CBSE Class 8 Mathematics Exponents And Powers Worksheet Set 03

Exponents and Powers

Question 1. Find the multiplicative inverse of the following:
(a) \(3^{-4}\)
(b) \(7^{-2}\)
(c) \(9^{-9}\)
(d) \(10^{-80}\)
Answer: The multiplicative inverse (or reciprocal) of a number \(a\) is \(\frac{1}{a}\), or using exponent rules, \(a^{-n}\) becomes \(a^n\).
(a) The multiplicative inverse of \(3^{-4}\) is \(3^4\) (or 81).
(b) The multiplicative inverse of \(7^{-2}\) is \(7^2\) (or 49).
(c) The multiplicative inverse of \(9^{-9}\) is \(9^9\).
(d) The multiplicative inverse of \(10^{-80}\) is \(10^{80}\).
In simple words: To find the multiplicative inverse of a number with a negative exponent, simply change the sign of the exponent from negative to positive.

Exam Tip: Remember that \(a^{-m} \times a^m = 1\), which is why the multiplicative inverse of \(a^{-m}\) is \(a^m\).

Question 2. Expand the following numbers using exponents:
(a) \(18964.63\)
(b) \(1064.373\)
Answer: (a) For \(18964.63\):
\(1 \times 10^4 + 8 \times 10^3 + 9 \times 10^2 + 6 \times 10^1 + 4 \times 10^0 + 6 \times 10^{-1} + 3 \times 10^{-2}\)
(b) For \(1064.373\):
\(1 \times 10^3 + 0 \times 10^2 + 6 \times 10^1 + 4 \times 10^0 + 3 \times 10^{-1} + 7 \times 10^{-2} + 3 \times 10^{-3}\)
In simple words: Write each digit multiplied by its corresponding power of 10 based on its place value.

Exam Tip: Units place is \(10^0\), tens place is \(10^1\), tenths place after the decimal is \(10^{-1}\), hundredths is \(10^{-2}\), and so on.

Question 3. Simplify and write in exponential form:
(a) \((-3)^{-3} \times (-3)^{-2} \times (-3)^{-5}\)
(b) \(a^{-8} \times a^{10} \times a^{-2}\)
(c) \((2^3 \times 2^6 \times 2^2) \div 2^{-6}\)
Answer: Using the law of exponents \(a^m \times a^n = a^{m+n}\) and \(a^m \div a^n = a^{m-n}\):
(a) \((-3)^{-3} \times (-3)^{-2} \times (-3)^{-5} = (-3)^{(-3) + (-2) + (-5)} = (-3)^{-10}\)
(b) \(a^{-8} \times a^{10} \times a^{-2} = a^{-8 + 10 + (-2)} = a^0 = 1\)
(c) \((2^3 \times 2^6 \times 2^2) \div 2^{-6} = (2^{3+6+2}) \div 2^{-6} = 2^{11} \div 2^{-6} = 2^{11 - (-6)} = 2^{17}\)
In simple words: When multiplying bases that are the same, add their exponents. When dividing, subtract the exponent of the denominator from the numerator.

Exam Tip: Any non-zero base raised to the power of 0 is always equal to 1, as seen in part (b).

Question 4. Express \(9^{-3}\) as a power with the base 3.
Answer: Substitute \(9 = 3^2\) into the given expression:
\(9^{-3} = (3^2)^{-3}\)
Using the power of a power rule \((a^m)^n = a^{m \times n}\):
\(= 3^{2 \times (-3)} = 3^{-6}\)
In simple words: Replace the base 9 with \(3^2\) and multiply the exponents together.

Exam Tip: Always convert the base to the prime factorization form (like 2, 3, or 5) before applying the power of a power rule.

Question 5. Find the value of:
(a) \(\left[\left(\frac{1}{2}\right)^0 + \left(\frac{1}{3}\right)^3 + \left(\frac{2}{3}\right)^2\right]\)
(b) \(\left[\left(\frac{1}{3}\right)^{-2} - \left(\frac{1}{2}\right)^{-3}\right] \div \left(\frac{1}{4}\right)^{-2}\)
(c) \(\left(\frac{9}{5}\right)^{-8} \times \left(\frac{5}{9}\right)^{-5}\)
(d) \((9^2 - 4^3) \times \left(\frac{-3}{17}\right)^2 \times \frac{3^4}{9}\)
Answer: (a) \(\left(\frac{1}{2}\right)^0 = 1\)
\(\left(\frac{1}{3}\right)^3 = \frac{1}{27}\)
\(\left(\frac{2}{3}\right)^2 = \frac{4}{9}\)
Sum = \(1 + \frac{1}{27} + \frac{4}{9} = \frac{27 + 1 + 12}{27} = \frac{40}{27}\)

