How do you solve these exponential equations?

#3+2(3^(2x+3))=57#

#e^(2x)-13e^(x)+42=0#

1 Answer
Gió
Jan 6, 2017

I tried this:

Explanation:

1) let us manipulate our expression to isolate the exponential:
#3^(2x+3)=(57-3)/2#
#3^(2x+3)=27#
We see that #27=3^3#, so:
#3^(2x+3)=3^3#
The exponentials will be equal if the exponents are the same, or:
#2x+3=3#
#2x=3-3#
#x=0#

2) let us use a trick and write instead of #e^x# the new incognita #t#, so our equation becomes:
#t^2-13t+42=0#
That is a quadratic equation that we solve using the Quadratic Formula to get:
#t_(1,2)=(13+-sqrt(169-168))/2=(13+-1)/2#
Giving:
#t=14/2=7#
#t_2=12/2=6#
BUT
We said previously that #t=e^x#
So:

For #t=7# then #e^x=7#
And so (using the definition of natural log):
#x=ln(7)#

For #t=6# then #e^x=6#
And so (using the definition of natural log):
#x=ln(6)#

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