CBSE Class 8 Mental Maths Exponents And Powers Worksheet - Free Printable

Let's solve each question step by step with clear explanations.

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1. 2 multiplied eleven times is written as _________.

When a number is multiplied by itself multiple times, we use exponents.

So, $ 2 \times 2 \times 2 \times \ldots $ (11 times) = $ 2^{11} $

✔ Answer: $ 2^{11} $

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2. What is the base of the exponent $ 9^6 $?

In an expression like $ a^n $, $ a $ is the base, and $ n $ is the exponent.

Here, $ 9^6 $ → base is $ 9 $

✔ Answer: 9

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3. Simplify and give the answer: $ \left(\frac{2}{5}\right)^5 \times \left(\frac{2}{5}\right)^{-3} $

Use the rule: $ a^m \times a^n = a^{m+n} $

So:
$$
\left(\frac{2}{5}\right)^{5 + (-3)} = \left(\frac{2}{5}\right)^2 = \frac{4}{25}
$$

✔ Answer: $ \frac{4}{25} $

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4. Find the missing number: $ 7^5 = \frac{1}{7^\square} $

We know that $ \frac{1}{a^n} = a^{-n} $

So:
$$
7^5 = \frac{1}{7^{-5}} \Rightarrow \text{But here it says } 7^5 = \frac{1}{7^x}
\Rightarrow 7^5 = 7^{-x} \Rightarrow 5 = -x \Rightarrow x = -5
$$

Wait — but $ \frac{1}{7^x} = 7^{-x} $, so:
$$
7^5 = 7^{-x} \Rightarrow 5 = -x \Rightarrow x = -5
$$

But the blank is in the denominator: $ \frac{1}{7^\square} $, so:

$ 7^5 = \frac{1}{7^{-5}} $ → So $ \square = -5 $

But that would mean $ \frac{1}{7^{-5}} = 7^5 $, which is correct.

✔ Answer: $ -5 $

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5. Find $ \left(\frac{-2}{5}\right)^2 \times \left(\frac{5}{2}\right)^2 $

First, square both fractions:

$$
\left(\frac{-2}{5}\right)^2 = \frac{4}{25}, \quad \left(\frac{5}{2}\right)^2 = \frac{25}{4}
$$

Now multiply:
$$
\frac{4}{25} \times \frac{25}{4} = \frac{100}{100} = 1
$$

✔ Answer: 1

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6. Find the value of $ (3^2)^2 $

First compute $ 3^2 = 9 $, then $ 9^2 = 81 $

Alternatively, use power of a power: $ (a^m)^n = a^{m \cdot n} $

So $ (3^2)^2 = 3^{2 \times 2} = 3^4 = 81 $

✔ Answer: 81

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7. Calculate $ 2^5 + 5^2 $

$ 2^5 = 32 $, $ 5^2 = 25 $, so $ 32 + 25 = 57 $

✔ Answer: 57

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8. Find the value of $ x $, when $ 2^x = 4^3 $

Note: $ 4 = 2^2 $, so $ 4^3 = (2^2)^3 = 2^{6} $

So $ 2^x = 2^6 \Rightarrow x = 6 $

✔ Answer: 6

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9. How much is $ (\sqrt{5})^5 \div (\sqrt{5})^3 $?

Use rule: $ \frac{a^m}{a^n} = a^{m-n} $

So:
$$
(\sqrt{5})^{5-3} = (\sqrt{5})^2 = 5
$$

✔ Answer: 5

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10. Find the value of $ 3^2 \times 2^3 $

$ 3^2 = 9 $, $ 2^3 = 8 $, so $ 9 \times 8 = 72 $

✔ Answer: 72

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11. If $ (x^5 \times x^{-2})^2 = 64 $, what is $ x $?

First simplify inside the parentheses:

$ x^5 \times x^{-2} = x^{5-2} = x^3 $

Then $ (x^3)^2 = x^6 $

So:
$$
x^6 = 64
$$

Now $ 64 = 2^6 $, so $ x^6 = 2^6 \Rightarrow x = 2 $ or $ x = -2 $

But since $ x^6 $ is even power, both positive and negative work.

