What is the inverse of f(x)= 1/(log_{x+1)(2)} ?

The inverse of f(x)= 1/(log_{x+1)(2)} is 2^x-1

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inverse of f(x)= 1/(log_{x+1)(2)}

inverse

f (x) = \frac{1}{lo g _{x + 1} ( 2 )}

2^{x} - 1

Solution steps

y = \frac{1}{lo g _{x + 1} ( 2 )}

Replace

x

with

y

x = \frac{1}{lo g _{y + 1} ( 2 )}

Solve for

y, x = \frac{1}{lo g _{y + 1} ( 2 )}

2^{x} - 1

e^{x - 2} + 3

\frac{1}{s - 1}

\frac{1}{6} x - 3

5 x - 9

5 x + 9

What is the inverse of f(x)= 1/(log_{x+1)(2)} ?
The inverse of f(x)= 1/(log_{x+1)(2)} is 2^x-1