How do you find the inverse of $f(x)=log[x+1]$ ?

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Answer 1 · 2018-06-26T21:04:24

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

$y = log[ x+1 ]$

Swap $x$ for the $y$

$x = log[y+1]$

Then rearrange for $y$

$e^x = e^log[y+1]$

$e^x = y +1$

$e^x -1 = y$

$=> f^(-1) (x) = e^x -1$

If you take $logx -= log_e x -= lnx$

If you work with $logx -= log_10 x$

The answer would just be $= 10^x -1$

How do you find the inverse of f(x)=log[x+1]f(x)=log[x+1]?