Natural Logarithm - solve the equation I am having problems understanding how to solve $e^{4x}+4e^{2x}-21 = 0$.

Try temporarily setting $u=e^{2x}$ so it becomes $u^2+4u-21=0.$ Then solve the quadratic, discarding any negative solutions since $u=e^{2x}>0.$ After that use the $\ln$ function to get $x$ back.
For example if it came out one value of $u$ was $u=5$ then from $e^{2x}=5$ comes $2x=\ln 5$ and then $x=(1/2) \ln 5.$
