CBSE Class 7 Exponents and Powers Worksheets - Free Printable

Problem Set: Exponents (Powers)

Below, I will solve each problem step by step, explaining the reasoning and applying the rules of exponents.

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#### 1. Simplify and write in exponential form:

a. \( (-5)^3 \times (-5)^4 \)
Using the rule \( a^m \times a^n = a^{m+n} \):
\[
(-5)^3 \times (-5)^4 = (-5)^{3+4} = (-5)^7
\]

b. \( P^3 \times P^{-10} \)
Using the same rule:
\[
P^3 \times P^{-10} = P^{3 + (-10)} = P^{-7}
\]

c. \( 3^5 \times 3^{-10} \times 3^6 \)
Combine the exponents:
\[
3^5 \times 3^{-10} \times 3^6 = 3^{5 + (-10) + 6} = 3^{1}
\]

d. \( (2^5 \div 2^8)^5 \times 2^{-5} \)
First, simplify inside the parentheses using \( a^m \div a^n = a^{m-n} \):
\[
2^5 \div 2^8 = 2^{5-8} = 2^{-3}
\]
Then raise to the power of 5:
\[
(2^{-3})^5 = 2^{-3 \times 5} = 2^{-15}
\]
Finally, multiply by \( 2^{-5} \):
\[
2^{-15} \times 2^{-5} = 2^{-15 + (-5)} = 2^{-20}
\]

e. \( (-4)^{-3} \times (5)^3 \times (-5)^{-3} \)
Simplify each term:
\[
(-4)^{-3} = \frac{1}{(-4)^3}, \quad (5)^3 = 5^3, \quad (-5)^{-3} = \frac{1}{(-5)^3}
\]
Combine them:
\[
(-4)^{-3} \times (5)^3 \times (-5)^{-3} = \frac{1}{(-4)^3} \times 5^3 \times \frac{1}{(-5)^3}
\]
Group the bases:
\[
= \frac{5^3}{(-4)^3 \times (-5)^3} = \frac{5^3}{(-1)^3 \times 4^3 \times 5^3} = \frac{5^3}{-1 \times 4^3 \times 5^3} = \frac{1}{-4^3} = -\frac{1}{4^3}
\]

f. \( (-3)^4 \times \left(\frac{2}{3}\right)^4 \)
Use the property \( (ab)^n = a^n \times b^n \):
\[
(-3)^4 \times \left(\frac{2}{3}\right)^4 = \left((-3) \times \frac{2}{3}\right)^4 = \left(-2\right)^4 = 2^4
\]

g. \( \frac{1}{8} \times 3^{-5} \)
Rewrite \( \frac{1}{8} \) as \( 2^{-3} \):
\[
\frac{1}{8} \times 3^{-5} = 2^{-3} \times 3^{-5}
\]
This is already in exponential form.

h. \( (-4)^5 \div (4)^8 \)
Simplify the division:
\[
(-4)^5 \div (4)^8 = \frac{(-4)^5}{4^8}
\]
Since \( (-4)^5 = -4^5 \):
\[
= \frac{-4^5}{4^8} = -4^{5-8} = -4^{-3}
\]

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#### 2. Find the value of:

a. \( \left(\frac{2}{3}\right)^{-2} \)
Using \( \left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n \):
\[
\left(\frac{2}{3}\right)^{-2} = \left(\frac{3}{2}\right)^2 = \frac{3^2}{2^2} = \frac{9}{4}
\]

b. \( \left[\left(\frac{1}{3}\right)^{-2} - \left(\frac{1}{2}\right)^{-3}\right] \div \left(\frac{1}{4}\right)^{-2} \)
Simplify each term:
\[
\left(\frac{1}{3}\right)^{-2} = 3^2 = 9, \quad \left(\frac{1}{2}\right)^{-3} = 2^3 = 8, \quad \left(\frac{1}{4}\right)^{-2} = 4^2 = 16
\]
Substitute back:
\[
\left[9 - 8\right] \div 16 = 1 \div 16 = \frac{1}{16}
\]

