Question

Find the values of $x$ for which
$\cos hx=\frac{13}{5}$ expressing your answers as natural logarithms.

Answer 1

We know that;

$coshx=\frac{1}{2}(e^x+e^{-x})$

$\to$$\frac{1}{2}(e^x+e^{-x})=\frac{13}{5}$

$\to$$5(e^{2x}+1)=26e^x$

$\to$$5e^{2x}-26e^x+5=0$

$\to$$e^x=25or{e}^x=1$

.

$x=ln25$ or $x=ln1=0$