CBSE Class 8 Mathematics Exponents And Powers Worksheet Set 02

Mathematics - Exponents and Powers

Question 1. Find the multiplicative inverse of the following: (a) \(3^{-4}\) (b) \(7^{-2}\) (c) \(a^{-9}\) (d) \(10^{-80}\)
Answer: The multiplicative inverse of a number \(a\) is \(a^{-1}\) or \(\frac{1}{a}\) such that their product is 1. Using the law of negative exponents \(a^{-n} = \frac{1}{a^n}\), the multiplicative inverse of \(a^{-n}\) is \(a^n\).
(a) Multiplicative inverse of \(3^{-4}\) is \(3^4\) (or 81).
(b) Multiplicative inverse of \(7^{-2}\) is \(7^2\) (or 49).
(c) Multiplicative inverse of \(a^{-9}\) is \(a^9\).
(d) Multiplicative inverse of \(10^{-80}\) is \(10^{80}\).
In simple words: The multiplicative inverse of a number with a negative exponent is the same number with a positive exponent.

Exam Tip: Remember that the reciprocal of \(a^{-n}\) is \(a^n\). Do not change the base or the sign of the base.

Question 2. Expand the following numbers using exponents: (a) 18964.63 (b) 7064.373
Answer: To expand a number using exponents, multiply each digit by the corresponding power of 10 based on its place value.
(a) \(18964.63\)
\(= (1 \times 10^4) + (8 \times 10^3) + (9 \times 10^2) + (6 \times 10^1) + (4 \times 10^0) + (6 \times 10^{-1}) + (3 \times 10^{-2})\)
(b) \(7064.373\)
\(= (7 \times 10^3) + (0 \times 10^2) + (6 \times 10^1) + (4 \times 10^0) + (3 \times 10^{-1}) + (7 \times 10^{-2}) + (3 \times 10^{-3})\)
In simple words: Break down each digit according to its place value and write it as a product of the digit and a power of 10.

Exam Tip: For digits after the decimal point, use negative powers of 10 starting from \(10^{-1}\) moving right.

Question 3. Simplify and write in exponential form: (a) \((-3)^{-3} \times (-3)^{-2} \times (-3)^5\) (b) \(a^{-8} \times a^{10} \times a \times a^{-2}\) (c) \((2^3 \times 2^6 \times 2^2) \div 2^{-6}\)
Answer: Use the laws of exponents: \(a^m \times a^n = a^{m+n}\) and \(a^m \div a^n = a^{m-n}\).
(a) \((-3)^{-3} \times (-3)^{-2} \times (-3)^5\)
\(= (-3)^{-3 + (-2) + 5}\)
\(= (-3)^{-5 + 5}\)
\(= (-3)^0 = 1\)
(b) \(a^{-8} \times a^{10} \times a \times a^{-2}\)
\(= a^{-8 + 10 + 1 + (-2)}\)
\(= a^{3 - 2} = a^1 = a\)
(c) \((2^3 \times 2^6 \times 2^2) \div 2^{-6}\)
\(= (2^{3 + 6 + 2}) \div 2^{-6}\)
\(= 2^{11} \div 2^{-6}\)
\(= 2^{11 - (-6)}\)
\(= 2^{11 + 6} = 2^{17}\)
In simple words: When multiplying terms with the same base, add their powers. When dividing, subtract the power of the divisor.

Exam Tip: Note that a standalone variable \(a\) has an exponent of 1. Be careful with subtraction of negative exponents like \(11 - (-6) = 11 + 6\).

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

Exam Tip: Always prime factorize the base first when asked to change the base of an exponential expression.

Question 5. Find the value of:
(a) \(\left[\left(\frac{1}{2}\right)^0 + \left(\frac{1}{5}\right)^3 + \left(\frac{2}{3}\right)^2\right]\)
Answer: Evaluate each term separately: any non-zero number raised to 0 is 1.
\(\left(\frac{1}{2}\right)^0 = 1\)
\(\left(\frac{1}{5}\right)^3 = \frac{1}{125}\)
\(\left(\frac{2}{3}\right)^2 = \frac{4}{9}\)
Summing the terms:
\(= 1 + \frac{1}{125} + \frac{4}{9}\)
Find the common denominator for 125 and 9, which is \(1125\):
\(= \frac{1125 + 9 + 500}{1125}\)
\(= \frac{1634}{1125}\)
In simple words: Calculate each bracket value individually and add them using a common denominator.

Exam Tip: Remember that \(a^0 = 1\) for any \(a \neq 0\).

(b) \(\left[\left(\frac{1}{3}\right)^{-2} - \left(\frac{1}{2}\right)^{-3}\right] \div \left(\frac{1}{4}\right)^{-2}\)
Answer: Apply negative exponent rule \(\left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n\):
\(\left(\frac{1}{3}\right)^{-2} = 3^2 = 9\)
\(\left(\frac{1}{2}\right)^{-3} = 2^3 = 8\)
\(\left(\frac{1}{4}\right)^{-2} = 4^2 = 16\)
Substitute these values back into the expression:
\(= (9 - 8) \div 16\)
\(= 1 \div 16 = \frac{1}{16}\)
In simple words: Flip the fractions to make the negative exponents positive, simplify, and then divide.

