Difference between revisions of "2013 AMC 12A Problems/Problem 14"

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<math>4d</math> = <math>\log_{12}{1250}</math> - <math>\log_{12}{162}</math> = <math>\log_{12}{(1250/162)}</math>, and
 
<math>4d</math> = <math>\log_{12}{1250}</math> - <math>\log_{12}{162}</math> = <math>\log_{12}{(1250/162)}</math>, and
  
<math>d</math> = (1/4)(<math>\log_{12}{(1250/162)}</math>) = <math>\log_{12}{(1250/162)^{1/4}}</math>
+
<math>d</math> = <math>\frac{1}{4}</math>(<math>\log_{12}{(1250/162)}</math>) = <math>\log_{12}{(1250/162)^{1/4}}</math>
  
  
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<math>\log_{12}{162}</math> + <math>\log_{12}{(1250/162)^{1/4}}</math> = <math>\log_{12}{((162)(1250/162)^{1/4})}</math>
 
<math>\log_{12}{162}</math> + <math>\log_{12}{(1250/162)^{1/4}}</math> = <math>\log_{12}{((162)(1250/162)^{1/4})}</math>
  
<math>x</math> = (162)<math>(1250/162)^{1/4}</math> = (162)<math>(625/81)^{1/4}</math> = <math>(162)(5/3)</math> = <math>270</math>, which is <math>B</math>
+
<math>x</math> = <math>(162)</math><math>(1250/162)^{1/4}</math> = <math>(162)</math><math>(625/81)^{1/4}</math> = <math>(162)(5/3)</math> = <math>270</math>, which is <math>B</math>

Revision as of 23:49, 6 February 2013

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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