So I ran into some confusion while doing this problem, and I won't bore you with the details, but it comes down to trying to solve $e^x – e^{-x} = 0$.
I know to solve it, we can rewrite it as $e^x – \frac{1}{e^x} = 0$ and then get LCD so form $\frac{e^{2x} – 1}{e^x} = 0$ and then rewrite it as $e^{2x} = 1$, take the natural logarithm of both sides and it becomes $2x = \log(1)$ or when $x = 0$ (if anything up there is wrong, please tell me)
My problem is when I try to do an alternative. Starting with $e^x – e^{-x} = 0$, I try to add to both sides to get $e^x = e^{-x}$, and then take the natural logarithm of both sides to get $x = -x$, which is not a true statement. Could someone explain to me what I'm doing wrong, please?
Thanks in advance
Best Answer
If $$x = -x$$
Then we can say that $$x = 0$$
Because
$$ x + x = -x + x $$
$$ 2x = 0 $$
$$ x = 0 $$