Question

Find the value of x, if log (9/32) to the base x = -(1/8)​

Answer 1

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

$\mathsf{log\,_{x}\left(\dfrac{9}{32}\right)=-\dfrac{1}{8}}$

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

$\textsf{The value of x}$

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

$\underline{\textbf{Concept used:}}$

$\boxed{\mathsf{N=a^x\;\;\iff\;\;log\,_{a}N=x}}$

$\mathsf{Consider,}$

$\mathsf{log\,_{x}\left(\dfrac{9}{32}\right)=-\dfrac{1}{8}}$

$\textsf{Convert this into exponential form, we get}$

$\mathsf{(x)^{\dfrac{-1}{8}}=\dfrac{9}{32}}$

$\implies\mathsf{\dfrac{1}{x^{\dfrac{1}{8}}}=\dfrac{9}{32}}$

$\implies\mathsf{x^{\dfrac{1}{8}}=\dfrac{32}{9}}$

$\implies\boxed{\mathsf{x=\left(\dfrac{32}{9}\right)^8}}$

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

Logx+1(x^2+x-6)^2 =4

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