What is the inverse of f(x)=log_{10}(2/(x+1)) ?

The inverse of f(x)=log_{10}(2/(x+1)) is (2^{1-x})/(5^x)-1

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inverse of f(x)=log_{10}(2/(x+1))

inverse

f (x) = lo g_{10} (\frac{2}{x + 1})

\frac{2 ^{1 - x}}{5 ^{x}} - 1

Solution steps

y = lo g_{10} (\frac{2}{x + 1})

Replace

x

with

y

x = lo g_{10} (\frac{2}{y + 1})

Solve for

y, x = lo g_{10} (\frac{2}{y + 1})

\frac{2 ^{1 - x}}{5 ^{x}} - 1

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

e^{2 y}

f (x) = \frac{4 x + 3}{5}

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

g (x) = \frac{6 - x}{8}

What is the inverse of f(x)=log_{10}(2/(x+1)) ?
The inverse of f(x)=log_{10}(2/(x+1)) is (2^{1-x})/(5^x)-1