Exam Tip: Reciprocal of a fraction raised to a negative power simply inverts the fraction and changes the exponent to positive.

(c) \(\left(\frac{9}{5}\right)^{-8} \times \left(\frac{5}{9}\right)^{-5}\)
Answer: Rewrite the second term by making the base identical to the first term using \(\left(\frac{b}{a}\right)^{-n} = \left(\frac{a}{b}\right)^n\):
\(\left(\frac{5}{9}\right)^{-5} = \left(\frac{9}{5}\right)^5\)
Now substitute back into the expression:
\(= \left(\frac{9}{5}\right)^{-8} \times \left(\frac{9}{5}\right)^5\)
Using \(a^m \times a^n = a^{m+n}\):
\(= \left(\frac{9}{5}\right)^{-8 + 5}\)
\(= \left(\frac{9}{5}\right)^{-3}\)
\(= \left(\frac{5}{9}\right)^3 = \frac{125}{729}\)
In simple words: Make the bases the same by reciprocating one fraction, then add the exponents.

Exam Tip: Aligning bases is the key to solving multiplicative exponent problems.

(d) \((9^2 - 4^3) \times \left(\frac{-3}{17}\right) \times \frac{3^4}{9}\)
Answer: Evaluate each part:
\(9^2 = 81\)
\(4^3 = 64\)
\(9^2 - 4^3 = 81 - 64 = 17\)
Substitute into the expression:
\(= 17 \times \left(\frac{-3}{17}\right) \times \frac{81}{9}\)
Simplify the terms:
\(= 17 \times \left(\frac{-3}{17}\right) \times 9\)
\(= -3 \times 9 = -27\)
In simple words: Calculate values inside brackets, cancel common terms in numerators and denominators, and multiply.

Exam Tip: Cancel 17 in the numerator and denominator carefully to avoid calculation errors.

Question 6. Simplify: \(\left[\left(\frac{1}{4}\right)^4 \times \left(\frac{1}{4}\right)^3\right]^3 \times \left[\left(\frac{3}{5}\right)^{12} \div \left(\frac{3}{5}\right)^5\right]\)
Answer: Simplify the first bracket using product rule \(a^m \times a^n = a^{m+n}\):
\(\left(\frac{1}{4}\right)^4 \times \left(\frac{1}{4}\right)^3 = \left(\frac{1}{4}\right)^{4+3} = \left(\frac{1}{4}\right)^7\)
Apply power of a power rule \((a^m)^n = a^{m \times n}\):
\(\left[\left(\frac{1}{4}\right)^7\right]^3 = \left(\frac{1}{4}\right)^{21}\)
Simplify the second bracket using quotient rule \(a^m \div a^n = a^{m-n}\):
\(\left(\frac{3}{5}\right)^{12} \div \left(\frac{3}{5}\right)^5 = \left(\frac{3}{5}\right)^{12-5} = \left(\frac{3}{5}\right)^7\)
Combine both parts:
\(= \left(\frac{1}{4}\right)^{21} \times \left(\frac{3}{5}\right)^7\)
In simple words: Simplify inside the brackets first using exponent laws, then apply powers, and multiply the final results.

Exam Tip: Always follow the order of operations (PEMDAS/BODMAS) by simplifying inner brackets first.

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 left-hand side using the product rule \(a^m \times a^n = a^{m+n}\):
\(\left(\frac{2}{3}\right)^{-5 + 12} = \left(\frac{2}{3}\right)^{3x-2}\)
\(\left(\frac{2}{3}\right)^7 = \left(\frac{2}{3}\right)^{3x-2}\)
Since the bases are equal on both sides, equate the exponents:
\(7 = 3x - 2\)
Add 2 to both sides:
\(3x = 7 + 2\)
\(3x = 9\)
Divide by 3:
\(x = \frac{9}{3} = 3\)
In simple words: Simplify the LHS into a single exponent, then set the powers equal to each other since the bases are identical.

Exam Tip: When bases are identical in an equation (\(a^m = a^n\)), you can directly set \(m = n\).

Question 8. Simplify: \(-\left(\frac{9}{b}\right)^{-4} \times \left(\frac{4ab}{3a}\right)^2 \times \left(\frac{b}{2a}\right)^3\)
Answer: Simplify each term:
\(-\left(\frac{9}{b}\right)^{-4} = -\left(\frac{b}{9}\right)^4 = -\frac{b^4}{6561}\)
\(\left(\frac{4ab}{3a}\right)^2 = \left(\frac{4b}{3}\right)^2 = \frac{16b^2}{9}\)
\(\left(\frac{b}{2a}\right)^3 = \frac{b^3}{8a^3}\)
Multiply all terms together:
\(= -\frac{b^4}{6561} \times \frac{16b^2}{9} \times \frac{b^3}{8a^3}\)
Cancel 8 and 16 (leaving 2 in the numerator):
\(= -\frac{2 \cdot b^{4+2+3}}{6561 \times 9 \times a^3}\)
\(= -\frac{2b^9}{59049a^3}\)
In simple words: Expand each bracket term by applying the exponents to numerators and denominators, then multiply and combine like variables.

