Question

How do you solve `e^(2x)-4e^x-5=0`?

Answer 1

See explanation.

Explanation:

First step is to substitute `e^x` with a new variable:

`t=e^x`

This makes the equation a quadratic one: `t^2-4t-5=0`

Since `e^x` is never negative we have to choose only positive solutions of the equation.

`Delta=(-4)^2-4*1*(-5)=16+2=36`

`sqrt(Delta)=6`

`t_1=(4-6)/2=-1`

We cannot take this solution because (as I wrote earlier) `e^x` is never negative.

`t_2=(4+6)/2=10/2=5`

This solution is a valid one, so we can now calculate the oroginal variable `x`:

`e^x=5 =>x=ln5`

Answer: This equation has one solution `x=ln5`