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

Revision as of 22:47, 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

$d$ = (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 $B$

@@ Line 1: / Line 1: @@
 Since the sequence is arithmetic,
-<math>\log_{12}{162}</math> + 4d = <math>\log_{12}{1250}</math>, where d is the common difference.
+<math>\log_{12}{162}</math> + <math>4d</math> = <math>\log_{12}{1250}</math>, where <math>d</math> is the common difference.
 Therefore,
-d = <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
-d = (1/4)(<math>\log_{12}{(1250/162)}</math>) = <math>\log_{12}{(1250/162)^{1/4}}</math>
+<math>d</math> = (1/4)(<math>\log_{12}{(1250/162)}</math>) = <math>\log_{12}{(1250/162)^{1/4}}</math>
-Now that we found d, we just add it to the first term to find x:
+Now that we found <math>d</math>, we just add it to the first term to find <math>x</math>:
 <math>\log_{12}{162}</math> + <math>\log_{12}{(1250/162)^{1/4}}</math> = <math>\log_{12}{((162)(1250/162)^{1/4})}</math>
-x = (162)<math>(1250/162)^{1/4}</math> = (162)<math>(625/81)^{1/4}</math> = (162)(5/3) = 270, which is B
+<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>