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 1==
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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.
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<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
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<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>
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<math>d</math> = <math>\frac{1}{4}</math>(<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:
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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
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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>
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==Solution 2==
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As the sequence <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> is an arithmetic progression, the sequence <math>162,x,y,z,1250</math> must be a geometric progression.
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If we factor the two known terms we get <math>162=2\cdot 3^4</math> and <math>1250=2\cdot 5^4</math>, thus the quotient is obviously <math>5/3</math> and therefore <math>x=162\cdot(5/3) = 270</math>.
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== Video Solution ==
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https://youtu.be/RdIIEhsbZKw?t=944
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~ pi_is_3.14
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== See also ==
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{{AMC12 box|year=2013|ab=A|num-b=13|num-a=15}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Revision as of 04:32, 21 January 2021

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 1

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 $B$

Solution 2

As the sequence $\log_{12}{162}$, $\log_{12}{x}$, $\log_{12}{y}$, $\log_{12}{z}$, $\log_{12}{1250}$ is an arithmetic progression, the sequence $162,x,y,z,1250$ must be a geometric progression.

If we factor the two known terms we get $162=2\cdot 3^4$ and $1250=2\cdot 5^4$, thus the quotient is obviously $5/3$ and therefore $x=162\cdot(5/3) = 270$.

Video Solution

https://youtu.be/RdIIEhsbZKw?t=944

~ pi_is_3.14

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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