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

Question

5419 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-14T20:33:11

See 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#