How do you find the inverse of $1-ln(x-2)=f(x)$ ?

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Answer 1 · 2016-01-14T03:37:31

Inverse x and y.

$f^-1(x)=e^(1-x)+2$

The least formal way, (but easier in my opinion) is replacing x and y, where $y=f(x)$ . Therefore, the function:

$f(x)=1-ln(x-2)$

$y=1-ln(x-2)$

Has an inverse function of:

$x=1-ln(y-2)$

Now solve for y:

$ln(y-2)=1-x$

$ln(y-2)=lne^(1-x)$

Logarithmic function $ln$ is 1-1 for any $x>0$

$y-2=e^(1-x)$

$y=e^(1-x)+2$

Which gives the inverse function:

$f^-1(x)=e^(1-x)+2$

How do you find the inverse of 1-ln(x-2)=f(x)1-ln(x-2)=f(x)?