What is the inverse of log_{2}((-x-1)/(x-1)) ?

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

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inverse of log_{2}((-x-1)/(x-1))

inverse

lo g_{2} (\frac{- x - 1}{x - 1})

\frac{- 1 + 2 ^{x}}{2 ^{x} + 1}

Solution steps

y = lo g_{2} (\frac{- x - 1}{x - 1})

Replace

x

with

y

x = lo g_{2} (\frac{- y - 1}{y - 1})

Solve for

y, x = lo g_{2} (\frac{- y - 1}{y - 1})

\frac{- 1 + 2 ^{x}}{2 ^{x} + 1}

f (x) = \frac{2 x + 1}{5 - 3 x}

f (x) = 14 x^{3} - 5

f (x) = 7 x + 24

\frac{x ^{2}}{- x ^{2} + 4}

f (x) = 1 + lo g_{\frac{1}{e}} (2 - x)

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