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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

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    $\begingroup$ First, you should use MathJax. See meta.math.stackexchange.com/questions/5020/… for a summary. In your final paragraph, you say $x=-x$ is not a true statement. But if $x=0$, it is true. $\endgroup$
    – rogerl
    Jul 14, 2014 at 21:09
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    $\begingroup$ Your fist approach leads to $2x=\ln{1}=0$ then $x=0$. Your second approach leads to $x=-x$ then $x=0$ $\endgroup$
    – rlartiga
    Jul 14, 2014 at 21:11
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    $\begingroup$ Not to be a snob about it, but in mathematics, $\log x$ refers to the natural logarithm, and the common logarithm is then $\log_{10} x$. $\endgroup$
    – James47
    Jul 14, 2014 at 21:14
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    $\begingroup$ @James47: Any mathematician would understand $\ln(x)$ to be the natural logarithm of $x$. On the other hand, using $\log(x)$ can sometimes introduce ambiguity. $\endgroup$
    – Deathkamp Drone
    Jul 14, 2014 at 21:22
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    $\begingroup$ In fact, $e^x-e^{-x}$ is precisely $2\sinh x$. So you are really solving $\sinh x = 0$. The solutions are the complex numbers $i\pi k$ for integral $k$; the only real-valued solution among these, if that's what you need, is $0$. $\endgroup$
    – MPW
    Jul 14, 2014 at 22:19

6 Answers 6

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If $$x = -x$$

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

Because

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

$$ 2x = 0 $$

$$ x = 0 $$

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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.

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    $\begingroup$ Interesting. I didn't think of that. Thanks! $\endgroup$
    – Astro
    Jul 15, 2014 at 14:15
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Both are correct. $x=-x\Leftrightarrow 2x=0\Leftrightarrow x=0$.

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  • $\begingroup$ Simple and to the point. Thank you $\endgroup$
    – Astro
    Jul 15, 2014 at 14:15
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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 $k=0$).

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    $\begingroup$ I think the phrasing of the question makes it clear that the OP is asking from the perspective of real-valued functions only. This is swatting a fly with a trebuchet. $\endgroup$
    – mweiss
    Jul 15, 2014 at 2:59
  • $\begingroup$ Yeah mweiss is right. I didn't ask for that because I didn't know such a thing existed. Thanks for the extra info though. It's to be exposed to different things. $\endgroup$
    – Astro
    Jul 15, 2014 at 14:16
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    $\begingroup$ @mweiss, "swatting a fly with a trebuchet" I need to remember this phrase for use in the future. $\endgroup$
    – Brian S
    Jul 15, 2014 at 15:02
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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.

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  • $\begingroup$ I forgot taking the natural log of a negative is a no-no. Thanks for pointing that out $\endgroup$
    – Astro
    Jul 15, 2014 at 14:17
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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. $$\ln(e^x)=\ln(e^{-x})$$ $$x=-x$$ $$x=0$$

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