$\tanh x = \dfrac{(e^x-e^{-x})}{(e^x+e^{-x})}$ Let $a=e^x $ $b=\dfrac {1}{e^x }$ $\therefore \tanh x = \dfrac{a-b}{a+b}$ $\dfrac23 = \dfrac{a-b}{a+b}$ $2a+2b=3a-3b $ $a=5b $ $\dfrac ab=e^{2x}$ $2x=\log(a/b)$ $\therefore x=0.5\log(5)$ $\therefore \cosh 2x=\cosh(\log5)$ $\therefore \cosh 2x=2.6$