c. \( \left(\frac{5}{8}\right)^{-7} \times \left(\frac{8}{5}\right)^{-5} \)
Using \( \left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n \):
\[
\left(\frac{5}{8}\right)^{-7} = \left(\frac{8}{5}\right)^7, \quad \left(\frac{8}{5}\right)^{-5} = \left(\frac{5}{8}\right)^5
\]
Combine:
\[
\left(\frac{8}{5}\right)^7 \times \left(\frac{5}{8}\right)^5 = \left(\frac{8}{5} \times \frac{5}{8}\right)^5 \times \left(\frac{8}{5}\right)^2 = 1^5 \times \left(\frac{8}{5}\right)^2 = \left(\frac{8}{5}\right)^2 = \frac{64}{25}
\]

d. \( (3^0 + 4^{-1}) \times 2^2 \)
Simplify each term:
\[
3^0 = 1, \quad 4^{-1} = \frac{1}{4}, \quad 2^2 = 4
\]
Combine:
\[
(1 + \frac{1}{4}) \times 4 = \left(\frac{4}{4} + \frac{1}{4}\right) \times 4 = \frac{5}{4} \times 4 = 5
\]

e. \( (2^{-1} \times 4^{-1}) \div 2^{-2} \)
Simplify each term:
\[
2^{-1} = \frac{1}{2}, \quad 4^{-1} = \frac{1}{4}, \quad 2^{-2} = \frac{1}{4}
\]
Combine:
\[
\left(\frac{1}{2} \times \frac{1}{4}\right) \div \frac{1}{4} = \frac{1}{8} \div \frac{1}{4} = \frac{1}{8} \times 4 = \frac{4}{8} = \frac{1}{2}
\]

f. \( (3^{-1} + 4^{-1} + 5^{-1})^0 \)
Any non-zero number raised to the power of 0 is 1:
\[
(3^{-1} + 4^{-1} + 5^{-1})^0 = 1
\]

g. \( \frac{8^{-1} \times 5^3}{2^{-4}} \)
Simplify each term:
\[
8^{-1} = \frac{1}{8}, \quad 5^3 = 125, \quad 2^{-4} = \frac{1}{16}
\]
Combine:
\[
\frac{\frac{1}{8} \times 125}{\frac{1}{16}} = \frac{125}{8} \times 16 = \frac{125 \times 16}{8} = \frac{2000}{8} = 250
\]

h. \( \left\{ \left(\frac{-2}{3}\right)^{-2} \right\}^2 \)
Simplify inside the parentheses:
\[
\left(\frac{-2}{3}\right)^{-2} = \left(\frac{3}{-2}\right)^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4}
\]
Raise to the power of 2:
\[
\left(\frac{9}{4}\right)^2 = \frac{81}{16}
\]

i. \( \left(\frac{1}{3}\right)^{-1} - \left(\frac{1}{4}\right)^{-1} \)
Simplify each term:
\[
\left(\frac{1}{3}\right)^{-1} = 3, \quad \left(\frac{1}{4}\right)^{-1} = 4
\]
Combine:
\[
3 - 4 = -1
\]

j. \( (5^{-1} \times 2^{-1}) \times 6^{-1} \)
Simplify each term:
\[
5^{-1} = \frac{1}{5}, \quad 2^{-1} = \frac{1}{2}, \quad 6^{-1} = \frac{1}{6}
\]
Combine:
\[
\left(\frac{1}{5} \times \frac{1}{2}\right) \times \frac{1}{6} = \frac{1}{10} \times \frac{1}{6} = \frac{1}{60}
\]

k. \( \frac{25 \times t^{-4}}{5^{-3} \times 10 \times t^{-8}} \)
Simplify each term:
\[
25 = 5^2, \quad 10 = 2 \times 5
\]
Combine:
\[
\frac{5^2 \times t^{-4}}{5^{-3} \times 2 \times 5 \times t^{-8}} = \frac{5^2 \times t^{-4}}{5^{-2} \times 2 \times t^{-8}} = \frac{5^{2 - (-2)}}{2} \times \frac{t^{-4 - (-8)}}{1} = \frac{5^4}{2} \times t^4 = \frac{625}{2} \times t^4
\]

l. \( \frac{3^{-5} \times 10^{-5} \times 125}{6^{-5} \times 6^{-5}} \)
Simplify each term:
\[
10^{-5} = (2 \times 5)^{-5} = 2^{-5} \times 5^{-5}, \quad 125 = 5^3, \quad 6^{-5} = (2 \times 3)^{-5} = 2^{-5} \times 3^{-5}
\]
Combine:
\[
\frac{3^{-5} \times 2^{-5} \times 5^{-5} \times 5^3}{(2^{-5} \times 3^{-5}) \times (2^{-5} \times 3^{-5})} = \frac{3^{-5} \times 2^{-5} \times 5^{-2}}{2^{-10} \times 3^{-10}} = 3^{-5 - (-10)} \times 2^{-5 - (-10)} \times 5^{-2} = 3^5 \times 2^5 \times 5^{-2} = \frac{3^5 \times 2^5}{5^2} = \frac{243 \times 32}{25} = \frac{7776}{25}
\]

