Q: Starting from the definition of $\displaystyle \mathrm{cosh}x$ in terms of $\displaystyle e^x$, find, in terms of natural logarithms, the value for $\displaystyle x$ for which $\displaystyle 5 = 3 \mathrm{cosh}x$.

I did:

$\displaystyle 5 = 3 \mathrm{cosh}x$

$\displaystyle 5 = 3 \left( \frac{e^x + e^{-x}}{2} \right)$

$\displaystyle 10 = 3e^x + 3e^{-x}$

$\displaystyle 0 = 3e^x + e^{-x} - 10$

$\displaystyle 0 = 3e^{2x} - 10e^x + 1$

However, that does not factorise. The answer is $\displaystyle \pm \ln 3$. Can someone do this question to see if they can get the answer. If possible, please show me your steps. Thanks in advance. (Smile)