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

Revision as of 09:51, 4 July 2013

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 $b \Rightarrow \mathrm {(D)}$

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 ==See Also==
 {{AMC12 box|year=2007|ab=B|num-b=16|num-a=18}}
+{{MAA Notice}}

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

Revision as of 09:51, 4 July 2013

Problem 17

Solution

See Also