How do you solve these exponential equations?

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

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

1 Answer
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)