Iflogx(916)=−12,thenxisequalto:Iflogx(916)=−12,thenxisequalto:{\rm{If}}\,{\log _x}\left( {{9 \over {16}}} \right) = - {1 \over 2},\,{\rm{then}}\,x\,{\rm{is}}\,{\rm{equal}}\,{\rm{to:}}" /> Iflogx(916)=−12,thenxisequalto:Iflogx(916)=−12,thenxisequalto:{\rm{If}}\,{\log _x}\left( {{9 \over {16}}} \right) = - {1 \over 2},\,{\rm{then}}\,x\,{\rm{is}}\,{\rm{equal}}\,{\rm{to:}}" /> Iflogx(916)=−12,thenxisequalto:Iflogx(916)=−12,thenxisequalto:{\rm{If}}\,{\log _x}\left( {{9 \over {16}}} \right) = - {1 \over 2},\,{\rm{then}}\,x\,{\rm{is}}\,{\rm{equal}}\,{\rm{to:}}" />

India’s No.1 Educational Platform For UPSC,PSC And All Competitive Exam

Download Our App

$I f \log_{x} (\frac{9}{16}) = - \frac{1}{2}, t h e n x i s e q u a l t o :$ ${\rm{If}}\,{\log _x}\left( {{9 \over {16}}} \right) = - {1 \over 2},\,{\rm{then}}\,x\,{\rm{is}}\,{\rm{equal}}\,{\rm{to:}}$

1- ³/₄
2³/₄
3⁸¹/₂₅₆
4²⁵⁶/₈₁

Answer:- 1

Explanation:-

Solution:

\begin{aligned} \log_{x} (\frac{9}{16}) = - \frac{1}{2} \\ \Rightarrow x^{- \frac{1}{2}} = \frac{9}{16} \\ \Rightarrow \frac{1}{\sqrt{x}} = \frac{9}{16} \\ \Rightarrow \sqrt{x} = \frac{16}{9} \\ \Rightarrow x = {(\frac{16}{9})}^{2} \\ \Rightarrow x = \frac{256}{81} \end{aligned}

$\eqalign{ & {\log _x}\left( {{9 \over {16}}} \right) = - {1 \over 2} \cr & \Rightarrow {x^{ - {1 \over 2}}} = {9 \over {16}} \cr & \Rightarrow {1 \over {\sqrt x }} = {9 \over {16}} \cr & \Rightarrow \sqrt x = {{16} \over 9} \cr & \Rightarrow x = {\left( {{{16} \over 9}} \right)^2} \cr & \Rightarrow x = {{256} \over {81}} \cr}$

Workspace Report

Post your Comments

", "text": "

I f \log_{x} (\frac{9}{16}) = - \frac{1}{2}, t h e n x i s e q u a l t o :

${\rm{If}}\,{\log _x}\left( {{9 \over {16}}} \right) = - {1 \over 2},\,{\rm{then}}\,x\,{\rm{is}}\,{\rm{equal}}\,{\rm{to:}}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" }, "answerCount": "4", "acceptedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} \log_{x} (\frac{9}{16}) = - \frac{1}{2} \\ \Rightarrow x^{- \frac{1}{2}} = \frac{9}{16} \\ \Rightarrow \frac{1}{\sqrt{x}} = \frac{9}{16} \\ \Rightarrow \sqrt{x} = \frac{16}{9} \\ \Rightarrow x = {(\frac{16}{9})}^{2} \\ \Rightarrow x = \frac{256}{81} \end{aligned}

$\eqalign{ & {\log _x}\left( {{9 \over {16}}} \right) = - {1 \over 2} \cr & \Rightarrow {x^{ - {1 \over 2}}} = {9 \over {16}} \cr & \Rightarrow {1 \over {\sqrt x }} = {9 \over {16}} \cr & \Rightarrow \sqrt x = {{16} \over 9} \cr & \Rightarrow x = {\left( {{{16} \over 9}} \right)^2} \cr & \Rightarrow x = {{256} \over {81}} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} \log_{x} (\frac{9}{16}) = - \frac{1}{2} \\ \Rightarrow x^{- \frac{1}{2}} = \frac{9}{16} \\ \Rightarrow \frac{1}{\sqrt{x}} = \frac{9}{16} \\ \Rightarrow \sqrt{x} = \frac{16}{9} \\ \Rightarrow x = {(\frac{16}{9})}^{2} \\ \Rightarrow x = \frac{256}{81} \end{aligned}

$\eqalign{ & {\log _x}\left( {{9 \over {16}}} \right) = - {1 \over 2} \cr & \Rightarrow {x^{ - {1 \over 2}}} = {9 \over {16}} \cr & \Rightarrow {1 \over {\sqrt x }} = {9 \over {16}} \cr & \Rightarrow \sqrt x = {{16} \over 9} \cr & \Rightarrow x = {\left( {{{16} \over 9}} \right)^2} \cr & \Rightarrow x = {{256} \over {81}} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

Test

Classes

E-Book