Selina Concise Solutions for ICSE Class 8 Mathematics Chapter 2 Exponents

Question 1. Evaluate:
(i) \( (3^{-1} \times 9^{-1}) \div 3^{-2} \)
(ii) \( (3^{-1} \times 4^{-1}) \div 6^{-1} \)
(iii) \( (2^{-1} + 3^{-1})^3 \)
(iv) \( (3^{-1} \div 4^{-1})^2 \)
(v) \( (2^2 + 3^2) \times \left(\frac{1}{2}\right)^2 \)
(vi) \( (5^2 - 3^2) \times \left(\frac{2}{3}\right)^{-3} \)
(vii) \( \left[\left(\frac{1}{4}\right)^{-3} - \left(\frac{1}{3}\right)^{-3}\right] \div \left(\frac{1}{6}\right)^{-3} \)
(viii) \( \left[\left(-\frac{3}{4}\right)^{-2}\right]^2 \)
(ix) \( \left\{\left(\frac{3}{5}\right)^{-2}\right\}^{-2} \)
(x) \( (5^{-1} \times 3^{-1}) \div 6^{-1} \)
Answer:
(i) \( (3^{-1} \times 9^{-1}) \div 3^{-2} \)
\( = \left(\frac{1}{3} \times \frac{1}{9}\right) \div \frac{1}{3^2} \)
\( = \frac{1}{27} \div \frac{1}{9} \)
\( = \frac{1}{27} \times 9 = \frac{1}{3} \)

(ii) \( (3^{-1} \times 4^{-1}) \div 6^{-1} \)
\( = \left(\frac{1}{3} \times \frac{1}{4}\right) \div \frac{1}{6} \)
\( = \frac{1}{12} \div \frac{1}{6} \)
\( = \frac{1}{12} \times 6 = \frac{1}{2} \)

(iii) \( (2^{-1} + 3^{-1})^3 \)
\( = \left(\frac{1}{2} + \frac{1}{3}\right)^3 \)
\( = \left(\frac{3 + 2}{6}\right)^3 \)
\( = \left(\frac{5}{6}\right)^3 \)
\( = \frac{5 \times 5 \times 5}{6 \times 6 \times 6} = \frac{125}{216} \)

(iv) \( (3^{-1} \div 4^{-1})^2 \)
\( = \left(\frac{1}{3} \div \frac{1}{4}\right)^2 \)
\( = \left(\frac{1}{3} \times 4\right)^2 \)
\( = \left(\frac{4}{3}\right)^2 \)
\( = \frac{16}{9} = 1\frac{7}{9} \)

(v) \( (2^2 + 3^2) \times \left(\frac{1}{2}\right)^2 \)
\( = (4 + 9) \times \frac{1}{4} \)
\( = 13 \times \frac{1}{4} = \frac{13}{4} = 3\frac{1}{4} \)

(vi) \( (5^2 - 3^2) \times \left(\frac{2}{3}\right)^{-3} \)
\( = (25 - 9) \times \left(\frac{3}{2}\right)^3 \)
\( = 16 \times \left(\frac{3 \times 3 \times 3}{2 \times 2 \times 2}\right) \)
\( = 16 \times \frac{27}{8} = 2 \times 27 = 54 \)

(vii) \( \left[\left(\frac{1}{4}\right)^{-3} - \left(\frac{1}{3}\right)^{-3}\right] \div \left(\frac{1}{6}\right)^{-3} \)
\( = \left[\left(\frac{4}{1}\right)^3 - \left(\frac{3}{1}\right)^3\right] \div \left(\frac{6}{1}\right)^3 \)
\( = [64 - 27] \div 216 \)
\( = 37 \div 216 = \frac{37}{216} \)

(viii) \( \left[\left(-\frac{3}{4}\right)^{-2}\right]^2 \)
\( = \left(-\frac{3}{4}\right)^{-4} \)
\( = \left(-\frac{4}{3}\right)^4 \)
\( = \frac{(-4) \times (-4) \times (-4) \times (-4)}{3 \times 3 \times 3 \times 3} \)
\( = \frac{256}{81} = 3\frac{13}{81} \)

(ix) \( \left\{\left(\frac{3}{5}\right)^{-2}\right\}^{-2} \)
\( = \left(\frac{3}{5}\right)^{-2 \times (-2)} \)
\( = \left(\frac{3}{5}\right)^4 \)
\( = \frac{3 \times 3 \times 3 \times 3}{5 \times 5 \times 5 \times 5} = \frac{81}{625} \)

(x) \( (5^{-1} \times 3^{-1}) \div 6^{-1} \)
\( = \left(\frac{1}{5} \times \frac{1}{3}\right) \div \frac{1}{6} \)
\( = \frac{1}{15} \div \frac{1}{6} \)
\( = \frac{1}{15} \times 6 = \frac{2}{5} \)
In simple words: First, change any negative powers to positive ones by flipping the fractions. Then, solve the math inside the brackets before doing the rest of the operations.

Exam Tip: Be careful with signs. Flipping a fraction only changes the sign of its exponent, not the sign of the base itself.

Question 2. If \( 1125 = 3^m \times 5^n \); find m and n.
Answer:
First, let us find the prime factorization of 1125 by dividing it: \[ \begin{array}{r|l} 3 & 1125 \\ \hline 3 & 375 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array} \] Thus, the factors of 1125 are:
\( 1125 = 3 \times 3 \times 5 \times 5 \times 5 \)
\( 1125 = 3^2 \times 5^3 \)
On comparing this to the given expression \( 3^m \times 5^n \), we get:
\( m = 2 \) and \( n = 3 \)
In simple words: Break down the number 1125 into its basic building blocks by dividing. Since there are two 3s and three 5s, the powers are 2 and 3.

