Question

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

Answer 1

`x_1~~0`, `x_2~~1.25`

Explanation:

I would solve this graphically, as algerbraically seems rather advanced. I would draw two graphs,
`f=e^(2x)-5`, i.e. the red graph on the figure
`g=ln(x)+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(x)=ln(x)+7=-4` so `ln(x)=-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(0)=e^(2*0)-5=1-5=4`

The other value is `x~~1.25` based on point E on the graph. Also this would be an approximation.