What is the inverse of f(x)=((a^x-1))/((a^x+1)) ?

The inverse of f(x)=((a^x-1))/((a^x+1)) is (ln(\frac{-1-x)/(x-1))}{ln(a)}

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inverse of f(x)=((a^x-1))/((a^x+1))

inverse

f (x) = \frac{( a ^{x} - 1 )}{( a ^{x} + 1 )}

\frac{ln ( \frac{- 1 - x}{x - 1} )}{ln ( a )}

Solution steps

y = \frac{a ^{x} - 1}{a ^{x} + 1}

Replace

x

with

y

x = \frac{a ^{y} - 1}{a ^{y} + 1}

Solve for

y, x = \frac{a ^{y} - 1}{a ^{y} + 1}

\frac{ln ( \frac{- 1 - x}{x - 1} )}{ln ( a )}

2 (x - 1)^{2} - 8

f (x) = \frac{( 4 x - 5 )}{6}

f (x) = 4 x + cos (π) x

f (x) = \frac{1}{3 x + 6}

f (x) = \frac{1}{3 x + 2}

What is the inverse of f(x)=((a^x-1))/((a^x+1)) ?
The inverse of f(x)=((a^x-1))/((a^x+1)) is (ln(\frac{-1-x)/(x-1))}{ln(a)}