Exam Tip: Cancel out common variables and numerical factors carefully before final multiplication.

Question 9. Evaluate: (a) \(\frac{4^{-1/2} \times 2^{1/2} \times 2}{8 \times 8^{-1/2}}\) (b) \(\frac{(48)^{-2} \times (64)^{1/2}}{(24)^{-1}}\)
Answer: (a) Express all numbers with base 2: \(4 = 2^2\) and \(8 = 2^3\).
\(4^{-1/2} = (2^2)^{-1/2} = 2^{-1}\)
\(8 = 2^3\)
\(8^{-1/2} = (2^3)^{-1/2} = 2^{-3/2}\)
Substitute into the expression:
\(\frac{2^{-1} \times 2^{1/2} \times 2^1}{2^3 \times 2^{-3/2}}\)
Simplify numerator and denominator using exponent laws:
Numerator: \(2^{-1 + 1/2 + 1} = 2^{1/2}\)
Denominator: \(2^{3 + (-3/2)} = 2^{3/2}\)
Divide numerator by denominator:
\(= 2^{1/2 - 3/2} = 2^{-2/2} = 2^{-1} = \frac{1}{2}\)

(b) Prime factorize 48, 64, and 24: \(48 = 16 \times 3 = 2^4 \times 3\), \(64 = 2^6\), \(24 = 8 \times 3 = 2^3 \times 3\).
\((48)^{-2} = (2^4 \times 3)^{-2} = 2^{-8} \times 3^{-2}\)
\((64)^{1/2} = (2^6)^{1/2} = 2^3\)
\((24)^{-1} = (2^3 \times 3)^{-1} = 2^{-3} \times 3^{-1}\)
Substitute into the expression:
\(\frac{2^{-8} \times 3^{-2} \times 2^3}{2^{-3} \times 3^{-1}} = \frac{2^{-8+3} \times 3^{-2}}{2^{-3} \times 3^{-1}} = \frac{2^{-5} \times 3^{-2}}{2^{-3} \times 3^{-1}}\)
\(= 2^{-5 - (-3)} \times 3^{-2 - (-1)}\)
\(= 2^{-5+3} \times 3^{-2+1} = 2^{-2} \times 3^{-1}\)
\(= \frac{1}{2^2} \times \frac{1}{3} = \frac{1}{4} \times \frac{1}{3} = \frac{1}{12}\)
In simple words: Convert all numbers to prime base factors (like 2 and 3), then apply rules of exponents to combine and simplify.

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

Question 10. Write the following in standard form: (a) 0.0000389 (b) 19280000 (c) \(\frac{0.000462}{10^7}\)
Answer: Standard form is expressed as \(k \times 10^n\) where \(1 \leq k < 10\) and \(n\) is an integer.
(a) \(0.0000389 = 3.89 \times 10^{-5}\)
(b) \(19280000 = 1.928 \times 10^7\)
(c) \(\frac{0.000462}{10^7} = \frac{4.62 \times 10^{-4}}{10^7} = 4.62 \times 10^{-4-7} = 4.62 \times 10^{-11}\)
In simple words: Move the decimal point so that there is only one non-zero digit to its left, and adjust the power of 10 accordingly.

Exam Tip: Moving the decimal to the right makes the exponent negative; moving it to the left makes the exponent 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) \(2.08 \times 10^{-5} = 0.0000208\)
(b) \(381624 \times 10^6 = 381624000000\)
(c) \(9 \times 10^{-7} = 0.0000009\)
In simple words: Shift the decimal point to the left for negative powers of 10 and to the right for positive powers of 10.

Exam Tip: Count the number of zeros carefully when expanding scientific notation into usual form.

Question 12. Simplify: \((2/3)^3 \times (2/3)^{-2} \div \left[\left(1/2\right)^2\right]^{-2} \times 1/24\)
Answer: Simplify the terms step by step:
\((2/3)^3 \times (2/3)^{-2} = (2/3)^{3+(-2)} = (2/3)^1 = \frac{2}{3}\)
\(\left[\left(\frac{1}{2}\right)^2\right]^{-2} = \left(\frac{1}{2}\right)^{-4} = 2^4 = 16\)
Substitute these into the expression:
\(= \left(\frac{2}{3} \div 16\right) \times \frac{1}{24}\)
\(= \left(\frac{2}{3} \times \frac{1}{16}\right) \times \frac{1}{24}\)
\(= \frac{2}{48} \times \frac{1}{24}\)
\(= \frac{1}{24} \times \frac{1}{24} = \frac{1}{576}\)
In simple words: Simplify the exponents first, convert division to multiplication by taking the reciprocal, and then multiply all fractions.

Exam Tip: Take care with nested brackets and negative powers when converting fractions.