence, proved.

In simple words: Equate the matching terms to a single constant \( k \) to express the variables \( a, b, c \) as fractional powers of \( k \). Plug these values into \( b^2 = ac \) to derive the final expression for \( y \).

Exam Tip: Using a constant parameter 'k' is a powerful method for multi-variable ratios. Clearly state this step at the beginning of your proof.

Question 10. If \( 5^{-p} = 4^{-q} = 20^r \), show that \( \frac{1}{p} + \frac{1}{q} + \frac{1}{r} = 0 \).
Answer:
Let:
\( 5^{-p} = 4^{-q} = 20^r = k \)

This gives us individual values in terms of the constant \( k \):
\( 5 = k^{-\frac{1}{p}} \)
\( 4 = k^{-\frac{1}{q}} \)
\( 20 = k^{\frac{1}{r}} \)

We know the mathematical identity:
\( 5 \times 4 = 20 \)

Substitute our constant expressions into this identity:
\( k^{-\frac{1}{p}} \times k^{-\frac{1}{q}} = k^{\frac{1}{r}} \)
\( \implies k^{-\frac{1}{p} - \frac{1}{q}} = k^{\frac{1}{r}} \)

Comparing the exponents:
\( -\frac{1}{p} - \frac{1}{q} = \frac{1}{r} \)
\( \implies \frac{1}{p} + \frac{1}{q} + \frac{1}{r} = 0 \)
Hence, proved.

In simple words: Express the bases 5, 4, and 20 as fractional exponents of a constant \( k \). Multiply the terms for 5 and 4 to equal 20, combine their exponent expressions, and rearrange to solve.

Exam Tip: Pay close attention to negative signs during reciprocal operations. A misplaced negative can change the final sum value from 0.

Question 11. If \( (m+n)^{-1}(m^{-1} + n^{-1}) = m^x n^y \), show that \( x + y + 2 = 0 \).
Answer:
Given relation:
\( (m+n)^{-1}(m^{-1} + n^{-1}) = m^x n^y \)
\( \implies \frac{1}{m+n} \left( \frac{1}{m} + \frac{1}{n} \right) = m^x n^y \)
\( \implies \frac{1}{m+n} \left( \frac{m+n}{mn} \right) = m^x n^y \)

Cancelling out the common term \( (m+n) \) from the numerator and denominator:
\( \implies \frac{1}{mn} = m^x n^y \)
\( \implies m^{-1}n^{-1} = m^x n^y \)

By matching corresponding exponents on both sides, we find:
\( x = -1 \) and \( y = -1 \)

Now, calculate the value of \( x + y + 2 \):
\( \text{L.H.S.} = x + y + 2 \)
\( = (-1) + (-1) + 2 \)
\( = -2 + 2 \)
\( = 0 = \text{R.H.S.} \)
Hence, proved.

In simple words: Simplify the negative exponents on the left side first. Once you cancel out the common terms, match the powers of \( m \) and \( n \) to find \( x \) and \( y \), and plug them in to prove the result is 0.

Exam Tip: Always look for factoring simplifications like cancelling the \( (m+n) \) binomial before carrying out lengthy expansions.

Question 12. If \( 5^{x+1} = 25^{x-2} \), find the value of \( 3^{x-3} \times 2^{3-x} \).
Answer:
Given equation:
\( 5^{x+1} = 25^{x-2} \)
\( \implies 5^{x+1} = (5^2)^{x-2} \)
\( \implies 5^{x+1} = 5^{2x-4} \)

Since the bases on both sides are identical, we equate the exponents:
\( x + 1 = 2x - 4 \)
\( \implies 2x - x = 4 + 1 \)
\( \implies x = 5 \)

Now, we substitute \( x = 5 \) into our second expression:
\( 3^{x-3} \times 2^{3-x} = 3^{5-3} \times 2^{3-5} \)
\( = 3^2 \times 2^{-2} \)
\( = 9 \times \frac{1}{2^2} \)
\( = 9 \times \frac{1}{4} \)
\( = \frac{9}{4} \)

In simple words: Solve the first exponential equation for \( x \) by writing both sides with a base of 5. Substitute this value of \( x \) into the second expression to get the final fraction.

