Solve the equation. e and natural logs

Question

$$e^x − 6e^{-x} − 1 = 0$$

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.

Answer 1

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.

Answer 2

HINT: multiply both sides by $e^x$ to turn it into a quadratic equation in $e^x$.

Answer 3

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

Answer 4

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

Answer 5

Hint:

$$e^x − 6e^{-x} − 1 = 0\Longleftrightarrow$$ $$e^{-x}(e^x-3)(e^x+2)=0\Longleftrightarrow$$ $$e^{-x}=0\vee(e^x-3)=0\vee(e^x+2)=0\Longleftrightarrow$$ $$e^{-x}=0\vee(e^x-3)=0\vee(e^x+2)=0\Longleftrightarrow$$ $$error\vee e^x=3\vee e^x=-2$$