Q.Solve the equation $\cosh x = 2$ for $x$, giving the answer in logarithmic form.

The inverse hyperbolic tangent has the logarithmic form $\tanh^{-1}x = \dfrac12\log_e\dfrac{1+x}{1-x}$; substituting $x=\dfrac12$ gives the required value directly.

Concept. For $|x| < 1$, $\tanh^{-1}x = \dfrac12\log_e\left(\dfrac{1+x}{1-x}\right)$.

(Derivation: if $y = \tanh^{-1}x$, then $x = \tanh y = \dfrac{e^y-e^{-y}}{e^y+e^{-y}}$. Solving for $e^{2y}$ gives $e^{2y} = \dfrac{1+x}{1-x}$, so $2y = \log_e\dfrac{1+x}{1-x}$.)

Step 1. Substitute $x = \dfrac12$: …

Set $\sinh x=\frac{e^x-e^{-x}}{2}=3$, solve the quadratic $t^2-6t-1=0$ with $t=e^x$, and take the positive root.

By definition $\sinh x = \dfrac{e^x-e^{-x}}{2} = 3$, so $e^x-e^{-x}=6$.

Let $t=e^x>0$. Then $t-\dfrac{1}{t}=6 \Rightarrow t^2-6t-1=0$.

Solve: $t = \dfrac{6\pm\sqrt{36+4}}{2} = \dfrac{6\pm\sqrt{40}}{2} = 3\pm\sqrt{10}$.

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