Solving base e equation $e^x - e^{-x} = 0$

Question

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

Answer 1

If $$x = -x$$

Then we can say that $$x = 0$$

Because

$$ x + x = -x + x $$

$$ 2x = 0 $$

$$ x = 0 $$

Answer 2

Both are correct. $x=-x\Leftrightarrow 2x=0\Leftrightarrow x=0$.

Answer 3

You are solving for $x$ when $e^x=e^{-x}$. It is not the case for all $x$ that $e^x=e^{-x}$; therefore, the statement $x=-x$ will not be true for all $x$, just the $x$ you are solving for.

Answer 4

When you take logs of both sides, you don't get $x=-x$. You get $x = -x +2\pi ik$ for integral $k$. This means $2x = 2\pi ik$, so $x = \pi ik$.

The only real-valued solution, then, is $x=0$ (by taking $k=0$).

Answer 5

and then take the natural logarithm of both sides to get $x=−x$, which is not a true statement

And right there is your mistake.

$x=-x $ is a true statement... but only for $x=0$. Just add $x$ to both sides.

Answer 6

Start off with $$e^x-e^{-x}=0$$ Add $e^{-x}$ to both sides of the equation. The equation becomes $$e^x=e^{-x}$$ Take the natural logarithm of both sides. $$\ln(e^x)=\ln(e^{-x})$$ $$x=-x$$ $$x=0$$