But unless restricted, both are valid.

However, let’s check if there's a unique solution.

But $ x = 2 $: $ (2^5 \cdot 2^{-2})^2 = (2^3)^2 = 8^2 = 64 $ ✔
$ x = -2 $: $ ((-2)^5 \cdot (-2)^{-2})^2 = (-32 \cdot 1/4)^2 = (-8)^2 = 64 $ ✔

So both work.

But typically, unless specified, we may accept $ x = 2 $ as principal root.

But technically, $ x = \pm 2 $

But let's see if the problem expects one answer.

Since $ x^6 = 64 $, $ x = \pm 2 $

✔ Answer: $ x = \pm 2 $ (or just $ 2 $ if only positive considered)

But safest to say $ x = \pm 2 $

Wait — actually, $ x^6 = 64 \Rightarrow |x|^6 = 64 \Rightarrow |x| = 2 \Rightarrow x = \pm 2 $

✔ Answer: $ x = \pm 2 $

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12. $ 1^3 + 1^{-3} = ? $

$ 1^3 = 1 $, $ 1^{-3} = \frac{1}{1^3} = 1 $

So $ 1 + 1 = 2 $

✔ Answer: 2

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13. Find the value $ (2^{11} + 3^2 - 5^1)^0 $

Any non-zero number raised to power 0 is 1.

So we need to check if the base is non-zero.

Compute inside:
$ 2^{11} = 2048 $, $ 3^2 = 9 $, $ 5^1 = 5 $

So $ 2048 + 9 - 5 = 2052 $, which is not zero.

So $ (2052)^0 = 1 $

✔ Answer: 1

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14. The value of $ \frac{x^{21}}{x^{17}} = ? $

Using $ \frac{x^m}{x^n} = x^{m-n} $

So $ x^{21-17} = x^4 $

✔ Answer: $ x^4 $

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15. Follow the pattern and complete:

Given:
- $ 121 = 11^2 $
- $ 12321 = 111^2 $
- $ 123454321 = ? $

Pattern: The numbers on the left are palindromes made from increasing digits up to a point and back down.

And they equal squares of numbers like 11, 111, 1111...

So:
- $ 11^2 = 121 $
- $ 111^2 = 12321 $
- $ 1111^2 = 1234321 $
- $ 11111^2 = 123454321 $

So $ 123454321 = 11111^2 $

✔ Answer: $ 11111^2 $

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16. Find the value of $ a^2b^3 $ when $ a = 2 $, $ b = 3 $

$ a^2 = 4 $, $ b^3 = 27 $, so $ 4 \times 27 = 108 $

✔ Answer: 108

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17. Find the value of $ x $, if $ 2^x = \frac{1}{32} $

Note: $ 32 = 2^5 $, so $ \frac{1}{32} = 2^{-5} $

So $ 2^x = 2^{-5} \Rightarrow x = -5 $

✔ Answer: -5

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18. If $ 2^x \times 5^x = 1000 $, then $ x = ? $

Note: $ 2^x \cdot 5^x = (2 \cdot 5)^x = 10^x $

So $ 10^x = 1000 $

But $ 1000 = 10^3 $, so $ x = 3 $

✔ Answer: 3

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19. If $ 5^{2x - 6} = 1 $, then $ x = ? $

Any number to the power 0 is 1.

So $ 5^{2x - 6} = 1 \Rightarrow 2x - 6 = 0 \Rightarrow 2x = 6 \Rightarrow x = 3 $

✔ Answer: 3

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20. Find the value of $ x $ if $ 2^{3x} = 64 $

$ 64 = 2^6 $, so:

$ 2^{3x} = 2^6 \Rightarrow 3x = 6 \Rightarrow x = 2 $

✔ Answer: 2

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✔ Final Answers:

1. $ 2^{11} $
2. 9
3. $ \frac{4}{25} $
4. $ -5 $
5. 1
6. 81
7. 57
8. 6
9. 5
10. 72
11. $ \pm 2 $
12. 2
13. 1
14. $ x^4 $
15. $ 11111^2 $
16. 108
17. -5
18. 3
19. 3
20. 2

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