## baldymcgee6 Group Title Solve for x: e^x + e^-x = 2.5 one year ago one year ago

1. lgbasallote Group Title

are you allowed to use trigonometry?

2. baldymcgee6 Group Title

it was a cosh function so yes

3. lgbasallote Group Title

is there no such thing as arc cosh?

4. baldymcgee6 Group Title

If there is I have not been taught it yet

5. baldymcgee6 Group Title

i'm sure there is though

6. baldymcgee6 Group Title

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

7. AccessDenied Group Title

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

8. lgbasallote Group Title

"progress"

9. baldymcgee6 Group Title

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

10. LolWolf Group Title

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

11. LolWolf Group Title

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

12. baldymcgee6 Group Title

where did the ln come from?

13. LolWolf Group Title

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

14. AccessDenied Group Title

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.

15. baldymcgee6 Group Title

k thanks

16. AccessDenied Group Title

You're welcome! :)

17. Hero Group Title

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