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Difference between revisions of "2009 AMC 12B Problems/Problem 15"

(New page: == Problem == Assume <math>0 < r < 3</math>. Below are five equations for <math>x</math>. Which equation has the largest solution <math>x</math>? <math>\textbf{(A)}\ 3(1 + r)^x = 7\qqua...)
 
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== Solution ==
 
== Solution ==
  
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Intuitivelly, <math>x</math> will be largest for that option for which the value in the parentheses is smallest.
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'''(B)''' Intuitively, <math>x</math> will be largest for that option for which the value in the parentheses is smallest.
  
 
Formally, first note that each of the values in parentheses is larger than <math>1</math>. Now, each of the options is of the form <math>3f(r)^x = 7</math>. This can be rewritten as <math>x\log f(r) = \log\frac 73</math>. As <math>f(r)>1</math>, we have <math>\log f(r)>0</math>. Thus <math>x</math> is the largest for the option for which <math>\log f(r)</math> is smallest. And as <math>\log f(r)</math> is an increasing function, this is the option for which <math>f(r)</math> is smallest.
 
Formally, first note that each of the values in parentheses is larger than <math>1</math>. Now, each of the options is of the form <math>3f(r)^x = 7</math>. This can be rewritten as <math>x\log f(r) = \log\frac 73</math>. As <math>f(r)>1</math>, we have <math>\log f(r)>0</math>. Thus <math>x</math> is the largest for the option for which <math>\log f(r)</math> is smallest. And as <math>\log f(r)</math> is an increasing function, this is the option for which <math>f(r)</math> is smallest.
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Thus the answer is <math>\boxed{\text{(B) } 3(1 + r/10)^x = 7}</math>.
 
Thus the answer is <math>\boxed{\text{(B) } 3(1 + r/10)^x = 7}</math>.
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== See Also ==
 
== See Also ==
  
 
{{AMC12 box|year=2009|ab=B|num-b=14|num-a=16}}
 
{{AMC12 box|year=2009|ab=B|num-b=14|num-a=16}}

Revision as of 17:23, 29 January 2011

Problem

Assume $0 < r < 3$. Below are five equations for $x$. Which equation has the largest solution $x$?

$\textbf{(A)}\ 3(1 + r)^x = 7\qquad \textbf{(B)}\ 3(1 + r/10)^x = 7\qquad \textbf{(C)}\ 3(1 + 2r)^x = 7$ $\textbf{(D)}\ 3(1 + \sqrt {r})^x = 7\qquad \textbf{(E)}\ 3(1 + 1/r)^x = 7$

Solution

(B) Intuitively, $x$ will be largest for that option for which the value in the parentheses is smallest.

Formally, first note that each of the values in parentheses is larger than $1$. Now, each of the options is of the form $3f(r)^x = 7$. This can be rewritten as $x\log f(r) = \log\frac 73$. As $f(r)>1$, we have $\log f(r)>0$. Thus $x$ is the largest for the option for which $\log f(r)$ is smallest. And as $\log f(r)$ is an increasing function, this is the option for which $f(r)$ is smallest.

We now get the following easier problem: Given that $0<r<3$, find the smallest value in the set $\{ 1+r, 1+r/10, 1+2r, 1+\sqrt r, 1+1/r\}$.

Clearly $1+r/10$ is smaller than the first and the third option.

We have $r^2 < 10$, dividing both sides by $10r$ we get $r/10 < 1/r$.

And finally, $r/100 < 1$, therefore $r^2/100 < r$, and as both sides are positive, we can take the square root and get $r/10 < \sqrt r$.

Thus the answer is $\boxed{\text{(B) } 3(1 + r/10)^x = 7}$.

See Also

2009 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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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