How do you solve #e^(2x)-5e^x+6=0#?

Question

15628 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-21T17:53:50

#x=ln3 or x=ln2#

Let # u=e^x#

Then we have

#u^2-5u+6=0#

#(u-3)(u-2)=0#

#u-3=0 or u-2=0#

#e^x-3=0 or e^x-2=0#

#e^x=3 or e^x=2#

#lne^x=ln3 or lne^x = ln 2#

#x ln e=ln3 or xlne=ln2#

#x=ln3 or x=ln2#