2007 AMC 12B Problems/Problem 17

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 $b=1$ , $ab^2 > \log_{10} b$ . 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 $b \Rightarrow \mathrm {(D)}$

Cheap Solution that most people probably used

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 $\mathrm {(D)}$ .

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

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2007 AMC 12B Problems/Problem 17

Contents

Problem

Solution

Cheap Solution that most people probably used

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