题目

Problem

Solve the equation

4\tanh x-\operatorname{sech}x=1

giving your answer in the form $x=\ln k$ where $k$ is a fully simplified rational number.

(6)

题目中文翻译

求解方程

4\tanh x-\operatorname{sech}x=1

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

解答

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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。

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Using

\begin{align*} \tanh x =&\,\frac{e^{2x}-1}{e^{2x}+1}, \\ \operatorname{sech}x =&\,\frac{2e^x}{e^{2x}+1}. \end{align*}

the equation becomes

\begin{align*} &\,4\biggl(\frac{e^{2x}-1}{e^{2x}+1}\biggr) \\ &\,\hspace{2pt}-\frac{2e^x}{e^{2x}+1}=1. \end{align*}

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

\begin{align*} 4(t^2-1)-2t =&\,t^2+1, \\ 3t^2-2t-5 =&\,0, \\ (3t-5)(t+1) =&\,0. \end{align*}

Thus

t=\frac53 \quad\text{or}\quad t=-1.

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

e^x=\frac53,

and therefore

\boxed{x=\ln\biggl(\frac53\biggr)}.

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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。

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

4\sinh x-1=\cosh x.

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

\begin{align*} (4\sinh x-1)^2 =&\,\cosh^2x, \\ 16\sinh^2x-8\sinh x+1 =&\,1+\sinh^2x, \\ 15\sinh^2x-8\sinh x =&\,0, \\ \sinh x(15\sinh x-8) =&\,0. \end{align*}

Hence

\sinh x=0 \quad\text{or}\quad \sinh x=\frac8{15}.

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

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

\begin{align*} x =&\,\operatorname{arsinh}\biggl(\frac8{15}\biggr) \\ =&\,\ln\biggl( \frac8{15} +\sqrt{1+\frac{64}{225}} \biggr) \\ =&\,\ln\biggl(\frac8{15}+\frac{17}{15}\biggr) \\ =&\,\ln\biggl(\frac53\biggr). \end{align*}

Moreover,

4\sinh x-1 =4\cdot\frac8{15}-1 =\frac{17}{15} =\cosh x,

so this value satisfies the equation before squaring. Therefore,

\boxed{x=\ln\biggl(\frac53\biggr)}.