(b) \(\left(\frac{1}{3}\right)^{-2} = 3^2 = 9\)
\(\left(\frac{1}{2}\right)^{-3} = 2^3 = 8\)
\(\left(\frac{1}{4}\right)^{-2} = 4^2 = 16\)
Expression = \((9 - 8) \div 16 = 1 \div 16 = \frac{1}{16}\)

(c) Rewrite \(\left(\frac{9}{5}\right)^{-8}\) as \(\left(\frac{5}{9}\right)^{8}\):
\(= \left(\frac{5}{9}\right)^{8} \times \left(\frac{5}{9}\right)^{-5} = \left(\frac{5}{9}\right)^{8 + (-5)} = \left(\frac{5}{9}\right)^3 = \frac{125}{729}\)

(d) \(9^2 - 4^3 = 81 - 64 = 17\)
\(\left(\frac{-3}{17}\right)^2 = \frac{9}{17^2} = \frac{9}{289}\)
\(\frac{3^4}{9} = \frac{81}{9} = 9\)
Expression = \(17 \times \frac{9}{289} \times 9 = \frac{17 \times 9 \times 9}{17 \times 17} = \frac{81}{17}\)
In simple words: Simplify each bracket separately using exponent rules, perform operations inside brackets first, and then combine terms.

Exam Tip: To clear a negative exponent on a fraction, flip the fraction upside down and make the exponent positive: \(\left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n\).

Question 6. Simplify: \(\left[\left(\frac{1}{4}\right)^4 \times \left(\frac{1}{4}\right)^3\right] \times \left[\left(\frac{3}{5}\right)^{12} \div \left(\frac{3}{5}\right)^5\right]\)
Answer: Simplify the first bracket using product rule \(a^m \times a^n = a^{m+n}\):
\(\left(\frac{1}{4}\right)^{4+3} = \left(\frac{1}{4}\right)^7\)
Simplify the second bracket using quotient rule \(a^m \div a^n = a^{m-n}\):
\(\left(\frac{3}{5}\right)^{12-5} = \left(\frac{3}{5}\right)^7\)
Combine both parts:
\(= \left(\frac{1}{4}\right)^7 \times \left(\frac{3}{5}\right)^7 = \left(\frac{1}{4} \times \frac{3}{5}\right)^7 = \left(\frac{3}{20}\right)^7\)
In simple words: Combine terms with identical bases first, then use the power distribution law when bases have identical exponents.

Exam Tip: When two different bases share the same exponent, you can multiply the bases together and keep the common exponent: \(a^n \times b^n = (ab)^n\).

Question 7. Find \(x\), if \(\left(\frac{2}{3}\right)^{-5} \times \left(\frac{2}{3}\right)^{12} = \left(\frac{2}{3}\right)^{3x-2}\)
Answer: Combine the exponents on the left-hand side using the product rule:
\(\left(\frac{2}{3}\right)^{-5 + 12} = \left(\frac{2}{3}\right)^{3x-2}\)
\(\left(\frac{2}{3}\right)^7 = \left(\frac{2}{3}\right)^{3x-2}\)
Since the bases are equal, equate the exponents:
\(7 = 3x - 2\)
\(3x = 7 + 2\)
\(3x = 9\)
\(x = 3\)
In simple words: Simplify the left side by adding the exponents, then set the exponents equal to each other because the bases are identical.

Exam Tip: Whenever \(a^m = a^n\) with \(a \neq 0, 1, -1\), you can directly set \(m = n\) to solve for unknown variables.

Question 8. Simplify: \(\left(\frac{9}{b}\right)^4 \times \left(\frac{4ab}{3a}\right)^2 \times \left(\frac{b}{2a}\right)^3\)
Answer: First, simplify the terms inside the brackets:
\(\left(\frac{4ab}{3a}\right)^2 = \left(\frac{4b}{3}\right)^2 = \frac{16b^2}{9}\)
Now rewrite the entire expression:
\(= \left(\frac{9^4}{b^4}\right) \times \left(\frac{16b^2}{9}\right) \times \left(\frac{b^3}{8a^3}\right)\)
Combine numerical and variable coefficients:
\(= \frac{9^4 \times 16 \times b^2 \times b^3}{9 \times 8 \times a^3 \times b^4}\)
Cancel common terms:
\(= \frac{9^3 \times 2 \times b^5}{a^3 \times b^4}\)
\(= \frac{729 \times 2 \times b}{a^3} = \frac{1458b}{a^3}\)
In simple words: Simplify inner fractions first, expand the powers, and then cancel out common factors in the numerator and denominator.

Exam Tip: Cancel common terms step-by-step to avoid calculation errors with large numbers and exponents.