Exam Tip: Always show the step-by-step division table for prime factorization to secure full marks for the working.

Question 3. Find x, if \( 9 \times 3^x = (27)^{2x-3} \)
Answer:
First, rewrite the terms 9 and 27 with base 3:
\( 9 \times 3^x = (27)^{2x-3} \)
\( 3^2 \times 3^x = (3^3)^{2x-3} \)
Apply the rules of exponents:
\( 3^{x+2} = 3^{3(2x-3)} \)

\( \implies 3^{x+2} = 3^{6x-9} \)
Since both sides share the same base, we can compare their powers directly:
\( x + 2 = 6x - 9 \)
Rearranging the variables and numbers:
\( 6x - x = 9 + 2 \)

\( \implies 5x = 11 \)

\( \implies x = \frac{11}{5} \)

\( \implies x = 2\frac{1}{5} \)
In simple words: Write both sides of the equation using the same base number, 3. Since the bases are the same, their powers must be equal, which lets us solve for x like a normal equation.

Exam Tip: When multiplying exponents with the same base, remember to add the powers together. When dealing with a power of a power, multiply the powers.

Exercise 2 (B)

Question 1. Compute:
(i) \( 1^8 \times 3^0 \times 5^3 \times 2^2 \)
(ii) \( (4^7)^2 \times (4^{-3})^4 \)
(iii) \( (2^{-9} \div 2^{-11})^3 \)
(iv) \( \left(\frac{2}{3}\right)^{-4} \times \left(\frac{27}{8}\right)^{-2} \)
(v) \( \left(\frac{56}{28}\right)^0 \div \left(\frac{2}{5}\right)^3 \times \frac{16}{25} \)
(vi) \( (12)^{-2} \times 3^3 \)
(vii) \( (-5)^4 \times (-5)^6 \div (-5)^9 \)
(viii) \( \left(-\frac{1}{3}\right)^4 \div \left(-\frac{1}{3}\right)^8 \times \left(-\frac{1}{3}\right)^5 \)
(ix) \( 9^0 \times 4^{-1} \div 2^{-4} \)
(x) \( (625)^{-\frac{3}{4}} \)
(xi) \( \left(\frac{27}{64}\right)^{-\frac{2}{3}} \)
(xii) \( \left(\frac{1}{32}\right)^{-\frac{2}{5}} \)
(xiii) \( (125)^{-\frac{2}{3}} \div (8)^{\frac{2}{3}} \)
(xiv) \( (243)^{\frac{2}{5}} \div (32)^{-\frac{2}{5}} \)
(xv) \( (-3)^4 - (\sqrt{3})^0 \times (-2)^5 \div (64)^{\frac{2}{3}} \)
(xvi) \( (27)^{\frac{2}{3}} \div \left(\frac{81}{16}\right)^{-\frac{1}{4}} \)
Answer:
(i) \( 1^8 \times 3^0 \times 5^3 \times 2^2 \)
Since \( 1^8 = 1 \) and \( 3^0 = 1 \):
\( = 1 \times 1 \times (5 \times 5 \times 5) \times (2 \times 2) \)
\( = 125 \times 4 = 500 \)

(ii) \( (4^7)^2 \times (4^{-3})^4 \)
Multiply the powers:
\( = 4^{14} \times 4^{-12} \)
Add the exponents:
\( = 4^{14 - 12} = 4^2 = 16 \)

(iii) \( (2^{-9} \div 2^{-11})^3 \)
Subtract the powers inside the bracket first:
\( = (2^{-9 - (-11)})^3 \)
\( = (2^{2})^3 \)
\( = 2^{6} = 64 \)

(iv) \( \left(\frac{2}{3}\right)^{-4} \times \left(\frac{27}{8}\right)^{-2} \)
Flip the fractions to make powers positive:
\( = \left(\frac{3}{2}\right)^4 \times \left(\frac{8}{27}\right)^2 \)
Write 8 and 27 as powers of 2 and 3:
\( = \frac{3^4}{2^4} \times \frac{(2^3)^2}{(3^3)^2} \)
\( = \frac{3^4}{2^4} \times \frac{2^6}{3^6} \)
\( = 2^{6 - 4} \times 3^{4 - 6} \)
\( = 2^2 \times 3^{-2} = \frac{4}{9} \)

(v) \( \left(\frac{56}{28}\right)^0 \div \left(\frac{2}{5}\right)^3 \times \frac{16}{25} \)
Since any non-zero number raised to 0 is 1:
\( = 1 \div \frac{2^3}{5^3} \times \frac{2^4}{5^2} \)
Convert division to multiplication by inverting the fraction:
\( = 1 \times \frac{5^3}{2^3} \times \frac{2^4}{5^2} \)
\( = 5^{3-2} \times 2^{4-3} \)
\( = 5^1 \times 2^1 = 10 \)

(vi) \( (12)^{-2} \times 3^3 \)
Write 12 as product of its prime factors \( 2^2 \times 3 \):
\( = (2^2 \times 3)^{-2} \times 3^3 \)
\( = (2^{-4} \times 3^{-2}) \times 3^3 \)
\( = 2^{-4} \times 3^{3-2} \)
\( = 2^{-4} \times 3^1 = \frac{3}{16} \)

(vii) \( (-5)^4 \times (-5)^6 \div (-5)^9 \)
Combine exponents with the same base:
\( = (-5)^{4 + 6 - 9} \)
\( = (-5)^1 = -5 \)

