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!