Question

how do i solve for x? *you have to show work*
1.) 2e^2x + 4xe^2x = 0
2.) 2 + 3(4)^(2x-1) = 43

Answer 1

1) Factor the equation first:

$2e^{2x} + 4xe^{2x} = 0$
$e^{2x}(2 + 4x) = 0$

So either $e^{2x} = 0$ or $2 + 4x = 0$. But $e^{2x} = 0$ can never equal 0! So, we only need to solve for the second equation:

$2 + 4x = 0$
$\bf x = {-\frac{1}{2}}$

2) First, single out the term with the exponent:

$2 + 3(4^{2x - 1}) = 43$
$3(4^{2x - 1}) = 41$
$4^{2x - 1} = \frac{41}{3}$

Now, taking the log base 4 of each side gives us

$2x - 1 = \log_4 \frac{41}{3}$

and we can now solve for x:

$2x - 1 = \log_4 \frac{41}{3}$
$2x = 1 + \log_4 \frac{41}{3}$
$2x = \log_4 4 + \log_4 \frac{41}{3}$
$2x = \log_4 \frac{164}{3}$
$x = \frac{1}{2} \log_4 \frac{164}{3}$
$x = \log_4 \sqrt \frac{164}{3}$
$\bf x = \log_4 2 \sqrt \frac{41}{3}$