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

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

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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