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lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
are you allowed to use trigonometry?
 2 years ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
it was a cosh function so yes
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
is there no such thing as arc cosh?
 2 years ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
If there is I have not been taught it yet
 2 years ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
i'm sure there is though
 2 years ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
I found the arc cosh function on my calculator, but how do I do this by hand?
 2 years ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.3
I think if we make it into a quadratic equation with \(e^x\), you could make some progress here.
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
"progress"
 2 years ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
so if we multiply everything by e^x to make it a quadratic
 2 years ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
Well, we know the definition of \(\cosh x\) is: \[ \cosh x=\frac{e^x+e^{x}}{2} \]So: \[ 2\cosh x=2.5\implies\\ x=\pm\ln(2) \]
 2 years ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
(Where \(x\in\mathbb{R}\))
 2 years ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
where did the ln come from?
 2 years ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
We have: \[ \cosh^{1}z=\ln\left(z+\sqrt{(z+1)(z1)}\right) \]So: \[ \cosh^{1}1.25=\ln\left(1.25+\sqrt{(2.25)(.25)}\right)=\ln(2) \]
 2 years ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.3
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.
 2 years ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
k thanks
 2 years ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.3
You're welcome! :)
 2 years ago

Hero Group TitleBest ResponseYou've already chosen the best response.0
I was wondering if something like that could be done. Now I have my answer.
 2 years ago
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