Logarithms with same base, do signs change when you drop the base and solve?

Question

$$e^{x^2}=e^{18x} \cdot 1/e^{80}$$

I dropped the e since it was the same base and solved:

$$x^2=18x-80$$

I then moved $ 18x$ and $80$ to the other side and got: $$x^2-18x+80=0$$

The roots are $ 8 $ and $ 10 $ but its not possible with the signs I got. Where did I go wrong with the signs?

Thank you

Sorry, this is the final form of what I meant. I am new here

Answer 1

$$e^{x^2}=e^{18x} \cdot 1/e^{80} \Rightarrow x^2=18x+(-80) \Rightarrow x^2-18x+80=0$$ what give us $x=10$ or $x=8$

Answer 2

Okey, there are a couple of mistakes on your procedure: $$e^{18x}/e^{80}=e^{(18x-80)}$$, so, when you "drop the $e$", that is to say, when you take the logarhythm of both sides, you have: $$x^2=18x-80$$. Then you only have to be carefull with your rearrangement of the equation.