(viii) \( \left(-\frac{1}{3}\right)^4 \div \left(-\frac{1}{3}\right)^8 \times \left(-\frac{1}{3}\right)^5 \)
Add and subtract exponents since bases are identical:
\( = \left(-\frac{1}{3}\right)^{4 - 8 + 5} \)
\( = \left(-\frac{1}{3}\right)^1 = -\frac{1}{3} \)

(ix) \( 9^0 \times 4^{-1} \div 2^{-4} \)
Since \( 9^0 = 1 \):
\( = 1 \times \frac{1}{4} \div \frac{1}{2^4} \)
\( = \frac{1}{4} \times 2^4 = \frac{1}{2^2} \times 2^4 \)
\( = 2^{4 - 2} = 2^2 = 4 \)

(x) \( (625)^{-\frac{3}{4}} \)
Write 625 as \( 5^4 \):
\( = (5^4)^{-\frac{3}{4}} \)
Multiply the powers:
\( = 5^{-3} = \frac{1}{5^3} = \frac{1}{125} \)

(xi) \( \left(\frac{27}{64}\right)^{-\frac{2}{3}} \)
Flip the fraction to make the power positive:
\( = \left(\frac{64}{27}\right)^{\frac{2}{3}} \)
Express terms as powers of 4 and 3:
\( = \left(\frac{4^3}{3^3}\right)^{\frac{2}{3}} \)
\( = \frac{4^2}{3^2} = \frac{16}{9} = 1\frac{7}{9} \)

(xii) \( \left(\frac{1}{32}\right)^{-\frac{2}{5}} \)
Write 32 as \( 2^5 \):
\( = \left(2^{-5}\right)^{-\frac{2}{5}} \)
Multiply exponents:
\( = 2^2 = 4 \)

(xiii) \( (125)^{-\frac{2}{3}} \div (8)^{\frac{2}{3}} \)
Write 125 as \( 5^3 \) and 8 as \( 2^3 \):
\( = (5^3)^{-\frac{2}{3}} \div (2^3)^{\frac{2}{3}} \)
\( = 5^{-2} \div 2^2 \)
\( = \frac{1}{25} \div 4 = \frac{1}{25 \times 4} = \frac{1}{100} \)

(xiv) \( (243)^{\frac{2}{5}} \div (32)^{-\frac{2}{5}} \)
Express 243 as \( 3^5 \) and 32 as \( 2^5 \):
\( = (3^5)^{\frac{2}{5}} \div (2^5)^{-\frac{2}{5}} \)
\( = 3^2 \div 2^{-2} \)
Convert the negative exponent:
\( = 3^2 \times 2^2 = 9 \times 4 = 36 \)

(xv) \( (-3)^4 - (\sqrt{3})^0 \times (-2)^5 \div (64)^{\frac{2}{3}} \)
Solve each term:
\( (-3)^4 = 81 \)
\( (\sqrt{3})^0 = 1 \)
\( (-2)^5 = -32 \)
\( (64)^{\frac{2}{3}} = (4^3)^{\frac{2}{3}} = 4^2 = 16 \)
Substitute these values back:
\( = 81 - 1 \times (-32) \div 16 \)
\( = 81 - (-2) = 81 + 2 = 83 \)

(xvi) \( (27)^{\frac{2}{3}} \div \left(\frac{81}{16}\right)^{-\frac{1}{4}} \)
Write 27 as \( 3^3 \), 81 as \( 3^4 \), and 16 as \( 2^4 \):
\( = (3^3)^{\frac{2}{3}} \div \left(\frac{3^4}{2^4}\right)^{-\frac{1}{4}} \)
\( = 3^2 \div \left(\frac{3}{2}\right)^{-1} \)
\( = 9 \div \frac{2}{3} \)
\( = 9 \times \frac{3}{2} = \frac{27}{2} = 13\frac{1}{2} \)
In simple words: Break down large numbers into basic bases like 2, 3, or 5. Use the rules of powers to combine them and simplify the calculations step-by-step.

Exam Tip: Be careful when simplifying negative powers inside division. Changing division to multiplication requires inverting the divisor, which is a common area where simple mistakes occur.

Question 2. Simplify:
(i) \( 8^{\frac{4}{3}} + 25^{\frac{3}{2}} - \left(\frac{1}{27}\right)^{-\frac{2}{3}} \)
(ii) \( [(64)^{-2}]^{-3} \div [\{(-8)^2\}^3]^2 \)
(iii) \( (2^{-3} - 2^{-4})(2^{-3} + 2^{-4}) \)
Answer:
(i) First, we rewrite the bases as powers of prime numbers:
\( 8^{\frac{4}{3}} + 25^{\frac{3}{2}} - \left(\frac{1}{27}\right)^{-\frac{2}{3}} \)
\( = (2^3)^{\frac{4}{3}} + (5^2)^{\frac{3}{2}} - \left(\frac{1}{3^3}\right)^{-\frac{2}{3}} \)
Applying the power rule \( (a^m)^n = a^{m \times n} \):
\( = 2^{3 \times \frac{4}{3}} + 5^{2 \times \frac{3}{2}} - \frac{1}{3^{3 \times \left(-\frac{2}{3}\right)}} \)
\( = 2^4 + 5^3 - \frac{1}{3^{-2}} \)
Since \( \frac{1}{a^{-n}} = a^n \), we can simplify further:
\( = 16 + 125 - 3^2 \)
\( = 141 - 9 \)
\( = 132 \)

