Question

How do you solve `4(3^2x) = e^x`?

Answer 1

Took it to a point and found by iteration 1 of the two solutions as
`~~5.23980.....` Perhaps some one else could take the method further to help you.

Explanation:

Given: `4(3^2x) = e^x`

~~~~~~~~~~~~~~ Pre amble ~~~~~~~~~~~~~~~
The key here is than `log_e (e)=1`

`log_e` is normally written as ln. So `log_e(e) =ln(e)=1`

The same way that `log_10(10)=1`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Write as `" "36x =e^x`

Taking logs

`ln(36) + ln(x) =xln(e)`

`ln(36) +ln(x) =x`

By the way; `36 = 6^2 -> ln(32) =ln(6^2) = 2 ln(6)`

Using iteration with a seed value of 3 gives 1 potential solution of

`5.23980.....`

Plotting this as `y= ln(36)+ln(x)-x` shows that there are two solutions when y=0. I could not find the seed value for iteration solution at `x~~`0.0285...#

![http://www.wolframalpha.com/input/?i=2ln%286%29%2Bln%28x%29-x%3D0]()