Ex.12.1 Q7 Exponents and Powers Solution - NCERT Maths Class 8


Question

Simplify.

(i) \(\begin{align}\frac{{25 \times {t^{ - 4}}}}{{{5^{ - 3}} \times 10 \times {t^{ - 8}}}} \end{align}\)

(ii) \(\begin{align} \frac{{{3^{ - 5}} \times {{10}^{ - 5}} \times 125}}{{{5^{ - 7}} \times {6^{ - 5}}}}\end{align}\)

 Video Solution
Exponents And Powers
Ex 12.1 | Question 7

Text Solution

(i) Evaluate \(\begin{align}\frac{{25 \times {t^{ - 4}}}}{{{5^{ - 3}} \times 10 \times {t^{ - 8}}}} \end{align}\)

What is known?

Expression in exponential form

What is unknown?

Value of the expression

Reasoning:

\[\begin{align}{a^{{m}}}{{ \times}}{a^{{n}}}{{ =}}{a^{{{m + n}}}} \quad \text{and} \quad \frac{{{a^{{m}}}}}{{{a^{{n}}}}}{{=}}{a^{{{m - n}}}} \end{align}\]

Steps:

\[\begin{align}\frac{{25 \times {t^{ - 4}}}}{{{5^{ - 3}} \times 10 \times {t^{ - 8}}}} &= \frac{{{5^2} \times {t^{ - 4}}}}{{{5^{ - 3}} \times 5 \times 2 \times {t^{ - 8}}}} \\\\& \left[ {{a^m} \times {a^n} = {a^{m + n}}} \right]\\\\
&= \frac{{{5^2} \times {t^{ - 4}}}}{{{5^{ - 3 + 1}} \times 2 \times {t^{ - 8}}}}\\&= \frac{{{5^2} \times {t^{ - 4}}}}{{{5^{ - 2}} \times 2 \times {t^{ - 8}}}}\\&= \frac{{{5^{2 - ( - 2)}} \times {t^{ - 4 - ( - 8)}}}}{2} \\\\& \left[ {\frac{{{a^m}}}{{{a^n}}} = {a^{m - n}}} \right]\\\\&= \frac{{{5^4} \times {t^{ - 4 + 8}}}}{2}\\&= \frac{{625{t^4}}}{2}\end{align}\]

(ii) Evaluate \(\begin{align} \frac{{{3^{ - 5}} \times {{10}^{ - 5}} \times 125}}{{{5^{ - 7}} \times {6^{ - 5}}}}\end{align}\)

What is known?

Expression in exponential form

What is unknown?

Value of the expression

Reasoning:

\[\frac{{{a^m}}}{{{a^n}}} = {a^{m - n}}\text{ and }{a^0} = 1\]

Steps:

\[\begin{align}&\frac{{{3}^{-5}}\times {{10}^{-5}}\times 125}{{{5}^{-7}}\times {{6}^{-5}}}\\&=\frac{{{3}^{-5}}\times {{\left( 2\times 5 \right)}^{-5}}\times {{5}^{3}}}{{{5}^{-7}}\times {{(2\times 3)}^{-5}}} \\& =\frac{{{3}^{-5}}\times {{2}^{-5}}\times {{5}^{-5}}\times {{5}^{3}}}{{{5}^{-7}}\times {{2}^{-5}}\times {{3}^{-5}}} \\& ={{3}^{-5-(-5)}}\times {{2}^{-5-(-5)}}\times {{5}^{-5-(-7)+3}} \\& ={{3}^{0}}\times {{2}^{0}}\times {{5}^{5}}\qquad\left[ \because {{a}^{0}}=1 \right] \\& =1\times 1\times {{5}^{5}}={{5}^{5}}\end{align}\]

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