AMC12 2021 B

AMC12 2021 B · Q9

AMC12 2021 B · Q9. It mainly tests Logarithms (rare), Manipulating equations.

What is the value of\[\frac{\log_2 80}{\log_{40}2}-\frac{\log_2 160}{\log_{20}2}?\]

求\[\frac{\log_2 80}{\log_{40}2}-\frac{\log_2 160}{\log_{20}2}\]的值。

(A) 0 0

(B) 1 1

(D) 2 2

(E) \log_2 5 \log_2 5

Answer

Correct choice: (D)

正确答案：（D）

Solution

\[\frac{\log_{2}{80}}{\log_{40}{2}}-\frac{\log_{2}{160}}{\log_{20}{2}}\] Note that $\log_{40}{2}=\frac{1}{\log_{2}{40}}$, and similarly $\log_{20}{2}=\frac{1}{\log_{2}{20}}$ \[= \log_{2}{80}\cdot \log_{2}{40}-\log_{2}{160}\cdot \log_{2}{20}\] \[=(\log_{2}{4}+\log_{2}{20})(\log_{2}{2}+\log_{2}{20})-(\log_{2}{8}+\log_{2}{20})\log_{2}{20}\] \[=(2+\log_{2}{20})(1+\log_{2}{20})-(3+\log_{2}{20})\log_{2}{20}\] Expanding, \[2+2\log_{2}{20}+\log_{2}{20}+(\log_{2}{20})^2-3\log_{2}{20}-(\log_{2}{20})^2\] All the log terms cancel, so the answer is $2\implies\boxed{\text{(D)}}$.

\[\frac{\log_{2}{80}}{\log_{40}{2}}-\frac{\log_{2}{160}}{\log_{20}{2}}\] 注意到$\log_{40}{2}=\frac{1}{\log_{2}{40}}$，类似地$\log_{20}{2}=\frac{1}{\log_{2}{20}}$ \[= \log_{2}{80}\cdot \log_{2}{40}-\log_{2}{160}\cdot \log_{2}{20}\] \[=( \log_{2}{4}+\log_{2}{20})( \log_{2}{2}+\log_{2}{20})-( \log_{2}{8}+\log_{2}{20})\log_{2}{20}\] \[=(2+\log_{2}{20})(1+\log_{2}{20})-(3+\log_{2}{20})\log_{2}{20}\] 展开后，$\[2+2\log_{2}{20}+\log_{2}{20}+(\log_{2}{20})^2-3\log_{2}{20}-(\log_{2}{20})^2\]$ 所有对数项抵消，答案为$2\implies\boxed{\text{(D)}}$。

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