Exponents and Powers – Worksheet featuring ten multiple-choice questions on mathematical operations involving exponents and powers.

Let's solve each problem step by step:

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Problem 1:

The value of \(3^5 \div 3^{-6}\) is:
(a) \(3^5\)
(b) \(3^{-6}\)
(c) \(3^{11}\)
(d) \(3^{-11}\)

Solution:
Using the property of exponents \(a^m \div a^n = a^{m-n}\):
\[
3^5 \div 3^{-6} = 3^{5 - (-6)} = 3^{5 + 6} = 3^{11}
\]

Answer: (c) \(3^{11}\)

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Problem 2:

The multiplicative inverse of \(10^{-100}\) is:
(a) 10
(b) 100
(c) \(10^{100}\)
(d) \(10^{-100}\)

Solution:
The multiplicative inverse of a number \(a\) is \(1/a\). For \(10^{-100}\):
\[
\text{Multiplicative inverse} = \frac{1}{10^{-100}} = 10^{100}
\]

Answer: (c) \(10^{100}\)

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Problem 3:

If \(x\) be any non-zero integer and \(m, n\) be negative integers, then \(x^m \times x^n\) is equal to:
(a) \(x^m\)
(b) \(x^{m+n}\)
(c) \(x^n\)
(d) \(x^{m-n}\)

Solution:
Using the property of exponents \(a^m \times a^n = a^{m+n}\):
\[
x^m \times x^n = x^{m+n}
\]

Answer: (b) \(x^{m+n}\)

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Problem 4:

If \(x\) be any integer different from zero and \(m\) be any positive integer, then \(x^{-m}\) is equal to:
(a) \(x^m\)
(b) \(-x^m\)
(c) \(\frac{1}{x^m}\)
(d) \(\frac{-1}{x^m}\)

Solution:
Using the property of exponents \(a^{-n} = \frac{1}{a^n}\):
\[
x^{-m} = \frac{1}{x^m}
\]

Answer: (c) \(\frac{1}{x^m}\)

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Problem 5:

The value of \((7^{-1} - 8^{-1})^{-1} - (3^{-1} - 4^{-1})^{-1}\) is:
(a) 44
(b) 56
(c) 68
(d) 12

Solution:
First, simplify \(7^{-1}\) and \(8^{-1}\):
\[
7^{-1} = \frac{1}{7}, \quad 8^{-1} = \frac{1}{8}
\]
So,
\[
7^{-1} - 8^{-1} = \frac{1}{7} - \frac{1}{8} = \frac{8 - 7}{56} = \frac{1}{56}
\]
Thus,
\[
(7^{-1} - 8^{-1})^{-1} = \left(\frac{1}{56}\right)^{-1} = 56
\]

Next, simplify \(3^{-1}\) and \(4^{-1}\):
\[
3^{-1} = \frac{1}{3}, \quad 4^{-1} = \frac{1}{4}
\]
So,
\[
3^{-1} - 4^{-1} = \frac{1}{3} - \frac{1}{4} = \frac{4 - 3}{12} = \frac{1}{12}
\]
Thus,
\[
(3^{-1} - 4^{-1})^{-1} = \left(\frac{1}{12}\right)^{-1} = 12
\]

Now, compute the expression:
\[
(7^{-1} - 8^{-1})^{-1} - (3^{-1} - 4^{-1})^{-1} = 56 - 12 = 44
\]

Answer: (a) 44

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Problem 6:

The standard form for \(0.000064\) is:
(a) \(64 \times 10^4\)
(b) \(64 \times 10^{-4}\)
(c) \(6.4 \times 10^5\)
(d) \(6.4 \times 10^{-5}\)

Solution:
To convert \(0.000064\) to standard form:
\[
0.000064 = 6.4 \times 10^{-5}
\]

Answer: (d) \(6.4 \times 10^{-5}\)

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Problem 7:

The usual form for \(2.03 \times 10^{-5}\) is:
(a) 0.203
(b) 0.000203
(c) 203000
(d) 0.0000203

Solution:
To convert \(2.03 \times 10^{-5}\) to usual form:
\[
2.03 \times 10^{-5} = 0.0000203
\]

Answer: (d) 0.0000203

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Problem 8:

\((1/10)^0\) is equal to:
(a) 0
(b) \(1/10\)
(c) 1
(d) 10

Solution:
Any non-zero number raised to the power of 0 is 1:
\[
(1/10)^0 = 1
\]

Answer: (c) 1

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Problem 9:

\((-9)^3 \div (-9)^8\) is equal to:
(a) \(9^5\)
(b) \(9^{-5}\)
(c) \((-9)^5\)
(d) \((-9)^{-5}\)

Solution:
Using the property of exponents \(a^m \div a^n = a^{m-n}\):
\[
(-9)^3 \div (-9)^8 = (-9)^{3-8} = (-9)^{-5}
\]

Answer: (d) \((-9)^{-5}\)

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Problem 10:

If \(x\) be any integer different from zero and \(m, n\) be any integers, then \((x^m)^n\) is equal to:
(a) \(x^{m+n}\)
(b) \(x^{mn}\)
(c) \(x^{m/n}\)
(d) \(x^{m-n}\)

Solution:
Using the property of exponents \((a^m)^n = a^{mn}\):
\[
(x^m)^n = x^{mn}
\]

Answer: (b) \(x^{mn}\)

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Final Answers:

1. (c) \(3^{11}\)
2. (c) \(10^{100}\)
3. (b) \(x^{m+n}\)
4. (c) \(\frac{1}{x^m}\)
5. (a) 44
6. (d) \(6.4 \times 10^{-5}\)
7. (d) 0.0000203
8. (c) 1
9. (d) \((-9)^{-5}\)
10. (b) \(x^{mn}\)

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