Exam Tip: Be sure to keep negative exponent fractions in the denominator correctly (e.g., \( 2^{-2} = 1/4 \)) when carrying out the final valuation steps.

Question 13. If \( 4^{x+3} = 112 + 8 \times 4^x \), find the value of \( (18x)^{3x} \).
Answer:
Given equation:
\( 4^{x+3} = 112 + 8 \times 4^x \)
\( \implies 4^x \times 4^3 = 112 + 8 \times 4^x \)
\( \implies 64 \times 4^x = 112 + 8 \times 4^x \)

Let \( 4^x = y \). This gives:
\( 64y = 112 + 8y \)
\( \implies 64y - 8y = 112 \)
\( \implies 56y = 112 \)
\( \implies y = 2 \)

Substitute back \( y = 4^x \):
\( 4^x = 2 \)
\( \implies (2^2)^x = 2^1 \)
\( \implies 2^{2x} = 2^1 \)

Comparing the exponents:
\( 2x = 1 \)
\( \implies x = \frac{1}{2} \)

Now, substitute \( x = \frac{1}{2} \) into the target expression:
\( (18x)^{3x} = \left( 18 \times \frac{1}{2} \right)^{3 \times \frac{1}{2}} \)
\( = (9)^{\frac{3}{2}} \)
\( = (3^2)^{\frac{3}{2}} \)
\( = 3^3 \)
\( = 27 \)

In simple words: Use a substitute variable like \( y \) for \( 4^x \) to solve the initial equation for \( x \). Substitute \( x = 1/2 \) into the target expression and solve the fractional exponent.

Exam Tip: Splitting terms like \( 4^{x+3} \) to \( 4^x \times 4^3 \) is the crucial first step. Avoid attempting to divide the constants before splitting exponents.

Question 14(i). Solve for \( x \): \( 4^{x-1} \times (0.5)^{3-2x} = \left(\frac{1}{8}\right)^{-x} \)
Answer:
We can rewrite the equation terms using a common base of 2:
\( 4^{x-1} = (2^2)^{x-1} = 2^{2x-2} \)
\( (0.5)^{3-2x} = \left(\frac{1}{2}\right)^{3-2x} = (2^{-1})^{3-2x} = 2^{2x-3} \)
\( \left(\frac{1}{8}\right)^{-x} = (2^{-3})^{-x} = 2^{3x} \)

Substitute these base-2 expressions back into our main equation:
\( 2^{2x-2} \times 2^{2x-3} = 2^{3x} \)
\( \implies 2^{(2x-2) + (2x-3)} = 2^{3x} \)
\( \implies 2^{4x-5} = 2^{3x} \)

Since the bases are now identical, we compare exponents:
\( 4x - 5 = 3x \)
\( \implies 4x - 3x = 5 \)
\( \implies x = 5 \)

In simple words: Express all numbers in the equation with a base of 2. Add the power expressions on the left together, set them equal to the right power, and solve for \( x \).

Exam Tip: Recognizing that decimals like 0.5 can be written as \( 2^{-1} \) is a very useful skill for working with base-2 equations.

Question 14(ii). Solve for \( x \): \( a^{2(3x+5)} \times a^{4x} = a^{8x+12} \)
Answer:
Using exponential rules for multiplying identical bases, we add the exponents:
\( a^{2(3x+5) + 4x} = a^{8x+12} \)
\( \implies a^{6x+10+4x} = a^{8x+12} \)
\( \implies a^{10x+10} = a^{8x+12} \)

Since the bases on both sides are identical, their exponents must be equal:
\( 10x + 10 = 8x + 12 \)
\( \implies 10x - 8x = 12 - 10 \)
\( \implies 2x = 2 \)
\( \implies x = 1 \)

In simple words: Combine the powers on the left-hand side through addition. Set the simplified left power equal to the right-hand side power and solve the simple equation.

Exam Tip: Be sure to distribute the outer multiplier correctly to all terms inside the parentheses (e.g., \( 2(3x+5) = 6x + 10 \)).

