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

1 Answer
May 9, 2018

#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

enter image source here

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.