Question

If $x$ = $\frac{a^n-a^{-n}}{a^n+a^{-n}}$, then the value of $n$ is equal to

• A. $\frac{1}{2}{\log }_a\left(\frac{1+x}{1-x}\right)$
• B. $\frac{1}{2}{\log }_a\left(\frac{1}{1-x}\right)$
• C. $\frac{1}{2}{\log }_a\left(\frac{1+x}{x-1}\right)$
• D. None of these

Answer 1

The correct option is A $\frac{1}{2}{\log }_a\left(\frac{1+x}{1-x}\right)$

$x$ $=\frac{a^n-a^{-n}}{a^n+a^{-n}}$

$x$ $=\frac{a^n-\frac{1}{a^n}}{a^n+\frac{1}{a^n}}$

$x$ $=\frac{a^{2n}-1}{a^{2n}+1}$

$x{a}^{2n}+x=a^{2n}-1$

$x+1=a^{2n}(1-x)$

$\frac{1+x}{1-x}=a^{2n}$

Taking log on both side,

$\log \frac{1+x}{1-x}=\log a^{2n}$

$\log \frac{1+x}{1-x}=2n\log a\dots ({\log }_y{x}^n=n{\log }_{y}x)$

$n=\frac{1}{2}\left(\frac{\log \frac{1+x}{1-x}}{\log a}\right)$

By change base rule, $[{\log }_{b}a=\frac{{\log }_{c}a}{{\log }_{c}b}]$

$n=\frac{1}{2}{\log }_a\left(\frac{1+x}{1-x}\right)$