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

(Problem 17)
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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>?
  
<math>\mathrm {(A)} 0</math>  <math>\mathrm {(B)} 1</math>  <math>\mathrm {(C)} a</math>  <math>\mathrm {(D)} b</math>  <math>\mathrm {(E)} \frac{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>
  
 
==Solution==
 
==Solution==

Revision as of 22:19, 18 December 2017

Problem 17

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 $\{0,1,a,b,1/b\}$?

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

Solution

Note that if $a$ is positive, then, the equation will have no solutions for $b$. This becomes more obvious by noting that at $b=1$, $ab^2 > \log_{10} b$. 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 $b \Rightarrow \mathrm {(D)}$

See Also

2007 AMC 12B (ProblemsAnswer KeyResources)
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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