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If $n = \frac{a^{x} - a^{- x}}{a^{x} + a^{- x}}$ $n=(a^x-a^(-x))/(a^x+a^(-x))$ then find ${log}_{a} (\frac{1 + n}{1 - n})$ $log_a((1+n)/(1-n))$ . Ans=2n

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Answer 1 · 2018-07-10T11:35:30

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

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

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

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

${log}_{a} (\frac{1 + n}{1 - n}) = 2 x$ $log_a((1+n)/(1-n)) = 2 x$

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The question is below?

If $n = \frac{a^{x} - a^{- x}}{a^{x} + a^{- x}}$ $n=(a^x-a^(-x))/(a^x+a^(-x))$ then find ${log}_{a} (\frac{1 + n}{1 - n})$ $log_a((1+n)/(1-n))$ . Ans=2n

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If n=ax−a−xax+a−xn=(a^x-a^(-x))/(a^x+a^(-x)) then find loga(1+n1−n)log_a((1+n)/(1-n)). Ans=2n

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If $n = \frac{a^{x} - a^{- x}}{a^{x} + a^{- x}}$ $n=(a^x-a^(-x))/(a^x+a^(-x))$ then find ${log}_{a} (\frac{1 + n}{1 - n})$ $log_a((1+n)/(1-n))$ . Ans=2n