What is the inverse of (2^x)/(7+2^x) ?

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

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inverse of (2^x)/(7+2^x)

inverse

\frac{2 ^{x}}{7 + 2 ^{x}}

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

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

f (x) = arccos (x^{2})

2 x - 3 + 5

f (x) = \frac{( - 2 x - 1 )}{x + 5}

\frac{8 x + 9}{5 x - 8}

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