Ex.13.2 Q1 Exponents and Powers Solution- NCERT Maths Class 7

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Question

Using laws of exponents, simplify and write the answer in exponential form:

(i) $${3^2} \times {3^4} \times {3^8}$$

(ii) $${6^{15}} \div {6^{10}}$$

(iii) $${{\rm{a}}^3} \times {{\rm{a}}^2}$$

(iv) $${{7}^{x}}\times {{7}^{2}}$$

(v) $${\left( {{5^2}} \right)^3} \div {5^3}$$

(vi) $${2^5} \times {\rm{ }}{5^5}$$

(vii) $${a^4} \times {b^4}$$

(viii) $${\left( {{3^4}} \right)^3}$$

(ix) $$( 2^{20} \div 2^{15}) \times 2^3$$

(x) $${8^t} \div {\rm{ }}{8^2}$$

Text Solution

What is known?

Exponential form of numbers.

What is unknown?

Exponential form by using laws of exponents.

(i)  $${3^2} \times {3^4} \times {3^8}$$

(ii)  $${6^{15}} \div {6^{10}}$$

(iii)  $$a^3 \times a^2$$

(iv)  $${{7}^{x}}\times {{7}^{2}}$$

(v) $${\left( {{5^2}} \right)^3} \div {5^3}$$

(vi) $${2^5} \times {\rm{ }}{5^5}$$

(vii)  $$a^4 \times b^4$$

(viii) $${\left( {{3^4}} \right)^3}$$

(ix) $$( 2^{20} \div 2^{15}) \times 2^3$$

(x) $${8^t} \div {\rm{ }}{8^2}$$

Reasoning:

To solve this question, you must remember the laws of exponents. Here are some important laws of exponents:

1. $$a^m \times a^n = a^{m+n}$$

2. $$a^m \times b^m = (ab)^m$$

3.  $$a^m \div a^n = a^{m-n}$$

4. $$\left( a^m \right)^n=a^{mn}$$

5. $$a^o =1$$

Steps:

(i)

$${3^2} \times {3^4} \times {3^8} = {3^{2 + 4 + 8}} = {3^{14}}$$

Using $$a^m \times a^n = a^{m+n}$$

(ii)

$${6^{15}} \div {6^{10}} = {6^{15 - 10}} = {6^5}$$

Using $$a^m \div a^n = a^{m-n}$$

(iii)

$$a^3 \times a^2 = a^{3 + 2} = a^5$$

Using $$a^m \times a^n = a^{m+n}$$

(iv)

$$7^x \times {7^2} = 7^{x + 2}$$

Using $$a^m \times a^n = a^{m+n}$$

(v)

\begin{align}(5^2)^3 \div 5^3 &= 5^6 \div 5^3\\&= 5^{6 - 3}\\&= 5^3\end{align}

Using $$\left( a^m \right)^n=a^{mn}$$  and  $$a^m \div a^n = a^{m-n}$$

(vi)

$${2^5} \times {5^5} = {(2 \times 5)^5} = {(10)^5}$$

Using $$a^m \times b^m = (ab)^m$$

(vii)

$$a^4 \times b^4 = (ab)^4$$

Using $$\left( a^m \right)^n=a^{mn}$$

(viii)

$${\left( {{3^4}} \right)^3} = {(3)^{12}}$$

Using $$\left( a^m \right)^n=a^{mn}$$

(ix)

\begin{align}\left( {{2}^{20}}\div {{2}^{15}} \right)\times {{2}^{3}} & ={{2}^{20-15}}\times{{2}^{3}} \\ & ={{2}^{5}}\times{{2}^{3}} \\& ={{2}^{8}} \\\end{align}

Using $$a^m \div a^n = a^{m-n}$$  and  $$a^m \times a^n = a^{m+n}$$

(x)

$${8^t} \div {8^2} = {8^{t - 2}}$$

Using $$a^m \div a^n = a^{m-n}$$

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