Finding inverse of a function.

Question

Could someone please help me finding the inverse of the following function: $$f(x)= \frac{x-1}{\ln(x)}$$ where $x>0$?

Thank you.

Answer 1

There is no solution in elementary functions but the inverse function is given by $$ x = -y \mathcal{W}\left( -\frac{e^{-1/y}}{y} \right) $$

where $\mathcal{W}$ is the product log function.

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step by step solution:

$$ y= \frac{x-1}{\ln{x}} \Longrightarrow -\ln{x} = - \frac{x-1}{y} \Longrightarrow \exp{\left(-\frac{x}{y}\right)} = \exp{\left(-\frac{1}{y}-\ln{x} \right)}$$

Now multiply both sides with $\frac{x}{y}$ and get:

$$\frac{x}{y}\exp{\left(-\frac{x}{y}\right)} = \frac{\exp{\left(-1/y\right)}}{y}\Longleftrightarrow -\frac{x}{y}\exp{\left(-\frac{x}{y}\right)} = -\frac{\exp{\left(-1/y\right)}}{y}$$

Now apply the definition of $\mathcal{W}$ and my answer mentioned before follows.

Answer 2

if you want to find inverse of function $f(x)=y$ ,try to replace places for $x$ and $y$ or use $f(y)=x$ and solve it to back to $y$ ,in your case set

$x=(y-1)/(ln(y)$ ,now interpret $y$ by $x$

now we have following ine

$x*ln(y)=(y-1)$

or

$ln(y)/(y-1)=1/x$

now please take on exponent form and also please note that $e^{ln(x)}=x$