TITLE: Solve the equation. e and natural logs
QUESTION [5 upvotes]: $$e^x − 6e^{-x} − 1 = 0$$
No idea how to solve this. If someone could show me the first one or two steps to push me in the right direction that would be great.

REPLY [6 votes]: HINT : Multiply the both sides by $e^x$ to get
$$(e^x)^2-6-e^x=0.$$
Now, let $e^x=t$.

REPLY [6 votes]: You can rewrite this as
$$(e^x)^2-6-e^x=0$$
$$(e^x+2)(e^x-3)=0$$
I'll let you take it from here

REPLY [5 votes]: let $e^x = a$
$$ a - \frac{6}{a} - 1 = 0$$
$$ a^2 - a-6 = 0$$
$$ a = 3 \ or \ -2 $$
$$ e^x = 3$$
Edit: for $x \in \mathbb{R}$
Can you find the value of x now?
Hint: take $\ln$ of both sides.