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#### 3. Evaluate:

a. \( 3^{-2} \)
\[
3^{-2} = \frac{1}{3^2} = \frac{1}{9}
\]

b. \( (-4)^{-2} \)
\[
(-4)^{-2} = \frac{1}{(-4)^2} = \frac{1}{16}
\]

c. \( \left(\frac{1}{2}\right)^{-5} \)
\[
\left(\frac{1}{2}\right)^{-5} = 2^5 = 32
\]

d. \( 1/3^{-2} \)
\[
\frac{1}{3^{-2}} = 3^2 = 9
\]

e. \( 2^{-3} \)
\[
2^{-3} = \frac{1}{2^3} = \frac{1}{8}
\]

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#### 4. Find the multiplicative inverse of the following:

The multiplicative inverse of \( a^n \) is \( a^{-n} \).

a. \( 2^{-4} \)
\[
\text{Inverse of } 2^{-4} \text{ is } 2^4 = 16
\]

b. \( 10^{-5} \)
\[
\text{Inverse of } 10^{-5} \text{ is } 10^5 = 100000
\]

c. \( 7^{-2} \)
\[
\text{Inverse of } 7^{-2} \text{ is } 7^2 = 49
\]

d. \( 5^{-3} \)
\[
\text{Inverse of } 5^{-3} \text{ is } 5^3 = 125
\]

e. \( 10^{-100} \)
\[
\text{Inverse of } 10^{-100} \text{ is } 10^{100}
\]

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#### 5. If \( \frac{m}{n} = \left(\frac{5}{7}\right)^4 \div \left(\frac{5}{7}\right)^0 \), find the value of \( \left(\frac{m}{n}\right)^2 \) if \( m = -3 \) and \( n = 4 \).

Simplify the given expression:
\[
\left(\frac{5}{7}\right)^4 \div \left(\frac{5}{7}\right)^0 = \left(\frac{5}{7}\right)^{4-0} = \left(\frac{5}{7}\right)^4
\]
So, \( \frac{m}{n} = \left(\frac{5}{7}\right)^4 \).
Given \( m = -3 \) and \( n = 4 \):
\[
\frac{m}{n} = \frac{-3}{4}
\]
Thus, \( \left(\frac{5}{7}\right)^4 = \frac{-3}{4} \).
Now, find \( \left(\frac{m}{n}\right)^2 \):
\[
\left(\frac{m}{n}\right)^2 = \left(\frac{-3}{4}\right)^2 = \frac{9}{16}
\]

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#### 6. Find the value of \( x^{-2} \) if \( x = \left(\frac{-2}{5}\right)^{-3} \div \left(\frac{5}{6}\right)^0 \).

Simplify \( x \):
\[
\left(\frac{-2}{5}\right)^{-3} = \left(\frac{5}{-2}\right)^3 = \frac{5^3}{(-2)^3} = \frac{125}{-8} = -\frac{125}{8}
\]
\[
\left(\frac{5}{6}\right)^0 = 1
\]
So,
\[
x = -\frac{125}{8} \div 1 = -\frac{125}{8}
\]
Now, find \( x^{-2} \):
\[
x^{-2} = \left(-\frac{125}{8}\right)^{-2} = \left(\frac{8}{-125}\right)^2 = \frac{8^2}{(-125)^2} = \frac{64}{15625}
\]

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#### 7. What should \( \left(\frac{7}{9}\right)^{-3} \) be divided so that the quotient becomes 9?

Let the divisor be \( y \). Then:
\[
\frac{\left(\frac{7}{9}\right)^{-3}}{y} = 9
\]
Simplify \( \left(\frac{7}{9}\right)^{-3} \):
\[
\left(\frac{7}{9}\right)^{-3} = \left(\frac{9}{7}\right)^3 = \frac{9^3}{7^3} = \frac{729}{343}
\]
So,
\[
\frac{\frac{729}{343}}{y} = 9 \implies \frac{729}{343y} = 9 \implies 729 = 9 \times 343y \implies y = \frac{729}{9 \times 343} = \frac{81}{343}
\]

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#### 8. What should \( \left(\frac{2}{5}\right)^4 \) be multiplied so that the product becomes 25?