Question 9. Evaluate:
(a) \(\frac{4^{-1/2} \times 2^{1/2}}{8 \times 8^{-1/2}}\)
(b) \(\frac{(48)^{-2} \times (64)^{7/2}}{(24)^{-1}}\)
Answer: (a) Rewrite base numbers using prime factors (\(4 = 2^2\) and \(8 = 2^3\)):
\(4^{-1/2} = (2^2)^{-1/2} = 2^{-1}\)
Denominator: \(8 \times 8^{-1/2} = 8^{1 - 1/2} = 8^{1/2} = (2^3)^{1/2} = 2^{3/2}\)
Numerator: \(2^{-1} \times 2^{1/2} = 2^{-1 + 1/2} = 2^{-1/2}\)
Expression = \(\frac{2^{-1/2}}{2^{3/2}} = 2^{-1/2 - 3/2} = 2^{-4/2} = 2^{-2} = \frac{1}{4}\)

(b) Prime factorize 48, 64, and 24:
\(48 = 16 \times 3 = 2^4 \times 3^1\)
\(64 = 2^6\)
\(24 = 8 \times 3 = 2^3 \times 3^1\)
Substitute into expression:
Numerator: \((2^4 \times 3)^{-2} \times (2^6)^{7/2} = 2^{-8} \times 3^{-2} \times 2^{21} = 2^{13} \times 3^{-2}\)
Denominator: \((2^3 \times 3)^{-1} = 2^{-3} \times 3^{-1}\)
Divide numerator by denominator:
\(= \frac{2^{13} \times 3^{-2}}{2^{-3} \times 3^{-1}} = 2^{13 - (-3)} \times 3^{-2 - (-1)} = 2^{16} \times 3^{-1} = \frac{2^{16}}{3} = \frac{65536}{3}\)
In simple words: Convert all numbers into prime bases with powers, then apply exponent laws for multiplication and division.

Exam Tip: Converting composite numbers to their prime factor form is the most reliable method for solving complex fractional exponent problems.

Question 10. Write the following in standard form:
(a) \(0.0000389\)
(b) \(19280000\)
(c) \(\frac{0.000462}{10^7}\)
Answer: (a) Move the decimal point 5 places to the right: \(3.89 \times 10^{-5}\)
(b) Move the decimal point 7 places to the left: \(1.928 \times 10^7\)
(c) Convert numerator to standard form: \(0.000462 = 4.62 \times 10^{-4}\)
Substitute back: \(\frac{4.62 \times 10^{-4}}{10^7} = 4.62 \times 10^{-4 - 7} = 4.62 \times 10^{-11}\)
In simple words: Standard form is expressed as \(k \times 10^n\), where \(1 \leq k < 10\) and \(n\) is an integer.

Exam Tip: If you move the decimal to the right, the exponent is negative. If you move it to the left, the exponent is positive.

Question 11. Express the following numbers in usual form:
(a) \(2.08 \times 10^{-5}\)
(b) \(381624 \times 10^6\)
(c) \(9 \times 10^{-7}\)
Answer: (a) Move the decimal point 5 places to the left: \(0.0000208\)
(b) Move the decimal point 6 places to the right: \(381624000000\)
(c) Move the decimal point 7 places to the left: \(0.0000009\)
In simple words: Usual form means writing the number out fully without using powers of 10.

Exam Tip: Count zeros carefully when shifting the decimal point for negative and positive powers of 10.

Question 12. Simplify: \(\left(\frac{2}{3}\right)^3 \times \left(\frac{2}{3}\right)^{-2} \div \left[\left(\frac{1}{2}\right)^2\right]^{-2} \times \frac{1}{24}\)
Answer: Simplify the first part with base \(\frac{2}{3}\):
\(\left(\frac{2}{3}\right)^{3 + (-2)} = \left(\frac{2}{3}\right)^1 = \frac{2}{3}\)
Simplify the bracket term: \(\left[\left(\frac{1}{2}\right)^2\right]^{-2} = \left(\frac{1}{2}\right)^{-4} = 2^4 = 16\)
Now substitute all simplified parts into the expression:
\(= \frac{2}{3} \div 16 \times \frac{1}{24}\)
\(= \frac{2}{3} \times \frac{1}{16} \times \frac{1}{24}\)
Cancel common factors:
\(= \frac{2}{3 \times 16 \times 24} = \frac{1}{3 \times 8 \times 24} = \frac{1}{576}\)
In simple words: Simplify each section using exponent laws first, then convert division into multiplication by taking the reciprocal.

Exam Tip: Always perform operations strictly according to the BODMAS rule, dealing with exponents and brackets before multiplication and division.