Question 14(iii). Solve for \( x \): \( (81)^{\frac{3}{4}} - \left(\frac{1}{32}\right)^{-\frac{2}{5}} + x\left(\frac{1}{2}\right)^{-1} \cdot 2^0 = 27 \)
Answer:
Let's simplify each part of our equation individually:
\( (81)^{\frac{3}{4}} = (3^4)^{\frac{3}{4}} = 3^3 = 27 \)
\( \left(\frac{1}{32}\right)^{-\frac{2}{5}} = (2^{-5})^{-\frac{2}{5}} = 2^2 = 4 \)
\( x\left(\frac{1}{2}\right)^{-1} \cdot 2^0 = x(2^1) \cdot 1 = 2x \)

Substitute these simplified expressions back into the main equation:
\( 27 - 4 + 2x = 27 \)
\( \implies 23 + 2x = 27 \)
\( \implies 2x = 27 - 23 \)
\( \implies 2x = 4 \)
\( \implies x = 2 \)

In simple words: Simplify each numerical index term first. Put the numerical results back into the equation to create a simple linear form, and solve for \( x \).

Exam Tip: Always remember that any base raised to the power of 0 equals 1 (e.g., \( 2^0 = 1 \)). Forgetting this basic rule is a common source of simple errors.

Question 14(iv). Solve for \( x \): \( 2^{3x} \times 2^3 = 2^{3x} \times 2 + 48 \)
Answer:
Let's simplify the exponential expressions:
\( 8 \times 2^{3x} = 2 \times 2^{3x} + 48 \)
\( \implies 8 \times 2^{3x} - 2 \times 2^{3x} = 48 \)
\( \implies 2^{3x}(8 - 2) = 48 \)
\( \implies 2^{3x} \times 6 = 48 \)
\( \implies 2^{3x} = \frac{48}{6} \)
\( \implies 2^{3x} = 8 \)
\( \implies 2^{3x} = 2^3 \)

Since the bases are identical, we equate the powers:
\( 3x = 3 \)
\( \implies x = 1 \)

In simple words: Treat \( 2^{3x} \) as a single algebraic variable block. Rearrange and combine the matching terms, divide by their coefficient, and solve for \( x \) using base 2.

Exam Tip: Treating terms like \( 2^{3x} \) as a variable block (e.g., \( 8A - 2A = 6A \)) makes regrouping operations intuitive and clean.

Question 14(v). Solve for \( x \): \( 3(2^x + 1) - 2^{x+2} + 5 = 0 \)
Answer:
Expand the brackets and rewrite the equation terms:
\( 3 \times 2^x + 3 - 2^x \times 2^2 + 5 = 0 \)
\( \implies 3 \times 2^x + 3 - 4 \times 2^x + 5 = 0 \)
\( \implies 2^x(3 - 4) + 8 = 0 \)
\( \implies 2^x(-1) + 8 = 0 \)
\( \implies -2^x = -8 \)
\( \implies 2^x = 8 \)
\( \implies 2^x = 2^3 \)

Comparing the exponents:
\( x = 3 \)

In simple words: Expand the equation to separate exponential and constant terms. Group them together, isolate \( 2^x \) on one side, and solve for \( x \) using power comparison.

Exam Tip: Ensure that \( 2^{x+2} \) is split into \( 2^x \times 2^2 \). This makes it easy to group and simplify matching exponential terms.

Exercise 7(C)

Question 1. Evaluate:
(i) \( 9^{\frac{5}{2}} - 3 \times 8^0 - \left(\frac{1}{81}\right)^{-\frac{1}{2}} \)
(ii) \( (64)^{\frac{2}{3}} - \sqrt[3]{125} - \frac{1}{2^{-5}} + (27)^{-\frac{2}{3}} \times \left(\frac{25}{9}\right)^{-\frac{1}{2}} \)
(iii) \( \left[ \left(-\frac{2}{3}\right)^{-2} \right]^3 \times \left(\frac{1}{3}\right)^{-4} \times 3^{-1} \times \frac{1}{6} \)
Answer:
(i)
\( 9^{\frac{5}{2}} - 3 \times 8^0 - \left(\frac{1}{81}\right)^{-\frac{1}{2}} \)
\( = (3^2)^{\frac{5}{2}} - 3 \times 1 - \left( 3^{-4} \right)^{-\frac{1}{2}} \)
\( = 3^{2 \times \frac{5}{2}} - 3 - 3^{-4 \times -\frac{1}{2}} \)
\( = 3^5 - 3 - 3^2 \)
\( = 243 - 3 - 9 \)
\( = 231 \)

