What is the inverse of f(x)=(1/2)^x+1,x<0 ?

The inverse of f(x)=(1/2)^x+1,x<0 is -(ln(x-1))/(ln(2))

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inverse of f(x)=(1/2)^x+1,x<0

inverse

f (x) = (\frac{1}{2})^{x} + 1, x < 0

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

Solution steps

y = (\frac{1}{2})^{x} + 1

Replace

x

with

y

x = (\frac{1}{2})^{y} + 1

Solve for

y, x = (\frac{1}{2})^{y} + 1

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

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

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

- \frac{1}{x} + 2

2 lo g_{10} (x + 3)

\frac{x ^{2} + 5 x + 6}{x + 2}

What is the inverse of f(x)=(1/2)^x+1,x<0 ?
The inverse of f(x)=(1/2)^x+1,x<0 is -(ln(x-1))/(ln(2))