Ex.13.2 Q4 Exponents and Powers Solutions - NCERT Maths Class 7

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Question

Express each of the following as a product of prime factors only in exponential form:

(i) \(108 × 192 \)

(ii) \( 270\)

(iii) \(729 × 64\)

(iv) \(768\)

 Video Solution
Exponents And Powers
Ex 13.2 | Question 4

Text Solution

What is Known?

(i) \(108 × 192 \)

(ii) \( 270\)

(iii) \(729 × 64\)

(iv) \(768\)

What is unknown?

Product of prime factors only in exponential form.

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.  \({{{a}}^{{m}}}{{ \div }}\,\,{{{a}}^{{n}}} = {{{a}}^{{{m - n}}}}\)

3.  \({{\left( {{{{a}}^{{m}}}} \right)}^{{n}}} = {{{a}}^{{{mn}}}}\)

4.  \({{{a}}^{{0}}}= 1\)

Steps:

(i) \(108 \times192\)

\(\begin{align} \\&=2\!\!\times\!\!2\!\! \times\!\! 2\!\! \times\!\! 2\!\!\times\!\!2\!\!\times\!\!2\!\!\times\!\!\,2\!\! \times\!\!2\!\!\times\!\!3\!\!\times\!\!3\!\!\times\!\!3\!\!\times\!\!3\\&\qquad {a^m}\!\!\times\!\!{a^n}\!\!=\!\!{a^{m + n}}\\&={2^8}\!\!\times\!\!{3^4}\end{align}\)

(ii) \(270\)

\(\begin{align}\\&=2 \!\!\times\!\!3\!\!\times\!\!3\!\!\times\!\!3\!\!\times\!\!5 \quad {{{a}}^{{m}}}\!\!{{ \times }}{{{a}}^{{n}}}\,\!\!{{ = }}\,{{{a}}^{{{m + n}}}}\\&= 2 \times {3^3} \times 5 \\ &= 10 \times 3^3\end{align}\)

(iii) \(\begin{align}729 \times 64\end{align}\)

\(\begin{align}\\&\!\!\!\!=\!\!3\!\!\times\!\!3\!\!\times\!\!3\!\!\times\!\!3\!\!\times\!\!3\!\!\times\!\!3\!\!\times\!\!2\!\!\times\!\!2\!\!\times\!\!2\!\!\times\!\!2\!\!\times\!\!2\!\!\times\!\!2\\&\qquad{a^m}\!\!\times\!\!{a^n}\!\!=\!\!{a^{m + n}}\\&=\!\!{3^6}\!\!\times\!\!{2^6}\\&= {\left( {3 \times 2} \right)^6}\end{align}\)

(iv)\(768\)

\(\begin{align}\\&= \!2 \!\times\! 2 \!\times\! 2 \!\times\! 2 \!\times\! 2 \!\times\! 2 \!\times\! 2 \!\times\! 2 \!\times\! 3\\&\qquad{{{a}}^{{m}}}{{ \times }}{{{a}}^{{n}}}{{ = }}\,{{{a}}^{{{m + n}}}}\\&= {2^8} \times 3\end{align}\)

  
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