(ii)
\( (64)^{\frac{2}{3}} - \sqrt[3]{125} - \frac{1}{2^{-5}} + (27)^{-\frac{2}{3}} \times \left(\frac{25}{9}\right)^{-\frac{1}{2}} \)
\( = (4^3)^{\frac{2}{3}} - \sqrt[3]{5^3} - 2^5 + (3^3)^{-\frac{2}{3}} \times \left( \left(\frac{5}{3}\right)^2 \right)^{-\frac{1}{2}} \)
\( = 4^{3 \times \frac{2}{3}} - 5 - 32 + 3^{3 \times -\frac{2}{3}} \times \left(\frac{5}{3}\right)^{2 \times -\frac{1}{2}} \)
\( = 4^2 - 5 - 32 + 3^{-2} \times \left(\frac{5}{3}\right)^{-1} \)
\( = 16 - 5 - 32 + \frac{1}{9} \times \frac{3}{5} \)
\( = -21 + \frac{3}{45} \)
\( = -21 + \frac{1}{15} \)
\( = \frac{-315 + 1}{15} \)
\( = \frac{-314}{15} \)
\( = -20\frac{14}{15} \)

(iii)
\( \left[ \left(-\frac{2}{3}\right)^{-2} \right]^3 \times \left(\frac{1}{3}\right)^{-4} \times 3^{-1} \times \frac{1}{6} \)
\( = \left(-\frac{2}{3}\right)^{-6} \times \left(3^{-1}\right)^{-4} \times 3^{-1} \times \frac{1}{6} \)
\( = \left(-\frac{3}{2}\right)^6 \times 3^4 \times \frac{1}{3} \times \frac{1}{6} \)
\( = \frac{3^6}{2^6} \times 3^3 \times \frac{1}{3 \times 2} \)
\( = \frac{3^6 \times 3^2}{2^6 \times 2} \)
\( = \frac{3^8}{2^7} \)

In simple words: Simplify each element separately by breaking composite numbers into prime bases. Apply index rules to multiply powers and combine terms at the end for the final evaluation.

Exam Tip: Improper fraction values should be converted into mixed fraction form for the final step to ensure standard presentation and score full marks.

Question 2. Simplify: \( \frac{3 \times 9^{n+1} - 9 \times 3^{2n}}{3 \times 3^{2n+3} - 9^{n+1}} \)
Answer:
Let us express all terms with base 3:
Numerator:
\( 3 \times 9^{n+1} - 9 \times 3^{2n} \)
\( = 3 \times (3^2)^{n+1} - 3^2 \times 3^{2n} \)
\( = 3^1 \times 3^{2n+2} - 3^{2n+2} \)
\( = 3^{2n+3} - 3^{2n+2} \)
\( = 3^{2n}(3^3 - 3^2) \)

Denominator:
\( 3 \times 3^{2n+3} - 9^{n+1} \)
\( = 3^1 \times 3^{2n+3} - (3^2)^{n+1} \)
\( = 3^{2n+4} - 3^{2n+2} \)
\( = 3^{2n}(3^4 - 3^2) \)

Now, assemble the simplified fraction:
\( \frac{3^{2n}(3^3 - 3^2)}{3^{2n}(3^4 - 3^2)} \)

Cancelling the common term \( 3^{2n} \) from both parts:
\( = \frac{3^3 - 3^2}{3^4 - 3^2} \)
\( = \frac{27 - 9}{81 - 9} \)
\( = \frac{18}{72} \)
\( = \frac{1}{4} \br />
In simple words: Rewrite composite numbers as powers of 3. Factor out and cancel the common base-3 index expression from both the numerator and denominator, leaving plain numbers to evaluate.

Exam Tip: Factoring out common bases like \( 3^{2n} \) is a very safe strategy to avoid complicated division and power subtractions.

