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So I was doing this problem. And I got to what I thought was a weird result. can anyone explain to me why this makes sense?

$$\begin{align} 5e^x &= 5e^{-x} \\ e^x &= e^{-x} \\ \ln (e^x) &= \ln (e^{-x}) \\ x &= -x \end{align}$$

This last step is what I'm wondering about. Although this is algebraically correct isn't it impossible for a number to be equal to its opposite? Don't both sides have to be equal? Note that I am not asking for help on how to solve it from here...

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    $\begingroup$ $x = -x \implies x(1) = x(-1) \implies x = 0$ $\endgroup$ – Felipe Jacob Feb 16 '14 at 16:46
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    $\begingroup$ x=-x holds only if x=0. So this must be an exercise and solving it means to find for which x that expression is true. $\endgroup$ – Emin Feb 16 '14 at 16:50
  • $\begingroup$ oh, that makes sense... I was thinking thinking along the lines of how can -5 = 5, but if its set to 0 it makes sense... I was wondering if there was any way to find it other than guessing though. $\endgroup$ – nachos Feb 16 '14 at 16:52
  • $\begingroup$ Clearly, it holds only for x=0, and also you can do this without taking log, by just transferring, so it will give you e^2x=1=>x=0 $\endgroup$ – peeyush.cray Feb 16 '14 at 16:53
  • $\begingroup$ alright thanks, i get it now. its a true statement but it needs qualifiers. $\endgroup$ – nachos Feb 16 '14 at 16:56
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Combine like terms: $$ x = -x \overset{+x}{\implies} 2x = 0 \overset{÷2}\implies x =0 $$

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Note that $e^{-x}=\cfrac 1{e^x}$ so you can also solve this by setting $y=e^x$ so that $$5y=\frac 5y$$ so that $$y^2=1 \text { or } (y+1)(y-1)=0$$ whence $e^x=\pm 1$ and then solve for $x$. It gives the same solution, but two things to remember - here we multiply by $y$, which avoids any accidental division by zero. The answer $\pm 1$ also reminds us that when we took logs we had better be sure that we were dealing with positive numbers. In simple situations these don't cause a problem, but more complex situations need more care.

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  • $\begingroup$ Talking about things to remember, you are multiplying by $y$ which might introduce false solutions. (Of course in this case it doesn't, but if you are being a nitpick then be a proper nitpick.) $\endgroup$ – TMM Feb 16 '14 at 17:36
  • $\begingroup$ @TMM Thanks for that. Squaring is more dangerous still. If $Ay=By$ then $A=B$ or $y=0$ - but in many cases where we clear fractions it is obvious (as here) that $y$ is non-zero. Which doesn't completely excuse the fault ... $\endgroup$ – Mark Bennet Feb 16 '14 at 17:50
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But why so difficult: $5e^x=5e^{-x} \Leftrightarrow e^x=e^{-x} \Leftrightarrow e^{2x}=1 \Leftrightarrow 2x=0 \Leftrightarrow x=0$

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