What is the inverse of (log_{10}(x))/(1-log_{10)(x)} ?

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

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inverse of (log_{10}(x))/(1-log_{10)(x)}

inverse

\frac{lo g _{10} ( x )}{1 - lo g _{10} ( x )}

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

Solution steps

y = \frac{lo g _{10} ( x )}{1 - lo g _{10} ( x )}

Replace

x

with

y

x = \frac{lo g _{10} ( y )}{1 - lo g _{10} ( y )}

Solve for

y, x = \frac{lo g _{10} ( y )}{1 - lo g _{10} ( y )}

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

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

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

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

10 x + x + 101

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

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