Solve algebraically: $e^x =e^{-x}$

Question

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.

Answer 1

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 $$e^x=1 \quad\text{or}\qquad e^x=-1.$$

Answer 2

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

Answer 3

Let $x=iy$. Then, $\cos(y)+i\sin(y)=\cos(y)-i\sin(y)$.

$\sin(y)=0$.

$y=n\pi, n=0,\pm1,\pm2, ...$

$x=i(n\pi)$

$x=0$ is only one of the many solutions.

Answer 4

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 $z=0$.

Answer 5

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

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

$$x = -x$$

$$x=0$$

Come on man.