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

Nov 9, 2015

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

Explanation:

Given: $4 \left({3}^{2} x\right) = {e}^{x}$

~~~~~~~~~~~~~~ Pre amble ~~~~~~~~~~~~~~~
The key here is than ${\log}_{e} \left(e\right) = 1$

${\log}_{e}$ is normally written as ln. So ${\log}_{e} \left(e\right) = \ln \left(e\right) = 1$

The same way that ${\log}_{10} \left(10\right) = 1$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Write as $\text{ } 36 x = {e}^{x}$

Taking logs

$\ln \left(36\right) + \ln \left(x\right) = x \ln \left(e\right)$

$\ln \left(36\right) + \ln \left(x\right) = x$

By the way; $36 = {6}^{2} \to \ln \left(32\right) = \ln \left({6}^{2}\right) = 2 \ln \left(6\right)$

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

$5.23980 \ldots . .$

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