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

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

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

inverse

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

- \frac{1 0 ^{x}}{1 0 ^{x} - 1}

Solution steps

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

Replace

x

with

y

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

Solve for

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

- \frac{1 0 ^{x}}{1 0 ^{x} - 1}

- 2 (x + 1)^{2} + 4

f (x) = - \frac{2}{3} x + 8

f (x) = \frac{x - 1}{x - 4}

\frac{3}{4 + x}

y = \frac{7 x - 9}{9 x + 4}

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