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AP calc final tomorrow, this was part of my review, I have no idea how to solve it. I know the answer but not how to get the answer, which is really important.

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    $\begingroup$ Draw the graphs of $y=e^x$ and $y=e^{-x}$ and look for where they cross. $\endgroup$ – Barry Cipra Dec 15 '16 at 2:23
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    $\begingroup$ $\exp$ is injective so $x = -x$. What can you conclude? $\endgroup$ – MathematicsStudent1122 Dec 15 '16 at 2:25
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    $\begingroup$ natural log of both sides, also. $\endgroup$ – The Count Dec 15 '16 at 2:28
  • $\begingroup$ Might become more obvious if you look at it as $e^x=\cfrac{1}{e^x}\,$. $\endgroup$ – dxiv Dec 15 '16 at 4:08
  • $\begingroup$ You might be able to find similar questions from the past using Approach0. For example, this related question was among the first hits: Solving base e equation $e^x - e^{-x} = 0$ $\endgroup$ – Martin Sleziak Dec 15 '16 at 6:05
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Let $t=e^x$. Then $e^{-x}=1/e^x=1/t$, so you have $$t=\frac1t$$ which means $$t^2=1.$$ This means $t=\pm1$. So it only remains all possibilities for $x$ such that $$e^x=1 \quad\text{or}\qquad e^x=-1.$$

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hint: multiply by $e^x$ on both sides. Then you get a constant on one side. Can you solve it now?

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  • $\begingroup$ This should be a comment, in my opinion. $\endgroup$ – The Count Dec 15 '16 at 15:19
  • $\begingroup$ @TheCount Sorry for that, but I see people giving hints as answers all the time. I will try to comment next time. Thank you! $\endgroup$ – Junkai Dong Dec 16 '16 at 0:52
  • $\begingroup$ Well it is just my opinion. No big deal. $\endgroup$ – The Count Dec 16 '16 at 1:33
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Let x=iy. Then, cos(y)+isin(y)=cos(y)-isin(y).

sin(y)=0.

y=n*pi, n=0, (+/-)1, (+/-)2, ...

x=i(n*pi)

x=0 is only one of the many solutions.

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$$e^x = e^{-x}$$

$$\ln(e^x) = \ln(e^{-x})$$

$$x = -x$$

$$x=0$$

Come on man.

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A more general question: for which $z \in \mathbb C$ is $e^z=e^{-z}$ ?

Answer: first recall that the complex solutions of the equation $e^w=1$ are given by

$$w=2 k \pi i, \quad k \in \mathbb Z.$$

Hence $e^z=e^{-z}$ iff $e^{2z}=1$ iff $z= k \pi i$ for some $k \in \mathbb Z$.

For real $z$ it follows:

$e^z=e^{-z}$ iff $z=0$.

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