Question

Question #3bdb8

Answer 1

`y=x^x*e^-x`

Explanation:

`e` is the inverse operation of `ln`, which means they undo or cancel each other out. For example, if you have `lnx`, you can take `e^lnx` 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:

`alnb=lnb^a`

`e^(a+b)=e^a*e^b`

First, use the log law to simplify:

`lny=xlnx-xrArrlny=lnx^x-x`

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

`rArrcancel(e)^(cancel(ln)y)=e^((ln(x^x)-x))`

Use the index law to separate the right hand side:

`y=e^ln(x^x)*e^-x`

`rArry=cancel(e)^(cancel(ln)(x^x))*e^-x`

`rArry=x^x*e^-x`