Question

Solve for x:
e^x + e^-x = 2.5

Answer 1

are you allowed to use trigonometry?

Answer 2

it was a cosh function so yes

Answer 3

is there no such thing as arc cosh?

Answer 4

If there is I have not been taught it yet

Answer 5

i'm sure there is though

Answer 6

I found the arc cosh function on my calculator, but how do I do this by hand?

Answer 7

I think if we make it into a quadratic equation with \(e^x\), you could make some progress here.

Answer 8

"progress"

Answer 9

so if we multiply everything by e^x to make it a quadratic

Answer 10

Well, we know the definition of \(\cosh x\) is:
\[
\cosh x=\frac{e^x+e^{-x}}{2}
\]So:
\[
2\cosh x=2.5\implies\\
x=\pm\ln(2)
\]

Answer 11

(Where \(x\in\mathbb{R}\))

Answer 12

where did the ln come from?

Answer 13

We have:
\[
\cosh^{-1}z=\ln\left(z+\sqrt{(z+1)(z-1)}\right)
\]So:
\[
\cosh^{-1}1.25=\ln\left(1.25+\sqrt{(2.25)(.25)}\right)=\ln(2)
\]

Answer 14

Yeah, for that quadratic method, move everything to left side and multiply by \(e^x\):
\((e^x)^2 - 2.5 e^x + 1 = 0\)
You can use quadratic formula:
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
\[ e^x = \frac{2.5 \pm \sqrt{(2.5)^2 - 4}}{2} \\
e^x = \frac{2.5 \pm \sqrt{2.25}}{2} \\
e^x = \frac{2.5 \pm 1.5}{2} \\

e^x = \frac{2.5 + 1.5}{2} \text{ or } e^x = \frac{2.5 - 1.5}{2} \\
e^x = 2 \text{ or } e^x = \frac{1}{2} \]
Then we can just take natural logarithms for the same as above answer.

Answer 15

k thanks

Answer 16

You're welcome! :)

Answer 17

I was wondering if something like that could be done. Now I have my answer.