Question 3. Solve for \( x \) and \( y \): \( 3^{x-1} \times 5^{2y-3} = 225 \).
Answer:
Let us find the prime factors of 225:
\( 225 = 9 \times 25 = 3^2 \times 5^2 \)

Now write the equation using these prime factors:
\( 3^{x-1} \times 5^{2y-3} = 3^2 \times 5^2 \)

By comparing corresponding exponents of matching bases on both sides, we get:
For base 3:
\( x - 1 = 2 \)
\( \implies x = 3 \)

For base 5:
\( 2y - 3 = 2 \)
\( \implies 2y = 5 \)
\( \implies y = \frac{5}{2} = 2\frac{1}{2} \)

Hence, \( x = 3 \) and \( y = 2\frac{1}{2} \).

In simple words: Break 225 down into its prime components, 3 and 5. Match the exponents of 3 and 5 separately on both sides to solve for \( x \) and \( y \).

Exam Tip: Clearly show the prime factorization step for composite constants to demonstrate your logical path to the examiner.

Question 4. If \( \left(\frac{a^{-1}b^2}{a^2b^{-4}}\right)^7 \div \left(\frac{a^3b^{-5}}{a^{-2}b^3}\right)^{-5} = a^x \cdot b^y \), find the value of \( x + y \).
Answer:
First, let us simplify the expressions inside the brackets:
First expression:
\( \left( \frac{a^{-1}b^2}{a^2b^{-4}} \right)^7 = \left( a^{-1-2}b^{2-(-4)} \right)^7 \)
\( = \left( a^{-3}b^6 \right)^7 \)
\( = a^{-21}b^{42} \)

Second expression:
\( \left( \frac{a^3b^{-5}}{a^{-2}b^3} \right)^{-5} = \left( a^{3-(-2)}b^{-5-3} \right)^{-5} \)
\( = \left( a^5b^{-8} \right)^{-5} \)
\( = a^{-25}b^{40} \)

Now perform the division of these simplified terms:
\( a^{-21}b^{42} \div a^{-25}b^{40} \)
\( = \frac{a^{-21}b^{42}}{a^{-25}b^{40}} \)
\( = a^{-21-(-25)}b^{42-40} \)
\( = a^4b^2 \)

Comparing this result with the Right Hand Side (R.H.S.):
\( a^4b^2 = a^x b^y \)

Matching the exponents gives:
\( x = 4 \) and \( y = 2 \)

Now compute the final sum:
\( x + y = 4 + 2 = 6 \)

In simple words: Simplify the internal base expressions first. Apply the outer exponents, perform index division subtraction, and match exponents to find the sum of \( x \) and \( y \).

Exam Tip: Watch out for negative sign collisions during exponent subtractions, e.g., \( -21 - (-25) = -21 + 25 = 4 \).

Question 5. If \( 3^{x+1} = 9^{x-3} \), find the value of \( 2^{1+x} \).
Answer:
Given equation:
\( 3^{x+1} = 9^{x-3} \)
\( \implies 3^{x+1} = (3^2)^{x-3} \)
\( \implies 3^{x+1} = 3^{2x-6} \)

Comparing the exponents:
\( x + 1 = 2x - 6 \)
\( \implies 2x - x = 1 + 6 \)
\( \implies x = 7 \)

Now substitute \( x = 7 \) into the target expression:
\( 2^{1+x} = 2^{1+7} = 2^8 = 256 \)

In simple words: Bring both sides of the equation to base 3 to calculate \( x \). Substitute that value of \( x \) into the target expression to evaluate the final value of 256.

Exam Tip: Memorizing powers of 2 up to \( 2^{10} \) is incredibly helpful for doing speedy evaluations in indexes.

Question 6. If \( 2^x = 4^y = 8^z \) and \( \frac{1}{2x} + \frac{1}{4y} + \frac{1}{8z} = 4 \), find the value of \( x \).
Answer:
Given equations:
\( 2^x = 4^y = 8^z \)
\( \implies 2^x = (2^2)^y = (2^3)^z \)
\( \implies 2^x = 2^{2y} = 2^{3z} \)

Equating powers gives:
\( x = 2y = 3z \)

Express \( y \) and \( z \) in terms of \( x \):
\( y = \frac{x}{2} \)
\( z = \frac{x}{3} \)

