3.

求解方程

并将答案写成 $x=\ln k$ 的形式，其中 $k$ 为完全化简的有理数。

把 $\tanh x$ 与 $\operatorname{sech}x$ 直接写成含 $e^x$ 的分式。令 $t=e^x$，由于 $t>0$，所得二次方程中的负根可立即排除，最后对正根取自然对数。

Using

the equation becomes

Let $t=e^x$, where $t>0$. Multiplying by $t^2+1$ gives

Thus

Since $t=e^x>0$, the root $t=-1$ is rejected. Hence

and therefore

先乘以恒正的 $\cosh x$，把方程化为 $4\sinh x-1=\cosh x$，再平方并使用 $\cosh^2x-\sinh^2x=1$。由于平方可能产生增根，求得候选值后必须逐一回到平方前的方程检验。

Since $\cosh x>0$, multiplying the original equation by $\cosh x$ gives

Squaring both sides and using $\cosh^2x=1+\sinh^2x$,

Hence

If $\sinh x=0$, then $x=0$, but the original left-hand side is $-1$, so this is an extraneous solution.

For $\sinh x=\frac8{15}$,

Moreover,

so this value satisfies the equation before squaring. Therefore,

先把方程写成 $4\tanh x-1=\operatorname{sech}x$，平方后用 $\operatorname{sech}^2x=1-\tanh^2x$。求出的候选值须回到平方前检验，最后用反双曲正切的对数形式求 $x$。

Let $t=\tanh x$. The equation is

Since $\operatorname{sech}x>0$, we require $t>\frac14$. Squaring and using $\operatorname{sech}^2x=1-t^2$,

Thus $t=0$ or $t=\frac8{17}$. The first fails $t>\frac14$. For $t=\frac8{17}$,

so the equation before squaring is satisfied. Hence