If $y=\frac{2^x-2^{-x}}{2^x+2^{-x}}$, where $0 < y < 1$, then the value of $x$ is....

$\frac{1}{2} \log _2\left(\frac{y}{1-y}\right)$

$\frac{1}{2} \log _2\left(\frac{1+y}{y}\right)$

$\frac{1}{2} \log _2\left(\frac{2+y}{2-y}\right)$

JEE MATHEMATICS MCQ: TRANSFORMATIONS

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Application +4 | -1

y=\frac{2^x-2^{-x}}{2^x+2^{-x}}

, where

0 < y < 1

, then the value of

x

is....

View answer and explanation

Correct option: B

\frac{1}{2} \log _2\left(\frac{1+y}{1-y}\right)

Option B (

\frac{1}{2} \log _2\left(\frac{1+y}{1-y}\right)

) is correct because it is the result of solving the given equation for

x

using algebraic manipulation and logarithms.

Practice All Chapter MCQs →

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\mathrm{\;sin}\,\,36^\circ \,\,\sin \,\,72^\circ \,\,\sin \,\,108^\circ \,\,\sin \,\,144^\circ \;