2013 AMC 12A Problem 14

Attempt Problem 14 of the 2013 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2013 AMC 12A solutions, or check the answer key.

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14.

The sequence log⁡12162, log⁡12x, log⁡12y, log⁡12z, log⁡121250 \begin{gathered} \log_{12} 162, \ \log_{12} x, \ \log_{12} y, \\ \ \log_{12} z, \ \log_{12} 1250 \end{gathered} is an arithmetic progression. What is x?x?
1253125\sqrt{3}
270270
1625162\sqrt{5}
434434
2256225\sqrt{6}
Answer: B
Concepts:logarithmgeometric sequencearithmetic sequence
Small hint:
Equal differences of logs mean 162,x,y,z,1250162, x, y, z, 1250 is a geometric sequence
Big hint:
Find the common ratio rr from 162r4=1250162r^4 = 1250
Solution:
Because the logarithms are in arithmetic progression, 162,x,y,z,1250162, x, y, z, 1250 is a geometric sequence. Its common ratio rr satisfies 162r4=1250,162 r^4 = 1250, so r4=62581r^4 = \tfrac{625}{81} and r=53.r = \tfrac53.
Therefore x=162⋅53=270.x = 162\cdot\tfrac53 = 270.
Thus, the correct answer is B.
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