What is the inverse of f(x)=((2^x+5))/(2^x-7) ?

The inverse of f(x)=((2^x+5))/(2^x-7) is (ln(\frac{5+7x)/(x-1))}{ln(2)}

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inverse of f(x)=((2^x+5))/(2^x-7)

inverse

f (x) = \frac{( 2 ^{x} + 5 )}{2 ^{x} - 7}

\frac{ln ( \frac{5 + 7 x}{x - 1} )}{ln ( 2 )}

Solution steps

y = \frac{2 ^{x} + 5}{2 ^{x} - 7}

Replace

x

with

y

x = \frac{2 ^{y} + 5}{2 ^{y} - 7}

Solve for

y, x = \frac{2 ^{y} + 5}{2 ^{y} - 7}

\frac{ln ( \frac{5 + 7 x}{x - 1} )}{ln ( 2 )}

f (x) = - tan (x + 7)

1 - 2 e^{- 2 x}

f (x) = (ln (x))^{8}

f (x) = (x - 18)^{2}

h (x) = (x - 2)^{3}

What is the inverse of f(x)=((2^x+5))/(2^x-7) ?
The inverse of f(x)=((2^x+5))/(2^x-7) is (ln(\frac{5+7x)/(x-1))}{ln(2)}