2013 AMC 12A Problems/Problem 14

Since the sequence is arithmetic,

$\log_{12}{162}$ + 4d = $\log_{12}{1250}$ , where d is the common difference.

Therefore,

4d = $\log_{12}{1250}$ - $\log_{12}{162}$ = $\log_{12}{(1250/162)}$ , and

d = (1/4)( $\log_{12}{(1250/162)}$ ) = $\log_{12}{(1250/162)^{1/4}}$

Now that we found d, we just add it to the first term to find x:

$\log_{12}{162}$ + $\log_{12}{(1250/162)^{1/4}}$ = $\log_{12}{((162)(1250/162)^{1/4})}$

x = (162) $(1250/162)^{1/4}$ = (162) $(625/81)^{1/4}$ = (162)(5/3) = 270, which is B

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