You are correct up to
$\displaystyle e^6 = 4 - x$
$\displaystyle e^6 - 4 = -x$
$\displaystyle x = 4 - e^6$.
For the second
$\displaystyle \displaystyle{e^{\frac{x}{4}} = 2^{5 - 9x}}$
$\displaystyle \displaystyle{\frac{x}{4} = \ln{(2^{5 - 9x})}}$
$\displaystyle \displaystyle{\frac{x}{4} = (5 - 9x)\ln{2}}$
$\displaystyle \displaystyle{x = 4(5 - 9x)\ln{2}}$.