How do I find a solution to #e^(2x) = 8e^(x) + 20# ?

Question

I've worked the equation to get to e^(x) = 10 and e^(x) = -2, but I don't understand how to advance it further

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Answer 1 · 2016-11-26T22:18:16

#x= ln10#

#e^(2x) - 8e^x - 20 = 0#

By properties of exponents, we have:

#(e^x)^2 - 8e^x - 20 = 0#

We let #t = e^x#.

#t^2 - 8t - 20 = 0#

#(t - 10)(t + 2) = 0#

#t= 10 and -2#

#e^x = 10 and e^x = -2#

Take the natural logarithm of both sides.

#ln(e^x) = ln10 and lne^x = ln-2#

Use the rule #loga^n= nloga#:

#xlne = ln10 and xlne = ln-2#

#ln# and #e# are opposites, so equal #1#.

#x = ln10# and #x = O/#, since #y= lnx# has a domain of #x > 0#.

Hopefully this helps!