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