Here's the question you clicked on:
baldymcgee6
Solve for x: e^x + e^-x = 2.5
are you allowed to use trigonometry?
it was a cosh function so yes
is there no such thing as arc cosh?
If there is I have not been taught it yet
i'm sure there is though
I found the arc cosh function on my calculator, but how do I do this by hand?
I think if we make it into a quadratic equation with \(e^x\), you could make some progress here.
so if we multiply everything by e^x to make it a quadratic
Well, we know the definition of \(\cosh x\) is: \[ \cosh x=\frac{e^x+e^{-x}}{2} \]So: \[ 2\cosh x=2.5\implies\\ x=\pm\ln(2) \]
(Where \(x\in\mathbb{R}\))
where did the ln come from?
We have: \[ \cosh^{-1}z=\ln\left(z+\sqrt{(z+1)(z-1)}\right) \]So: \[ \cosh^{-1}1.25=\ln\left(1.25+\sqrt{(2.25)(.25)}\right)=\ln(2) \]
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.
You're welcome! :)
I was wondering if something like that could be done. Now I have my answer.