# Question #3bdb8

Jun 17, 2017

$y = {x}^{x} \cdot {e}^{-} x$

#### Explanation:

$e$ is the inverse operation of $\ln$, which means they undo or cancel each other out. For example, if you have $\ln x$, you can take ${e}^{\ln} x$ and you are left with $x$:

I.e.

${\cancel{e}}^{\cancel{\ln} x} = x$

Also, there is a log law and an index law that will help:

$a \ln b = \ln {b}^{a}$

${e}^{a + b} = {e}^{a} \cdot {e}^{b}$

First, use the log law to simplify:

$\ln y = x \ln x - x \Rightarrow \ln y = \ln {x}^{x} - x$

Next, take e^ of each side of the equation:

$\Rightarrow {\cancel{e}}^{\cancel{\ln} y} = {e}^{\left(\ln \left({x}^{x}\right) - x\right)}$

Use the index law to separate the right hand side:

$y = {e}^{\ln} \left({x}^{x}\right) \cdot {e}^{-} x$

$\Rightarrow y = {\cancel{e}}^{\cancel{\ln} \left({x}^{x}\right)} \cdot {e}^{-} x$

$\Rightarrow y = {x}^{x} \cdot {e}^{-} x$