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

Epicwisdom (talk | contribs) m (moved 2013 AMC 12A Problems/Problems 14 to 2013 AMC 12A Problems/Problem 14: Incorrect title ("Problems 14")) |
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Since the sequence is arithmetic, | 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, | Therefore, | ||

− | 4d = <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> | <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> |

## Revision as of 23:47, 6 February 2013

Since the sequence is arithmetic,

+ = , where is the common difference.

Therefore,

= - = , and

= (1/4)() =

Now that we found , we just add it to the first term to find :

+ =

= (162) = (162) = = , which is