Question

e^(4x) + 5e^(2x) - 24 = 0. How do you solve this type of problem?

Answer 1

Let's start with this,
\[\large e^{4x} \qquad = \qquad (e^{2x})^2\]

This is established by using an exponent law.

Answer 2

So we currently have this.
\[\large (e^{2x})^2 + 5(e^{2x}) - 24 = 0\]

Answer 3

Make the substitution, \(\large u=(e^{2x})^2\)

That should simplify things down very nicely so you can solve it as a quadratic.
Lemme know if you're still confused.

Answer 4

Woops
I mean, let \(\large u=e^{2x}\)
Sorry typo.