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

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== Problem==
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The sequence
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<math>\log_{12}{162}</math>, <math>\log_{12}{x}</math>, <math>\log_{12}{y}</math>, <math>\log_{12}{z}</math>, <math>\log_{12}{1250}</math>
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is an arithmetic progression. What is <math>x</math>?
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<math> \textbf{(A)} \ 125\sqrt{3} \qquad \textbf{(B)} \ 270 \qquad \textbf{(C)} \ 162\sqrt{5} \qquad \textbf{(D)} \ 434 \qquad \textbf{(E)} \ 225\sqrt{6}</math>
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==Solution==
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Since the sequence is arithmetic,  
 
Since the sequence is arithmetic,  
  
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<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>
 
<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>
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== See also ==
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{{AMC12 box|year=2013|ab=A|num-b=13|num-a=15}}

Revision as of 18:41, 22 February 2013

Problem

The sequence

$\log_{12}{162}$, $\log_{12}{x}$, $\log_{12}{y}$, $\log_{12}{z}$, $\log_{12}{1250}$

is an arithmetic progression. What is $x$?

$\textbf{(A)} \ 125\sqrt{3} \qquad \textbf{(B)} \ 270 \qquad \textbf{(C)} \ 162\sqrt{5} \qquad \textbf{(D)} \ 434 \qquad \textbf{(E)} \ 225\sqrt{6}$

Solution

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 $B$

See also

2013 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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All AMC 12 Problems and Solutions
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