2013 AMC 12A Problems/Problem 14

Revision as of 23:04, 6 February 2013 by Indianninja707 (talk | contribs) (Created page with "Since the sequence is arithmetic, <math>\log_{12}{162}</math> + 4d = <math>\log_{12}{1250}</math>, where d is the common difference. Therefore, 4d = <math>\log_{12}{1250}</m...")
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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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