Now substitute these relations into the second equation:
\( \frac{1}{2x} + \frac{1}{4y} + \frac{1}{8z} = 4 \)
\( \implies \frac{1}{2x} + \frac{1}{4\left(\frac{x}{2}\right)} + \frac{1}{8\left(\frac{x}{3}\right)} = 4 \)
\( \implies \frac{1}{2x} + \frac{1}{2x} + \frac{3}{8x} = 4 \)

Convert these fractional terms to a common denominator of \( 8x \):
\( \implies \frac{4 + 4 + 3}{8x} = 4 \)
\( \implies \frac{11}{8x} = 4 \)
\( \implies 32x = 11 \)
\( \implies x = \frac{11}{32} \)

In simple words: Relate all variables to \( x \) using the first set of matching equations. Plug these equivalents into the second equation, combine terms, and solve.

Exam Tip: Take extra care when inverting complex denominators, e.g., \( \frac{1}{8(x/3)} = \frac{3}{8x} \). Avoid simple denominator division mistakes.

Question 7. If \( \frac{9^n \cdot 3^2 \cdot 3^n - (27)^n}{(3^m \cdot 2)^3} = 3^{-3} \), prove that \( m - n = 1 \).
Answer:
Express all terms using base 3:
Numerator:
\( 9^n \cdot 3^2 \cdot 3^n - (27)^n = (3^2)^n \cdot 3^2 \cdot 3^n - (3^3)^n \)
\( = 3^{2n} \cdot 3^2 \cdot 3^n - 3^{3n} \)
\( = 3^{3n+2} - 3^{3n} \)
\( = 3^{3n}\left( 3^2 - 1 \right) \)
\( = 3^{3n}(9 - 1) \)
\( = 3^{3n} \cdot 8 \)

Denominator:
\( \left( 3^m \cdot 2 \right)^3 = 3^{3m} \cdot 2^3 = 3^{3m} \cdot 8 \)

Substitute these simplified parts back into the main equation:
\( \frac{3^{3n} \cdot 8}{3^{3m} \cdot 8} = 3^{-3} \)
\( \implies \frac{3^{3n}}{3^{3m}} = 3^{-3} \)
\( \implies 3^{3n-3m} = 3^{-3} \)

Comparing the exponents:
\( 3n - 3m = -3 \)
\( \implies -3(m - n) = -3 \)
\( \implies m - n = 1 \)
Hence, proved.

In simple words: Convert all numbers to base 3. Factor out \( 3^{3n} \) to simplify the numerator, cancel out the coefficient of 8, and equate exponents to prove the relation.

Exam Tip: Avoid the mistake of attempting to combine terms like \( 3^{3n+2} - 3^{3n} \) into a single term without factoring out common factors first.

Question 8. Solve for \( x \): \( (13)^{\sqrt{x}} = 4^4 - 3^4 - 6 \)
Answer:
Let's simplify the constant numerical terms on the right-hand side:
\( 4^4 = 256 \)
\( 3^4 = 81 \)
\( \implies 4^4 - 3^4 - 6 = 256 - 81 - 6 \)
\( = 175 - 6 \)
\( = 169 \)

Now, formulate the equation with base 13:
\( (13)^{\sqrt{x}} = 169 \)
\( \implies (13)^{\sqrt{x}} = 13^2 \)

Since the bases match, equate the exponents:
\( \sqrt{x} = 2 \)
Squaring both sides of the equation:
\( x = 4 \)

In simple words: Evaluate the arithmetic on the right side to get 169, which is the square of 13. Equating powers shows that the square root of \( x \) is 2, meaning \( x \) is 4.

Exam Tip: Be sure to square both sides to remove radical elements like \( \sqrt{x} \). Do not make the mistake of taking roots again.

