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2007 AMC 12B Problems/Problem 17

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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 \mathrm {(B)} 1 \mathrm {(C)} a \mathrm {(D)} b \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)}

See Also

2007 AMC 12B (ProblemsResources)
Preceded by
Problem 16
Followed by
Problem 18
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