(ii) Let us simplify each term using exponent laws:
\( [(64)^{-2}]^{-3} \div [\{(-8)^2\}^3]^2 \)
\( = (2^6)^{-2 \times -3} \div (-8)^{2 \times 3 \times 2} \)
\( = 2^{36} \div (-8)^{12} \)
Express \( -8 \) with base \( -2 \):
\( = 2^{36} \div [(-2)^3]^{12} \)
\( = 2^{36} \div (-2)^{36} \)
\( = \frac{2^{36}}{(-2)^{36}} \)
Since the exponent 36 is an even number, the negative base becomes positive:
\( = \frac{2^{36}}{2^{36}} \)
\( = 2^{36-36} \)
\( = 2^0 \)
\( = 1 \) \( (\because a^0 = 1) \)

(iii) We can apply the identity \( (a-b)(a+b) = a^2 - b^2 \):
\( (2^{-3} - 2^{-4})(2^{-3} + 2^{-4}) \)
\( = (2^{-3})^2 - (2^{-4})^2 \)
\( = 2^{-6} - 2^{-8} \)
Now, rewrite with positive exponents:
\( = \frac{1}{2^6} - \frac{1}{2^8} \)
\( = \frac{1}{64} - \frac{1}{256} \)
Find the common denominator:
\( = \frac{4-1}{256} \)
\( = \frac{3}{256} \)
In simple words: To simplify expressions with exponents, express the bases as prime factors, apply the power of a power rule, and use standard algebraic identities like the difference of squares to solve.

Exam Tip: Be careful with signs when raising negative numbers to even powers, as the result always becomes positive.

Question 3. Evaluate:
(i) \( (-5)^0 \)
(ii) \( 8^0 + 4^0 + 2^0 \)
(iii) \( (8 + 4 + 2)^0 \)
(iv) \( 4x^0 \)
(v) \( (4x)^0 \)
(vi) \( [(10^3)^0]^5 \)
(vii) \( (7x^0)^2 \)
(viii) \( 9^0 + 9^{-1} - 9^{-2} + 9^{\frac{1}{2}} - 9^{-\frac{1}{2}} \)
Answer:
Recall the rule that any non-zero base raised to the power of zero equals 1, i.e., \( a^0 = 1 \).
(i) \( (-5)^0 = 1 \)

(ii) \( 8^0 + 4^0 + 2^0 = 1 + 1 + 1 = 3 \)

(iii) \( (8 + 4 + 2)^0 = (14)^0 = 1 \)

(iv) \( 4x^0 = 4 \times 1 = 4 \)

(v) \( (4x)^0 = 1 \)

(vi) \( [(10^3)^0]^5 = 10^{3 \times 0 \times 5} = 10^0 = 1 \)

(vii) \( (7x^0)^2 = 7^2 \times (x^0)^2 = 49 \times 1^2 = 49 \times 1 = 49 \)

(viii) Let us simplify each term step-by-step:
\( 9^0 + 9^{-1} - 9^{-2} + 9^{\frac{1}{2}} - 9^{-\frac{1}{2}} \)
\( = 1 + \frac{1}{9} - \frac{1}{9^2} + (3^2)^{\frac{1}{2}} - (3^2)^{-\frac{1}{2}} \)
\( = 1 + \frac{1}{9} - \frac{1}{81} + 3^{2 \times \frac{1}{2}} - 3^{2 \times \left(-\frac{1}{2}\right)} \)
\( = 1 + \frac{1}{9} - \frac{1}{81} + 3^1 - 3^{-1} \)
\( = 1 + \frac{1}{9} - \frac{1}{81} + 3 - \frac{1}{3} \)
Combine the numbers and fractions over a common denominator of 81:
\( = \frac{81}{81} + \frac{9}{81} - \frac{1}{81} + \frac{243}{81} - \frac{27}{81} \)
\( = \frac{81 + 9 - 1 + 243 - 27}{81} \)
\( = \frac{333 - 28}{81} \)
\( = \frac{305}{81} \)
Convert to a mixed fraction:
\( = 3\frac{62}{81} \)
In simple words: Remember that any base raised to the power of zero is always equal to 1. When dealing with fractional exponents, write the base as a power to simplify the exponents.

Exam Tip: Pay attention to the difference between \( 4x^0 \) and \( (4x)^0 \). In \( 4x^0 \), only \( x \) is raised to the power 0, while in \( (4x)^0 \), the entire term is raised to 0.

Question 4. Simplify:
(i) \( \frac{a^5 b^2}{a^2 b^{-3}} \)
(ii) \( 15y^8 \div 3y^3 \)
(iii) \( x^{10} y^6 \div x^3 y^{-2} \)
(iv) \( 5z^{16} \div 15z^{-11} \)
(v) \( (36x^2)^{\frac{1}{2}} \)
(vi) \( (125x^{-3})^{\frac{1}{3}} \)
(vii) \( (2x^2 y^{-3})^{-2} \)
(viii) \( (27x^{-3}y^6)^{\frac{2}{3}} \)
(ix) \( (-2x^{2/3} y^{-3/2})^6 \)
Answer:
(i) Using the quotient rule of exponents \( \frac{x^m}{x^n} = x^{m-n} \):
\( \frac{a^5 b^2}{a^2 b^{-3}} = a^{5-2} \cdot b^{2 - (-3)} \)
\( = a^3 b^5 \)

(ii) Divide the coefficients and apply the subtraction rule for exponents of the same base:
\( 15y^8 \div 3y^3 = \frac{15y^8}{3y^3} \)
\( = 5y^{8-3} \)
\( = 5y^5 \)

(iii) Simplify by subtracting the exponents of like bases:
\( x^{10} y^6 \div x^3 y^{-2} = \frac{x^{10}y^6}{x^3 y^{-2}} \)
\( = x^{10-3} \cdot y^{6 - (-2)} \)
\( = x^7 y^8 \)

