# Math Help - Two More Natural Log Problems

1. ## Two More Natural Log Problems

Help to get me in the right direction will be much appreciated.

2. You are correct up to

$e^6 = 4 - x$

$e^6 - 4 = -x$

$x = 4 - e^6$.

For the second

$\displaystyle{e^{\frac{x}{4}} = 2^{5 - 9x}}$

$\displaystyle{\frac{x}{4} = \ln{(2^{5 - 9x})}}$

$\displaystyle{\frac{x}{4} = (5 - 9x)\ln{2}}$

$\displaystyle{x = 4(5 - 9x)\ln{2}}$.