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

##
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

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

##### 1 Answer

Nov 26, 2016

#### Explanation:

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

By properties of exponents, we have:

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

We let

#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

#xlne = ln10 and xlne = ln-2#

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

Hopefully this helps!