1961 AMC 12 Problem 30

Attempt Problem 30 of the 1961 AMC 12 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 1961 AMC 12 solutions, or check the answer key.

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30.

If log⁡102=a\log_{10}2=a and log⁡103=b,\log_{10}3=b, then log⁡512\log_5 12 equals:
a+ba+1\dfrac{a+b}{a+1}
2a+ba+1\dfrac{2a+b}{a+1}
a+2b1+a\dfrac{a+2b}{1+a}
2a+b1−a\dfrac{2a+b}{1-a}
a+2b1−a\dfrac{a+2b}{1-a}
Answer: D
Concepts:logarithmalgebraic manipulation
Small hint:
Use the change-of-base formula with common logarithms
Big hint:
Write 12=22⋅312=2^2\cdot3 and 5=1025=\frac{10}{2}
Solution:
By change of base, log⁡512=log⁡12log⁡5=2log⁡2+log⁡3log⁡10−log⁡2=2a+b1−a. \begin{aligned} \log_5 12 &=\frac{\log12}{\log5}\\ &=\frac{2\log2+\log3} {\log10-\log2}\\ &=\frac{2a+b}{1-a}. \end{aligned}
Therefore, the correct answer is D.
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