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

 Video Solution
Exponents And Powers
Ex 13.2 | Question 1

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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