Difference between revisions of "2017 AMC 12A Problems/Problem 20"
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| − | Since <math>\log_b a=x</math> is equivalent to <math>a=b^x</math>, each possible value <math>x</math> yields exactly <math>199</math> solutions <math>(b,a)</math>, as we can assign <math>a=b^x</math> to each <math>b=2,3,\dots, | + | Since <math>\log_b a=x</math> is equivalent to <math>a=b^x</math>, each possible value <math>x</math> yields exactly <math>199</math> solutions <math>(b,a)</math>, as we can assign <math>a=b^x</math> to each <math>b=2,3,\dots,200</math>. In total, we have <math>3\cdot 199=\boxed{\textbf{(E) } 597}</math> solutions. |
==See Also== | ==See Also== | ||
{{AMC12 box|year=2017|ab=A|num-b=19|num-a=21}} | {{AMC12 box|year=2017|ab=A|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 08:56, 23 January 2018
Problem
How many ordered pairs
such that
is a positive real number and
is an integer between
and
, inclusive, satisfy the equation
Solution
By the properties of logarithms, we can rearrange the equation to read
with
. If
, we may divide by it and get
, which implies
. Hence, we have
possible values
, namely
Since
is equivalent to
, each possible value
yields exactly
solutions
, as we can assign
to each
. In total, we have
solutions.
See Also
| 2017 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 19 |
Followed by Problem 21 |
| 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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.