How do you solve these exponential equations?

$3 + 2 \left({3}^{2 x + 3}\right) = 57$ ${e}^{2 x} - 13 {e}^{x} + 42 = 0$

Jan 6, 2017

I tried this:

Explanation:

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

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

For $t = 7$ then ${e}^{x} = 7$
And so (using the definition of natural log):
$x = \ln \left(7\right)$

For $t = 6$ then ${e}^{x} = 6$
And so (using the definition of natural log):
$x = \ln \left(6\right)$