(iv) Simplify the fractional coefficient and subtract the exponents:
\( 5z^{16} \div 15z^{-11} = \frac{5z^{16}}{15z^{-11}} \)
\( = \frac{5}{15} z^{16 - (-11)} \)
\( = \frac{1}{3} z^{27} \)

(v) Distribute the fractional exponent to each factor inside the parentheses:
\( (36x^2)^{\frac{1}{2}} = (36)^{\frac{1}{2}} \cdot (x^2)^{\frac{1}{2}} \)
\( = (6^2)^{\frac{1}{2}} \cdot x^{2 \times \frac{1}{2}} \)
\( = 6x \)

(vi) Distribute the exponent \( \frac{1}{3} \):
\( (125x^{-3})^{\frac{1}{3}} = (125)^{\frac{1}{3}} \cdot (x^{-3})^{\frac{1}{3}} \)
\( = (5^3)^{\frac{1}{3}} \cdot x^{-3 \times \frac{1}{3}} \)
\( = 5x^{-1} \)
\( = \frac{5}{x} \)

(vii) Apply the exponent outside the parentheses to each term inside:
\( (2x^2 y^{-3})^{-2} = 2^{-2} \cdot (x^2)^{-2} \cdot (y^{-3})^{-2} \)
\( = \frac{1}{2^2} \cdot x^{-4} \cdot y^6 \)
\( = \frac{y^6}{4x^4} = \frac{1}{4} x^{-4} y^6 \)

(viii) Rewrite 27 as \( 3^3 \) and simplify:
\( (27x^{-3}y^6)^{\frac{2}{3}} = (27)^{\frac{2}{3}} \cdot (x^{-3})^{\frac{2}{3}} \cdot (y^6)^{\frac{2}{3}} \)
\( = (3^3)^{\frac{2}{3}} \cdot x^{-2} \cdot y^4 \)
\( = 3^2 \cdot x^{-2} \cdot y^4 \)
\( = 9x^{-2} y^4 = \frac{9y^4}{x^2} \)

(ix) Raise each term to the power of 6:
\( (-2x^{2/3} y^{-3/2})^6 = (-2)^6 \cdot (x^{2/3})^6 \cdot (y^{-3/2})^6 \)
\( = 64 \cdot x^{2/3 \times 6} \cdot y^{-3/2 \times 6} \)
\( = 64x^4 y^{-9} \)
\( = \frac{64x^4}{y^9} \)
In simple words: When simplifying terms with exponents, divide coefficients normally and subtract the exponents of identical bases. Always distribute a power outside of parentheses to all elements inside.

Exam Tip: Be mindful of sign changes when subtracting negative exponents, as subtracting a negative power becomes addition (e.g., \( 2 - (-3) = 5 \)).

Question 5. Simplify: \( (x^{a+b})^{a-b} \cdot (x^{b+c})^{b-c} \cdot (x^{c+a})^{c-a} \)
Answer:
We can multiply the exponents for each term using the rule \( (x^m)^n = x^{m \times n} \):
\( (x^{a+b})^{a-b} \cdot (x^{b+c})^{b-c} \cdot (x^{c+a})^{c-a} \)
\( = x^{(a+b)(a-b)} \cdot x^{(b+c)(b-c)} \cdot x^{(c+a)(c-a)} \)
Now, use the algebraic identity \( (m-n)(m+n) = m^2 - n^2 \) to expand the powers:
\( = x^{a^2-b^2} \cdot x^{b^2-c^2} \cdot x^{c^2-a^2} \)
Since the bases are identical, we add the exponents:
\( = x^{(a^2-b^2) + (b^2-c^2) + (c^2-a^2)} \)
\( = x^{a^2 - b^2 + b^2 - c^2 + c^2 - a^2} \)
All variables in the exponent cancel out to zero:
\( = x^0 \)
\( = 1 \)
In simple words: First multiply the powers inside and outside the parentheses using the difference of squares identity. Since the base is the same throughout, add all the resulting powers together, which simplifies to zero and gives 1.

Exam Tip: Always remember that any base raised to the power of zero is equal to 1. Show each algebraic expansion step clearly to secure full marks.

Question 6. Simplify:
(i) \( \sqrt[5]{x^{20} y^{-10} z^5} \div \frac{x^3}{y^3} \)
(ii) \( \left( \frac{256a^{16}}{81b^4} \right)^{-\frac{3}{4}} \)
Answer:
(i) Express the fifth root as a fractional exponent of \( \frac{1}{5} \):
\( \sqrt[5]{x^{20} y^{-10} z^5} \div \frac{x^3}{y^3} \)
\( = (x^{20}y^{-10}z^5)^{\frac{1}{5}} \div \frac{x^3}{y^3} \)
Distribute the exponent \( \frac{1}{5} \) to each term:
\( = x^{20 \times \frac{1}{5}} \cdot y^{-10 \times \frac{1}{5}} \cdot z^{5 \times \frac{1}{5}} \div \frac{x^3}{y^3} \)
\( = x^4 \cdot y^{-2} \cdot z^1 \times \frac{y^3}{x^3} \)
Group the like terms together and simplify:
\( = x^{4-3} \cdot y^{-2+3} \cdot z^1 \)
\( = x^1 \cdot y^1 \cdot z^1 \)
\( = xyz \)

(ii) Express the numerical bases 256 and 81 as fourth powers:
We know that \( 256 = 4^4 \) and \( 81 = 3^4 \). Substituting these into the expression gives:
\( \left[ \frac{256 a^{16}}{81 b^4} \right]^{-\frac{3}{4}} = \left[ \frac{4^4 a^{16}}{3^4 b^4} \right]^{-\frac{3}{4}} \)
Apply the power \( -\frac{3}{4} \) to both the numerator and denominator:
\( = \frac{4^{4 \times \left(-\frac{3}{4}\right)} a^{16 \times \left(-\frac{3}{4}\right)}}{3^{4 \times \left(-\frac{3}{4}\right)} b^{4 \times \left(-\frac{3}{4}\right)}} \)
\( = \frac{4^{-3} a^{-12}}{3^{-3} b^{-3}} \)
Use negative exponent rules \( x^{-n} = \frac{1}{x^n} \) to make the exponents positive:
\( = \frac{3^3 b^3}{4^3 a^{12}} \)
\( = \frac{27b^3}{64a^{12}} \)
\( = \frac{27}{64} a^{-12} b^3 \)
In simple words: To simplify complex fractional exponents, convert roots into fractional exponents, express large numbers as prime powers, and use negative exponent rules to move terms between the numerator and denominator.

