SOLUTION: Can you please help me solve this? I think I got the answer, but I am not sure. e^(x^2)=e^(x) * e^(6) I multiplied (e^x and e^6, which meant adding the exponents) e^(x^2)

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Can you please help me solve this? I think I got the answer, but I am not sure. e^(x^2)=e^(x) * e^(6) I multiplied (e^x and e^6, which meant adding the exponents) e^(x^2)      Log On


   

Question 923463: Can you please help me solve this? I think I got the answer, but I am not sure.
e^(x^2)=e^(x) * e^(6)
I multiplied (e^x and e^6, which meant adding the exponents)
e^(x^2)= e^(x+6)
Then got rid of the base e on both sides
x^2=x+6
(-x^2)+x+6=0
x=-2 x=3
Found 2 solutions by ewatrrr, josgarithmetic:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
Yes. Good Work.

Answer by josgarithmetic(39621) About Me  (Show Source):
You can put this solution on YOUR website!
The solution process and answer result are good.

A couple of formal differences could be made in place.

You would "take the log_base_e of both sides" at e%5E%28x%5E2%29=+e%5E%28x%2B6%29
ln%28e%5E%28x%5E2%29%29=ln%28e%5E%28x%2B6%29%29
%28x%5E2%29ln%28e%29=%28x%2B6%29%2Aln%28e%29
x%5E2=x%2B6

The next step will give a more convenient form to work with,
x%5E2=x%2B6
x%5E2-x-6=x%2B6-x-6
x%5E2-x-6=0

Easy enough to factor,
%28x%2B2%29%28x-3%29=0
Which give the solutions you found.