Question

Express 21^-9*(1/7) ^5÷21^-14 as a power of 3​

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{\dfrac{21^{-9}{\times}\left(\dfrac{1}{7}\right)^5}{21^{-14}}}$

$\underline{\textbf{To find:}}$

$\textsf{The Expression as power of 3}$

$\underline{\textbf{Solution:}}$

$\mathsf{Consider,}$

$\mathsf{\dfrac{21^{-9}{\times}\left(\dfrac{1}{7}\right)^5}{21^{-14}}}$

$\mathsf{Using,}$

$\mathsf{\dfrac{a^m}{a^n}=a^{m-n}}$

$\mathsf{=21^{-9-(-14)}{\times}\left(\dfrac{1}{7}\right)^5}$

$\mathsf{=21^{-9+14}{\times}\left(\dfrac{1}{7}\right)^5}$

$\mathsf{=21^5{\times}\left(\dfrac{1}{7^5}\right)}$

$\mathsf{=(3{\times}7)^5{\times}\left(\dfrac{1}{7^5}\right)}$

$\mathsf{=3^5{\times}7^5{\times}\left(\dfrac{1}{7^5}\right)}$

$\mathsf{=3^5}$

$\implies\boxed{\mathsf{\dfrac{21^{-9}{\times}\left(\dfrac{1}{7}\right)^5}{21^{-14}}=3^5}}$

$\underline{\textbf{Find more:}}$

(4.23)²-(2.71)²/4.23-2.71

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