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

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#### Explanation

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#### Explanation:

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9
Nov 16, 2015

${f}^{- 1} \left(x\right) = \frac{1}{2} \left(\ln x + 1\right)$ for $x > 0$

#### Explanation:

First, set plug $y$ for $f \left(x\right)$:

$y = {e}^{2 x - 1}$

Then, exchange $y$ and $x$ in the equation:

$x = {e}^{2 y - 1}$

Now, try to solve the equation for $y$ in terms of $x$.

To do so, we need to apply the logarithmic function $\ln$ to both sides of the equation. Please note that $x > 0$ needs to hold so that the logarithm is well defined.

$\ln \left(x\right) = \ln \left({e}^{2 y - 1}\right)$
$\iff \ln \left(x\right) = 2 y - 1$
$\iff \ln \left(x\right) + 1 = 2 y$
$\iff \frac{1}{2} \left(\ln \left(x\right) + 1\right) = y$

The only thing left to do is replacing $y$ with ${f}^{- 1} \left(x\right)$.
The inverse function is

${f}^{- 1} \left(x\right) = \frac{1}{2} \left(\ln x + 1\right)$ for $x > 0$

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