How do you solve #e^(2x)-6e^x+8=0#?

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Answer 1 · 2015-12-10T17:33:56

Use #e^(2x) = (e^x)^2# to see that this equation is quadratic in #e^x#.

#e^(2x)-6e^x+8=0#

#(e^x)^2-6e^x+8=0#

Factor without substituting

#(e^x-4)(e^x - 2 ) = 0#

So #e^x-4=0# #" or "# #e^x-2=0#

#e^x=4# #" or "# #e^x =2#

#x=ln4# #" or "# #x=ln2#

Substitute, then factor

Let #u = e^x#, then we want

#u^2-6u+8=0#

#(u-4)(u-2) = 0#

#u=4# #" or "# #u=2# Recall #u=e^x#, so

#e^x=4# #" or "# #e^x =2#

#x=ln4# #" or "# #x=ln2#