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

Revision as of 22:49, 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 $B$

@@ Line 8: / Line 8: @@
 <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>
@@ Line 16: / Line 16: @@
 <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>