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

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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>.
 
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 by OmegaLearn ==
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https://youtu.be/RdIIEhsbZKw?t=944
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 +
~ pi_is_3.14
  
 
== See also ==
 
== See also ==

Latest revision as of 03:18, 13 November 2022

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 by OmegaLearn

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

~ pi_is_3.14

See also

2013 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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