Exam Tip: Highlighting prime factorization of numbers like 256 and 81 makes it much simpler to cancel out fractional denominators in exponents.

Question 7. (i) \( (a^{-2}b)^{-2} \cdot (ab)^{-3} \)
(ii) \( (x^n y^{-m})^4 \times (x^3 y^{-2})^{-n} \)
(iii) \( \left( \frac{125 a^{-3}}{y^6} \right)^{-\frac{1}{3}} \)
(iv) \( \left( \frac{32 x^{-5}}{243 y^{-5}} \right)^{-\frac{1}{5}} \)
(v) \( (a^{-2} b)^{\frac{1}{2}} \times (ab^{-3})^{\frac{1}{3}} \)
(vi) \( (xy)^{m-n} \cdot (yz)^{n-l} \cdot (zx)^{l-m} \)
Answer:
(i) \( (a^{-2}b)^{-2} \cdot (ab)^{-3} \)
\( = (a^{-2 \times -2} \cdot b^{-2}) \cdot (a^{-3} \cdot b^{-3}) \)
\( = a^4 \cdot b^{-2} \cdot a^{-3} \cdot b^{-3} \)
\( = a^{4-3} \cdot b^{-2-3} \)
\( = a \cdot b^{-5} \)
\( = \frac{a}{b^5} \)

(ii) \( (x^n y^{-m})^4 \times (x^3 y^{-2})^{-n} \)
\( = x^{4n} y^{-4m} \times x^{-3n} y^{2n} \)
\( = x^{4n-3n} \cdot y^{-4m+2n} \)
\( = x^n y^{2n-4m} \)

(iii) \( \left( \frac{125 a^{-3}}{y^6} \right)^{-\frac{1}{3}} \)
\( = \left( \frac{5^3 a^{-3}}{y^6} \right)^{-\frac{1}{3}} \)
\( = \frac{5^{3 \times -\frac{1}{3}} \cdot a^{-3 \times -\frac{1}{3}}}{y^{6 \times -\frac{1}{3}}} \)
\( = \frac{5^{-1} \cdot a^1}{y^{-2}} \)
\( = \frac{a y^2}{5} \)

(iv) \( \left( \frac{32 x^{-5}}{243 y^{-5}} \right)^{-\frac{1}{5}} \)
\( = \left( \frac{2^5 x^{-5}}{3^5 y^{-5}} \right)^{-\frac{1}{5}} \)
\( = \frac{2^{5 \times -\frac{1}{5}} \cdot x^{-5 \times -\frac{1}{5}}}{3^{5 \times -\frac{1}{5}} \cdot y^{-5 \times -\frac{1}{5}}} \)
\( = \frac{2^{-1} \cdot x^1}{3^{-1} \cdot y^1} \)
\( = \frac{3x}{2y} \)

(v) \( (a^{-2} b)^{\frac{1}{2}} \times (ab^{-3})^{\frac{1}{3}} \)
\( = (a^{-2 \times \frac{1}{2}} \cdot b^{\frac{1}{2}}) \times (a^{\frac{1}{3}} \cdot b^{-3 \times \frac{1}{3}}) \)
\( = a^{-1} \cdot b^{\frac{1}{2}} \times a^{\frac{1}{3}} \cdot b^{-1} \)
\( = a^{-1 + \frac{1}{3}} \cdot b^{\frac{1}{2} - 1} \)
\( = a^{-\frac{2}{3}} \cdot b^{-\frac{1}{2}} \)
\( = \frac{1}{a^{2/3} b^{1/2}} \)

(vi) \( (xy)^{m-n} \cdot (yz)^{n-l} \cdot (zx)^{l-m} \)
\( = x^{m-n} \cdot y^{m-n} \cdot y^{n-l} \cdot z^{n-l} \cdot z^{l-m} \cdot x^{l-m} \)
\( = x^{m-n+l-m} \cdot y^{m-n+n-l} \cdot z^{n-l+l-m} \)
\( = x^{l-n} \cdot y^{m-l} \cdot z^{n-m} \)
In simple words: To simplify expressions with exponents, we multiply the inner exponents by outer exponents. Then, we combine terms with like bases by adding exponents during multiplication, or moving negative exponents to the other side of the fraction line.

Exam Tip: Be mindful of negative exponent rules. Whenever moving a term from numerator to denominator, or vice versa, always invert the sign of its exponent.

