How do you find the inverse of $y = e^{x - 1}$ $y=e^(x-1)$ ?

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Answer 1 · 2015-12-06T03:36:36

$f^{- 1} (x) = ln x + 1$ $f^(-1)(x) = ln x +1$

Given $y = e^{x - 1}$ $y= e^(x-1)$ (this is a one-to-one function)

Step 1: Switch $x$ $x$ for $y$ $y$ and $y$ $y$ for $x$ $x$ like this

$x = e^{y - 1}$ $color(red) x= e^(color(blue)(y)-1)$

Step 2: Begin to solve for $y$ $y$

Take $ln$ $ln$ of both side

$ln (x) = {ln e}^{y - 1}$ $ln(x) = lne^(y-1)$

Use the properties of log
$ln e = 1$ $ln e= 1$ ; ${log a}^{n} = n log a$ $log a^n = nlog a$

$ln x = y - 1$ $ln x= y-1$

$ln x + 1 = y$ $ln x + 1 = y$

Step 3: Change $y$ $y$ to $f^{- 1} (x)$ $f^(-1)(x)$

$f^{- 1} (x) = ln x + 1$ $f^(-1)(x) = ln x +1$

*Note: To solve exponential equation, we use logarithm, and vice versa

Remember: One-to-one function is the only function that have an i nverse that is an function.*

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