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Difference between revisions of "2007 AMC 12B Problems/Problem 17"

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==Problem 17==
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==Problem==
 
If <math>a</math> is a nonzero integer and <math>b</math> is a positive number such that <math>ab^2=\log_{10}b</math>, what is the median of the set <math>\{0,1,a,b,1/b\}</math>?
 
If <math>a</math> is a nonzero integer and <math>b</math> is a positive number such that <math>ab^2=\log_{10}b</math>, what is the median of the set <math>\{0,1,a,b,1/b\}</math>?
  
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<math>\mathrm{(A)}\ 0 \qquad \mathrm{(B)}\ 1 \qquad \mathrm{(C)}\ a \qquad \mathrm{(D)}\ b \qquad \mathrm{(E)}\ 1/b</math>
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<math>\mathrm{(A)}\ 0 \qquad \mathrm{(B)}\ 1 \qquad \mathrm{(C)}\ a \qquad \mathrm{(D)}\ b \qquad \mathrm{(E)}\ \frac{1}{b}</math>
  
 
==Solution==
 
==Solution==
  
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Note that if <math>a</math> is positive, then, the equation will have no solutions for <math>b</math>. This becomes more obvious by noting that at <math>b=1</math>, <math>ab^2 > \log_{10} b</math>. The LHS quadratic function will increase faster than the RHS logarithmic function, so they will never intersect.
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Note that if <math>a</math> is positive, then, the equation will have no solutions for <math>b</math>. This becomes more obvious by noting that at <math>a=1</math>, <math>ab^2 > \log_{10} b</math>. The LHS quadratic function will increase faster than the RHS logarithmic function, so they will never intersect.
  
 
This puts <math>a</math> as the smallest in the set since it must be negative.
 
This puts <math>a</math> as the smallest in the set since it must be negative.
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The median is <math>b \Rightarrow \mathrm {(D)}</math>
 
The median is <math>b \Rightarrow \mathrm {(D)}</math>
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==Solution 2==
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Let <math>b=0.1</math>. Then <math>a\cdot0.01 = -1,</math> giving <math>a=-100</math>. Then the ordered set is <math>\{-100, 0, 0.1, 1, 10\}</math> and the median is <math>0.1=b,</math> so the answer is <math>\mathrm {(D)}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2007|ab=B|num-b=16|num-a=18}}
 
{{AMC12 box|year=2007|ab=B|num-b=16|num-a=18}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 02:05, 19 May 2024

Problem

If $a$ is a nonzero integer and $b$ is a positive number such that $ab^2=\log_{10}b$, what is the median of the set $\{0,1,a,b,1/b\}$?

$\mathrm{(A)}\ 0 \qquad \mathrm{(B)}\ 1 \qquad \mathrm{(C)}\ a \qquad \mathrm{(D)}\ b \qquad \mathrm{(E)}\ \frac{1}{b}$

Solution

Note that if $a$ is positive, then, the equation will have no solutions for $b$. This becomes more obvious by noting that at $a=1$, $ab^2 > \log_{10} b$. The LHS quadratic function will increase faster than the RHS logarithmic function, so they will never intersect.

This puts $a$ as the smallest in the set since it must be negative.

Checking the new equation: $-b^2 = \log_{10}b$

Near $b=0$, $-b^2 > \log_{10} b$ but at $b=1$, $-b^2 < \log_{10} b$

This implies that the solution occurs somewhere in between: $0 < b < 1$

This also implies that $\frac{1}{b} > 1$

This makes our set (ordered) $\{a,0,b,1,1/b\}$

The median is $b \Rightarrow \mathrm {(D)}$

Solution 2

Let $b=0.1$. Then $a\cdot0.01 = -1,$ giving $a=-100$. Then the ordered set is $\{-100, 0, 0.1, 1, 10\}$ and the median is $0.1=b,$ so the answer is $\mathrm {(D)}$.

See Also

2007 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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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