Question 8. Show that:
\( \left( \frac{x^a}{x^{-b}} \right)^{a-b} \cdot \left( \frac{x^b}{x^{-c}} \right)^{b-c} \cdot \left( \frac{x^c}{x^{-a}} \right)^{c-a} = 1 \)
Answer:
L.H.S. \( = \left( \frac{x^a}{x^{-b}} \right)^{a-b} \cdot \left( \frac{x^b}{x^{-c}} \right)^{b-c} \cdot \left( \frac{x^c}{x^{-a}} \right)^{c-a} \)
\( = (x^{a-(-b)})^{a-b} \cdot (x^{b-(-c)})^{b-c} \cdot (x^{c-(-a)})^{c-a} \)
\( = (x^{a+b})^{a-b} \cdot (x^{b+c})^{b-c} \cdot (x^{c+a})^{c-a} \)
\( = x^{(a+b)(a-b)} \cdot x^{(b+c)(b-c)} \cdot x^{(c+a)(c-a)} \)
\( = x^{a^2 - b^2} \cdot x^{b^2 - c^2} \cdot x^{c^2 - a^2} \)
\( = x^{a^2 - b^2 + b^2 - c^2 + c^2 - a^2} \)
\( = x^0 \)
\( = 1 = \) R.H.S.
Hence proved.
In simple words: First, we change the negative power in the denominator into a positive power by bringing it to the top. Next, we multiply the inner and outer exponents using the difference of squares rule. Adding the exponents cancels everything out, leaving \(x^0\), which equals 1.

Exam Tip: Recognizing the algebraic identity \( (a+b)(a-b) = a^2 - b^2 \) early saves time and prevents algebraic mistakes in multi-step exponent calculations.

Question 9. Evaluate:
\( \frac{x^{5+n} \times (x^2)^{3n+1}}{x^{7n-2}} \)
Answer:
\( \frac{x^{5+n} \times (x^2)^{3n+1}}{x^{7n-2}} \)
\( = \frac{x^{5+n} \times x^{2(3n+1)}}{x^{7n-2}} \)
\( = \frac{x^{5+n} \times x^{6n+2}}{x^{7n-2}} \)
\( = \frac{x^{5+n+6n+2}}{x^{7n-2}} \)
\( = \frac{x^{7n+7}}{x^{7n-2}} \)
\( = x^{7n+7 - (7n-2)} \)
\( = x^{7n+7 - 7n+2} \)
\( = x^9 \)
In simple words: We expand the power of a power in the numerator, combine the bases by adding their exponents, and then subtract the exponent of the denominator. All the variables cancel out, leaving us with \(x^9\).

Exam Tip: When subtracting exponents, always put the denominator's exponent in parentheses, like \( -(7n-2) \), to make sure the signs are distributed correctly.

Question 10. Evaluate:
\( \frac{a^{2n+1} \times a^{(2n+1)(2n-1)}}{a^{n(4n-1)} \times (a^2)^{2n+3}} \)
Answer:
\( \frac{a^{2n+1} \times a^{(2n+1)(2n-1)}}{a^{n(4n-1)} \times (a^2)^{2n+3}} \)
\( = \frac{a^{2n+1} \times a^{4n^2-1}}{a^{4n^2-n} \times a^{2(2n+3)}} \)
\( = \frac{a^{2n+1+4n^2-1}}{a^{4n^2-n} \times a^{4n+6}} \)
\( = \frac{a^{4n^2+2n}}{a^{4n^2-n+4n+6}} \)
\( = \frac{a^{4n^2+2n}}{a^{4n^2+3n+6}} \)
\( = a^{4n^2+2n - (4n^2+3n+6)} \)
\( = a^{4n^2+2n - 4n^2-3n-6} \)
\( = a^{-n-6} \)
\( = a^{-(n+6)} \)
\( = \frac{1}{a^{n+6}} \)
In simple words: We simplify the exponents in the top and bottom parts first. Once we combine the terms, we subtract the bottom exponent from the top one, which leaves us with a simplified negative exponent.

Exam Tip: Identify algebraic identities like \( (2n+1)(2n-1) = 4n^2 - 1 \) right away to simplify your expressions more efficiently.

Question 11. \( (m+n)^{-1}(m^{-1}+n^{-1}) = (mn)^{-1} \)
Answer:
L.H.S. \( = (m+n)^{-1}(m^{-1}+n^{-1}) \)
\( = \frac{1}{m+n} \left( \frac{1}{m} + \frac{1}{n} \right) \)
\( = \frac{1}{m+n} \left( \frac{n+m}{mn} \right) \)
\( = \frac{1}{mn} \)
\( = (mn)^{-1} = \) R.H.S.
Hence proved.
In simple words: We convert negative exponents into fractions. After taking a common denominator for the sum inside the parenthesis, the \(m+n\) factors cancel out, leaving us with the desired result of \((mn)^{-1}\).

Exam Tip: Be careful not to expand \( (m+n)^{-1} \) as \( m^{-1} + n^{-1} \), which is a common algebraic error. The whole term must be written as \( \frac{1}{m+n} \).

Question 12. Prove that:
(i) \( \left( \frac{x^a}{x^b} \right)^{\frac{1}{ab}} \left( \frac{x^b}{x^c} \right)^{\frac{1}{bc}} \left( \frac{x^c}{x^a} \right)^{\frac{1}{ca}} = 1 \)
(ii) \( \frac{1}{1+x^{a-b}} + \frac{1}{1+x^{b-a}} = 1 \)
Answer:
(i) L.H.S. \( = \left( \frac{x^a}{x^b} \right)^{\frac{1}{ab}} \left( \frac{x^b}{x^c} \right)^{\frac{1}{bc}} \left( \frac{x^c}{x^a} \right)^{\frac{1}{ca}} \)
\( = (x^{a-b})^{\frac{1}{ab}} \cdot (x^{b-c})^{\frac{1}{bc}} \cdot (x^{c-a})^{\frac{1}{ca}} \)
\( = x^{\frac{a-b}{ab}} \cdot x^{\frac{b-c}{bc}} \cdot x^{\frac{c-a}{ca}} \)
\( = x^{\frac{a-b}{ab} + \frac{b-c}{bc} + \frac{c-a}{ca}} \)
Taking LCM of the denominators as \( abc \):
\( = x^{\frac{c(a-b) + a(b-c) + b(c-a)}{abc}} \)
\( = x^{\frac{ca - cb + ab - ac + bc - ba}{abc}} \)
\( = x^{\frac{0}{abc}} \)
\( = x^0 \)
\( = 1 = \) R.H.S.