Let the multiplier be \( z \). Then:
\[
\left(\frac{2}{5}\right)^4 \times z = 25
\]
Simplify \( \left(\frac{2}{5}\right)^4 \):
\[
\left(\frac{2}{5}\right)^4 = \frac{2^4}{5^4} = \frac{16}{625}
\]
So,
\[
\frac{16}{625} \times z = 25 \implies z = 25 \times \frac{625}{16} = \frac{15625}{16}
\]

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#### 9. Express each of the following rational numbers in exponential form:

a. \( \frac{1}{343} \)
\[
343 = 7^3 \implies \frac{1}{343} = 7^{-3}
\]

b. \( \frac{-25}{216} \)
\[
25 = 5^2, \quad 216 = 6^3 = (2 \times 3)^3 = 2^3 \times 3^3 \implies \frac{-25}{216} = \frac{-5^2}{2^3 \times 3^3}
\]

c. \( \frac{64}{27} \)
\[
64 = 4^3 = (2^2)^3 = 2^6, \quad 27 = 3^3 \implies \frac{64}{27} = \frac{2^6}{3^3}
\]

d. \( \frac{243000}{729000} \)
Simplify the fraction:
\[
\frac{243000}{729000} = \frac{243}{729} = \frac{3^5}{3^6} = 3^{-1}
\]

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#### 10. Simplify and express the result as powers of 2:

a. \( \left\{ \left(\frac{1}{2}\right)^3 \right\}^4 \div \left\{ \left(\frac{1}{2}\right)^4 \right\}^3 \)
Simplify each term:
\[
\left\{ \left(\frac{1}{2}\right)^3 \right\}^4 = \left(\frac{1}{2}\right)^{3 \times 4} = \left(\frac{1}{2}\right)^{12} = 2^{-12}
\]
\[
\left\{ \left(\frac{1}{2}\right)^4 \right\}^3 = \left(\frac{1}{2}\right)^{4 \times 3} = \left(\frac{1}{2}\right)^{12} = 2^{-12}
\]
Divide:
\[
\frac{2^{-12}}{2^{-12}} = 2^{-12 - (-12)} = 2^0 = 1
\]

b. \( (3^0 + 5^0) \div (4^0 + 2^0) \)
Simplify each term:
\[
3^0 = 1, \quad 5^0 = 1, \quad 4^0 = 1, \quad 2^0 = 1
\]
Combine:
\[
(1 + 1) \div (1 + 1) = 2 \div 2 = 1 = 2^0
\]

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Final Answers:

\[
\boxed{
\begin{array}{ll}
1. & \text{(a) } (-5)^7, \text{(b) } P^{-7}, \text{(c) } 3^1, \text{(d) } 2^{-20}, \text{(e) } -\frac{1}{4^3}, \text{(f) } 2^4, \text{(g) } 2^{-3} \times 3^{-5}, \text{(h) } -4^{-3} \\
2. & \text{(a) } \frac{9}{4}, \text{(b) } \frac{1}{16}, \text{(c) } \frac{64}{25}, \text{(d) } 5, \text{(e) } \frac{1}{2}, \text{(f) } 1, \text{(g) } 250, \text{(h) } \frac{81}{16}, \text{(i) } -1, \text{(j) } \frac{1}{60}, \text{(k) } \frac{625}{2} t^4, \text{(l) } \frac{7776}{25} \\
3. & \text{(a) } \frac{1}{9}, \text{(b) } \frac{1}{16}, \text{(c) } 32, \text{(d) } 9, \text{(e) } \frac{1}{8} \\
4. & \text{(a) } 16, \text{(b) } 100000, \text{(c) } 49, \text{(d) } 125, \text{(e) } 10^{100} \\
5. & \frac{9}{16} \\
6. & \frac{64}{15625} \\
7. & \frac{81}{343} \\
8. & \frac{15625}{16} \\
9. & \text{(a) } 7^{-3}, \text{(b) } \frac{-5^2}{2^3 \times 3^3}, \text{(c) } \frac{2^6}{3^3}, \text{(d) } 3^{-1} \\
10. & \text{(a) } 1, \text{(b) } 2^0 \\
\end{array}
}
\]