What is the inverse of f(x)=(2^x)/(2^x-1) ?

The inverse of f(x)=(2^x)/(2^x-1) is (ln(x/(x-1)))/(ln(2))

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inverse of f(x)=(2^x)/(2^x-1)

inverse

f (x) = \frac{2 ^{x}}{2 ^{x} - 1}

\frac{ln ( \frac{x}{x - 1} )}{ln ( 2 )}

Solution steps

y = \frac{2 ^{x}}{2 ^{x} - 1}

Replace

x

with

y

x = \frac{2 ^{y}}{2 ^{y} - 1}

Solve for

y, x = \frac{2 ^{y}}{2 ^{y} - 1}

\frac{ln ( \frac{x}{x - 1} )}{ln ( 2 )}

f (x) = 5 x^{\frac{1}{4}}

h (x) = \frac{2 x + 17}{3 x + 1}

f (x) = 2 (x + 1)^{2} - 3

6 x + 3

y = 3 x^{2}

What is the inverse of f(x)=(2^x)/(2^x-1) ?
The inverse of f(x)=(2^x)/(2^x-1) is (ln(x/(x-1)))/(ln(2))