(ii) L.H.S. \( = \frac{1}{1+x^{a-b}} + \frac{1}{1+x^{b-a}} \)
Since \( 1 = x^{a-a} = x^{b-b} \):
\( = \frac{1}{x^{a-a}+x^{a-b}} + \frac{1}{x^{b-b}+x^{b-a}} \)
\( = \frac{1}{x^a \cdot x^{-a} + x^a \cdot x^{-b}} + \frac{1}{x^b \cdot x^{-b} + x^b \cdot x^{-a}} \)
Taking common factors from the denominators:
\( = \frac{1}{x^a(x^{-a}+x^{-b})} + \frac{1}{x^b(x^{-b}+x^{-a})} \)
\( = \frac{1}{x^{-a}+x^{-b}} \left[ \frac{1}{x^a} + \frac{1}{x^b} \right] \)
\( = \frac{1}{x^{-a}+x^{-b}} [x^{-a}+x^{-b}] \)
\( = 1 = \) R.H.S.
In simple words: For the first part, we subtract denominator exponents from numerator exponents, distribute the fractional powers, and add them up. For the second part, we use algebraic tricks to rewrite the terms with similar denominators so they cancel out to 1.

Exam Tip: Finding a common denominator like \( abc \) is crucial in fractional exponent problems. Carefully track your additions to ensure everything cancels cleanly.

Question 13. Find the values of n, when:
(i) \( 12^{-5} \times 12^{2n+1} = 12^{13} \div 12^7 \)
(ii) \( \frac{a^{2n-3} \times (a^2)^{n+1}}{(a^4)^{-3}} = (a^3)^3 \div (a^6)^{-3} \)
Answer:
(i) \( 12^{-5} \times 12^{2n+1} = 12^{13} \div 12^7 \)
\( = 12^{-5 + 2n + 1} = \frac{12^{13}}{12^7} \)
\( = 12^{2n-4} = 12^{13-7} \)
\( = 12^{2n-4} = 12^6 \)
Comparing the exponents on both sides:
\( 2n - 4 = 6 \)
\(\implies 2n = 6 + 4\)
\(\implies 2n = 10\)
\(\implies n = 5\)

(ii) \( \frac{a^{2n-3} \times (a^2)^{n+1}}{(a^4)^{-3}} = (a^3)^3 \div (a^6)^{-3} \)
\( = \frac{a^{2n-3} \times a^{2n+2}}{a^{-12}} = a^9 \div a^{-18} \)
\( = \frac{a^{2n-3+2n+2}}{a^{-12}} = \frac{a^9}{a^{-18}} \)
\( = \frac{a^{4n-1}}{a^{-12}} = a^{9-(-18)} \)
\( = a^{4n-1 - (-12)} = a^{27} \)
\( = a^{4n+11} = a^{27} \)
Comparing the exponents on both sides:
\( 4n + 11 = 27 \)
\(\implies 4n = 27 - 11\)
\(\implies 4n = 16\)
\(\implies n = 4\)
In simple words: We simplify the exponents on both sides of the equation until we have a single base with one exponent on each side. Then, we set the exponents equal to each other and solve for \(n\).

Exam Tip: Remember to equate exponents only after simplifying each side of the equation to a single term with the same base.

Question 14. Simplify:
(i) \( \frac{a^{2n+3} \cdot a^{(2n+1)(n+2)}}{(a^3)^{2n+1} \cdot a^{n(2n+1)}} \)
(ii) \( \frac{x^{2n+7} \cdot (x^2)^{3n+2}}{x^{4(2n+3)}} \)
Answer:
(i) \( \frac{a^{2n+3} \cdot a^{(2n+1)(n+2)}}{(a^3)^{2n+1} \cdot a^{n(2n+1)}} \)
Numerator \( = a^{2n+3} \cdot a^{2n^2 + 4n + n + 2} \)
\( = a^{2n+3} \cdot a^{2n^2 + 5n + 2} \)
\( = a^{2n^2 + 7n + 5} \)
Denominator \( = a^{3(2n+1)} \cdot a^{2n^2 + n} \)
\( = a^{6n+3} \cdot a^{2n^2+n} \)
\( = a^{2n^2 + 7n + 3} \)
Therefore, the expression is:
\( = \frac{a^{2n^2 + 7n + 5}}{a^{2n^2 + 7n + 3}} \)
\( = a^{2n^2+7n+5 - (2n^2+7n+3)} \)
\( = a^2 \)

(ii) \( \frac{x^{2n+7} \cdot (x^2)^{3n+2}}{x^{4(2n+3)}} \)
\( = \frac{x^{2n+7} \cdot x^{6n+4}}{x^{8n+12} } \)
\( = \frac{x^{2n+7+6n+4}}{x^{8n+12}} \)
\( = \frac{x^{8n+11}}{x^{8n+12}} \)
\( = x^{8n+11 - (8n+12)} \)
\( = x^{-1} \)
\( = \frac{1}{x} \)
In simple words: We expand and simplify the exponents of the numerator and denominator separately. Once simplified, we subtract the bottom exponent from the top exponent to find the final simplified term.

Exam Tip: Be methodical when expanding binomials in exponent terms. Verify each multiplication step before combining like terms to avoid cumulative errors.