How do you solve #-3e^(2x) + 16=5# and find any extraneous solutions?

1 Answer
Alan P.
Apr 12, 2017

#x=ln(11/3)/2#
#color(white)("XXX")#...maybe I missed something, but I did not find any extraneous solutions.

Explanation:

#-3e^(2x)+16=5#
#color(white)("XXX")rarr -3e^(2x)=-11#

#color(white)("XXX")rarr e^(2x)=11/3#

#color(white)("XXX")rarr2x=ln(11/3)#

#color(white)("XXX")rarr x=ln(11/3)/2 ~~0.64964# (I used a calculator for this approximation)

Checking this solution back into the original equation demonstrates that the solution is not extraneous. (Again, I used a calculator for this).

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