\(2^a=3\)，\(3^b=2\)，求 \(\dfrac1{a+1}+\dfrac1{b+1}\) 的值 .
Solve
將 \(2^a=3\) 帶入 \(3^b=2\) 得 \(2^{ab}=2\)，即 \(ab=1\) .
於是
\[\begin{aligned}\dfrac1{a+1}+\dfrac1{b+1}&=\dfrac{a+b+2}{ab+a+b+1}\\&=\dfrac{a+b+2}{a+b+2}\\&=1\end{aligned} \]
\(2^a=3\) 故 \(2^{a+1}=6\)；\(3^b=2\) 故 \(3^{b+1}=6\) .
於是 \(2=(2^{a+1})^{\frac1{a+1}}=6^{\frac1{a+1}}\) .
同理可得 \(6^{\frac1{b+1}}=3\) .
於是
\[\begin{aligned}6^{\frac1{a+1}+\frac1{b+1}}&=6^{\frac1{b+1}}\cdot 6^{\frac1{a+1}}\\&=2\times 3\\&= 6\end{aligned} \]
也就是 \(\dfrac1{a+1}+\dfrac1{b+1}=1\) .
后面是非初一做法：
UPD1. APJ 的換底
\(2^a=3\Longrightarrow a=\log_23\)，\(3^b=2\Longrightarrow b=\log_32\) .
\[\begin{aligned} &\frac{1}{a+1}+\frac{1}{b+1}\\ =&\frac{1}{\log_23+1}+\frac{1}{\log_32+1}\\ =&\frac{1}{\frac{\ln3}{\ln2}+1}+\frac{1}{\frac{\ln2}{\ln3}+1}\\ =&\frac{\ln2}{\ln2+\ln3}+\frac{\ln3}{\ln3+\ln2}\\ =&1 \end{aligned} \]
UPD2. Python
