一道初一数学题

$2^a=3$，$3^b=2$，求 $\dfrac1{a+1}+\dfrac1{b+1}$ 的值 .

Solve

将 $2^a=3$ 带入 $3^b=2$ 得 $2^{ab}=2$，即 $ab=1$ .

于是

$$\begin{aligned}\dfrac1{a+1}+\dfrac1{b+1}&=\dfrac{a+b+2}{ab+a+b+1}\\&=\dfrac{a+b+2}{a+b+2}\\&=1\end{aligned}$$

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$2^a=3$ 故 $2^{a+1}=6$；$3^b=2$ 故 $3^{b+1}=6$ .

于是 $2=(2^{a+1})^{\frac1{a+1}}=6^{\frac1{a+1}}$ .

同理可得 $6^{\frac1{b+1}}=3$ .

于是

$$\begin{aligned}6^{\frac1{a+1}+\frac1{b+1}}&=6^{\frac1{b+1}}\cdot 6^{\frac1{a+1}}\\&=2\times 3\\&= 6\end{aligned}$$

也就是 $\dfrac1{a+1}+\dfrac1{b+1}=1$ .

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后面是非初一做法：

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UPD1. APJ 的换底

$2^a=3\Longrightarrow a=\log_23$，$3^b=2\Longrightarrow b=\log_32$ .

$$\begin{aligned} &\frac{1}{a+1}+\frac{1}{b+1}\\ =&\frac{1}{\log_23+1}+\frac{1}{\log_32+1}\\ =&\frac{1}{\frac{\ln3}{\ln2}+1}+\frac{1}{\frac{\ln2}{\ln3}+1}\\ =&\frac{\ln2}{\ln2+\ln3}+\frac{\ln3}{\ln3+\ln2}\\ =&1 \end{aligned}$$

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UPD2. Python