Can someone help me solve this please!?

Find all real values of x that satisfy the following equation:

e^ax = -e^-ax

Can you please show me the steps for this and determine the value of x? Thanks a lot! I don't think there IS a real solution.

Let Q = e^ax
the equation then becomes
Q = -1/Q
Q^2 = -1
Q = +/- i

OR
Multiply both sides by e^ax and you get
e^2ax = -e^0 = -1
there is no a that will do this; e^x is always positive for real x -e^-ax equals 1/-e^ax, so rewrite the question as:
e^ax=1/-e^ax
multiply by -e^ax:
e^ax(-e^ax)=0
either e^ax=0 or -e^ax=0
so, e=0