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 )}

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

f (x) = \frac{7}{x + 3}

f (x) = 7 x^{\frac{1}{5}} + 6

f (x) = 3 x^{2} - 6

f (x) = x^{7} + 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)}