Why does $e^x+e^{-x}-4$ equal $e^x-e^{-x}-4$ on this problem?

Question

I'm asked to find the intercepting points of two equations $y=4-\frac{1}{2}(e^x+e^{-x})$ and $y=\frac{1}{2}(e^x+e^{-x})$.

Putting them together and opening the brackets I get $e^x+e^{-x}=4$.
(My textbook gives $e^x-e^{-x}=4$ as an answer to this step.)

I continue to substitute $t=e^x$ and turn my own equation into $t+1/t=4$, then I proceed to solve for $t$ through expanding the first term by $t$ and creating a perfect square $(t-2)^2=3$, which in turn gives me $t=2\pm\sqrt{3}$. Now I can substitute $e^x$ back into the equation and get the points $x_1=\ln(2-3^{1/2})$ and $x_2=\ln(2+3^{1/2})$. To my surprise these are the same points that are listed in my textbook, even though the textbook uses $e^x-e^{-x}-4$ instead of $e^x+e^{-x}-4$.

Answer 1

If you want to solve $$ 4-\frac{e^x+e^{-x}}{2}=\frac{e^x+e^{-x}}{2} $$ the equation is indeed $e^x+e^{-x}=4$ and your solution seems correct; with $e^x=t$, it becomes $$ t+\frac{1}{t}=4 $$ or $$ t^2-4t+1=0 $$ whose roots are $2+\sqrt{3}$ and $2-\sqrt{3}$, as you found.

There is no way the equation can be rewritten as $4=e^x-e^{-x}$. There seems to be a typo in your textbook.

Answer 2

Note that following the answer posed in the title we have

$$e^x+e^{-x}-4=e^x-e^{-x}-4\iff e^{-x}=-e^{-x}\iff e^{-x}=0$$

which is never true.

Since we are dealing with

$$4-\frac{1}{2}(e^x+e^{-x})=\frac{1}{2}(e^x+e^{-x}) \iff 8-e^x-e^{-x}=e^x+e^{-x}\\\iff e^x+e^{-x}=4\iff e^{2x}-4e^x+1=0$$

and by $t=e^x$ we obtain

$$t^2-4t+1=0\implies t=e^x=2\pm\sqrt{3}$$

Answer 3

First, as you also figured out, $$e^{x}+e^{-x} = 4 \implies x = \ln(2 \pm \sqrt{3}).$$ Second, considering $x \in \mathbb{R}$, $$e^{x}-e^{-x} = 4 \implies x = \ln(2 + \sqrt{5}).$$ So there might be some typing mistake in the book solution.