Thread: Equation with 2 solutions

e^(2x)-7e^x+12=0 had two solutions

i got this (7(plus minus)sqrt(7^2-4*e*12))/2e now i replaced e with 2.718 because that is the value of e and I got 1.591 and 0.984. Is this hat I what's suppose to do?

Roots are:
x1=ln(3)
x2=ln(4)

Originally Posted by MaxJasper

Roots are:
x1=ln(3)
x2=ln(4)

How did you get that?

Let u=e^x then

u^2-7u+12=0

solve for u then plug into u=e^x and find x.

This is what I got when I solved for u:

U= (loge^x(-12)-2)/-7

The log is base e^x

Am I right?

Hello, RD33!

Factor: .

Then: .