Question 9. If \( 3^{4x} = (81)^{-1} \) and \( (10)^{\frac{1}{y}} = 0.0001 \), find the value of \( 2^{-x} \times 16^y \).
Answer:
Let us first solve for \( x \):
\( 3^{4x} = (81)^{-1} \)
\( \implies 3^{4x} = (3^4)^{-1} \)
\( \implies 3^{4x} = 3^{-4} \)
Comparing exponents:
\( 4x = -4 \)
\( \implies x = -1 \)

Now solve for \( y \):
\( (10)^{\frac{1}{y}} = 0.0001 = \frac{1}{10000} = 10^{-4} \)
Comparing exponents:
\( \frac{1}{y} = -4 \)
\( \implies y = -\frac{1}{4} \)

Now, substitute \( x = -1 \) and \( y = -\frac{1}{4} \) into our target expression:
\( 2^{-x} \times 16^y = 2^{-(-1)} \times 16^{-\frac{1}{4}} \)
\( = 2^1 \times (2^4)^{-\frac{1}{4}} \)
\( = 2^1 \times 2^{-1} \)
\( = 2^{1-1} \)
\( = 2^0 \)
\( = 1 \)

In simple words: Solve the separate power equations to calculate \( x = -1 \) and \( y = -1/4 \). Plug these results into the final expression and simplify to get 1.

Exam Tip: Writing decimal fractions in base-10 exponent forms makes determining the power values of \( y \) very simple and safe.

Question 10. Solve for \( x \): \( 3(2^x + 1) - 2^{x+2} + 5 = 0 \)
Answer:
Let us expand the bracket and rewrite exponential components:
\( 3 \times 2^x + 3 - 2^x \times 2^2 + 5 = 0 \)
\( \implies 3 \times 2^x + 3 - 4 \times 2^x + 5 = 0 \)

Combine the matching constant and exponential terms:
\( \implies 2^x(3 - 4) + 8 = 0 \)
\( \implies -2^x + 8 = 0 \)
\( \implies 2^x = 8 \)
\( \implies 2^x = 2^3 \)

Comparing the exponents:
\( x = 3 \)

In simple words: Open up the parentheses and split the compound power. Group the matching variables and constants together to find that \( 2^x = 8 \), meaning \( x = 3 \).

Exam Tip: Practice separating exponential index additions like \( 2^{x+2} = 2^x \times 2^2 \) to make factoring equations straightforward.

Question 11. If \( (a^m)^n = a^m \cdot a^n \), prove that \( m(n-1) - (n-1) = 1 \).
Answer:
Given equation:
\( (a^m)^n = a^m \cdot a^n \)
\( \implies a^{mn} = a^{m+n} \)
Comparing the exponents:
\( mn = m + n \) - (Equation 1)

Let us simplify the target Left Hand Side (L.H.S.) expression:
\( \text{L.H.S.} = m(n - 1) - (n - 1) \)
\( = mn - m - n + 1 \)
\( = mn - (m + n) + 1 \)

Substitute Equation (1) into this expression:
\( = mn - mn + 1 \)
\( = 1 = \text{R.H.S.} \)
Hence, proved.

In simple words: Equate the combined powers to find that \( mn = m + n \). Expand the algebraic target and substitute this value to show the expression equals 1.

Exam Tip: Grouping similar algebraic terms like \( -m-n \) into a factored bracket of \( -(m+n) \) makes substitution clear for exam evaluation.

Question 12. If \( m = \sqrt[3]{15} \) and \( n = \sqrt[3]{14} \), show that \( m - n - \frac{1}{m^2 + mn + n^2} = 0 \).
Answer:
Cubing both sides of our given parameters gives:
\( m^3 = 15 \) and \( n^3 = 14 \)

Let us simplify our target algebraic expression by bringing terms to a common denominator:
\( m - n - \frac{1}{m^2 + mn + n^2} = \frac{(m - n)(m^2 + mn + n^2) - 1}{m^2 + mn + n^2} \)

Using the standard algebraic identity \( (m-n)(m^2+mn+n^2) = m^3 - n^3 \):
\( = \frac{m^3 - n^3 - 1}{m^2 + mn + n^2} \)

Now, substitute the cubed values of \( m \) and \( n \):
\( = \frac{15 - 14 - 1}{m^2 + mn + n^2} \)
\( = \frac{0}{m^2 + mn + n^2} \)
\( = 0 \)
Hence, proved.

In simple words: Put the terms under a single common denominator. Use algebraic identities to simplify the top portion to \( m^3 - n^3 - 1 \), which becomes 0 upon substitution.

Exam Tip: Memorizing standard identities such as the difference of cubes is highly recommended to simplify multi-variable fractions rapidly.