Calculation with Hyperbolic Cosine

Question

Could you please check my work?

$\cosh \left(\ln \sqrt{5}\right) =\ ?$

\begin{align*}\cosh(x) &= \frac{e^x + e^{-x}}{2} \\ \\ \frac{e^{\ln \sqrt{5}} + e^{-\ln \sqrt{5}}}{2} &= \frac{\sqrt{5} + \frac{1}{\sqrt{5}}}{2}\\ &=\frac{\sqrt{5}}{2}+\frac{1}{2\sqrt{5}} \\ &= \frac{3}{\sqrt{5}} \end{align*}

Thanks.

Answer 1

Umm...this seems like it has an error to me, unless I misread something. (There appear to be some LaTeX problems, so it's entirely possible I've misread.)

But

\begin{align} \cosh \left( \ln \sqrt{5} \right) & = \frac{1}{2} \left( e^{\ln \sqrt{5}}+e^{-\ln \sqrt{5}} \right) \\ & = \frac{1}{2} \left( \sqrt{5} + \frac{1}{\sqrt{5}} \right) \\ & = \frac{1}{2} \left( \frac{5+1}{\sqrt{5}} \right) \\ & = \frac{3}{\sqrt{5}} = \frac{3\sqrt{5}}{5} \end{align}

Answer 2

That's correct, no errors.

As a note, its well worth learning how to present questions well (like using the LaTex capabilities of this website, as all the editors have been doing). It makes it much easier for viewers to read and understand your question. It also encourages you to think about what's important in your question, as this determines a lot about how you present it.