AMC12 2013 A

AMC12 2013 A · Q14

AMC12 2013 A · Q14. It mainly tests Logarithms (rare), Primes & prime factorization.

The sequence $ \log_{12} 162,\ \log_{12} x,\ \log_{12} y,\ \log_{12} z,\ \log_{12} 1250 $ is an arithmetic progression. What is $x$?

数列 $ \log_{12} 162,\ \log_{12} x,\ \log_{12} y,\ \log_{12} z,\ \log_{12} 1250 $ 是等差数列。求 $x$。

(A) 125\sqrt{3} 125\sqrt{3}

(B) 270 270

(D) 434 434

(E) 225\sqrt{6} 225\sqrt{6}

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Because the terms form an arithmetic sequence, $$\log_{12} y=\frac{1}{2}\left(\log_{12}162+\log_{12}1250\right)=\frac{1}{2}\log_{12}(162\cdot1250)$$ $$=\frac{1}{2}\log_{12}\left(2^2 3^4 5^4\right)=\log_{12}\left(2\cdot3^2 5^2\right).$$ Then $$\log_{12} x=\frac{1}{2}\left(\log_{12}162+\log_{12}y\right)=\frac{1}{2}\left(\log_{12}(2\cdot3^4)+\log_{12}(2\cdot3^2 5^2)\right)$$ $$=\frac{1}{2}\log_{12}\left(2^2 3^6 5^2\right)=\log_{12}\left(2\cdot3^3 5\right)=\log_{12}270.$$ Therefore $x=270$.

答案（B）：因为这些项构成等差数列， $$\log_{12} y=\frac{1}{2}\left(\log_{12}162+\log_{12}1250\right)=\frac{1}{2}\log_{12}(162\cdot1250)$$ $$=\frac{1}{2}\log_{12}\left(2^2 3^4 5^4\right)=\log_{12}\left(2\cdot3^2 5^2\right).$$ 然后 $$\log_{12} x=\frac{1}{2}\left(\log_{12}162+\log_{12}y\right)=\frac{1}{2}\left(\log_{12}(2\cdot3^4)+\log_{12}(2\cdot3^2 5^2)\right)$$ $$=\frac{1}{2}\log_{12}\left(2^2 3^6 5^2\right)=\log_{12}\left(2\cdot3^3 5\right)=\log_{12}270.$$ 因此 $x=270$。

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