2013 AMC 12A Problems/Problem 14

Revision as of 23:47, 6 February 2013 by Indianninja707 (talk | contribs)

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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