# Solve for x? e^(2x)-5=ln(x)+7 Please and thank you!!

May 9, 2018

${x}_{1} \approx 0$, ${x}_{2} \approx 1.25$

#### Explanation:

I would solve this graphically, as algerbraically seems rather advanced. I would draw two graphs,
$f = {e}^{2 x} - 5$, i.e. the red graph on the figure
$g = \ln \left(x\right) + 7$, i.e. the blue graph on the figure

the values of x that solve the equation, have to be on both graphs, i.e. must be where the graphs cross each other.

We find that there are two solutions, A=(0, -4) and E=(1.25, 7.22)

Now, ln(0) is undefined, but we see that
$f \left(x\right) = \ln \left(x\right) + 7 = - 4$ so $\ln \left(x\right) = - 11$ or x=e^-11=1.67017*10^-5  or basically 0. We might conclude from this that $x = {e}^{-} 11$, but it is based on the approximation that f(x)=-4 for this value of x, which would not be exact.

x=0 inserted in g:
$g \left(0\right) = {e}^{2 \cdot 0} - 5 = 1 - 5 = 4$

The other value is $x \approx 1.25$ based on point E on the graph. Also this would be an approximation.