Difference between revisions of "2017 AMC 12A Problems/Problem 20"
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==Solution== | ==Solution== | ||
− | By the properties of logarithms, we can rearrange the equation to read <math>2017 log_b a=(log_b a)^{2017}</math>. Then, subtracting <math> | + | By the properties of logarithms, we can rearrange the equation to read <math>2017 \log_b a=(\log_b a)^{2017}</math>. Then, subtracting <math>2017\log_b a</math> from each side yields <math>(\log_b a)^{2017}-2017\log_b a=0</math>. We then proceed to factor out the term <math>\log_b a</math> which results in <math>(\log_b a)(2016\log_b a -2017)=0</math>. Then, we set both factors equal to zero and solve. |
− | <math>log_b a=0</math> has exactly <math>199</math> solutions with the restricted domain of <math>[2,200]</math> since this equation will always have a solution in the form of <math>(1, b)</math>, and there are <math>199</math> possible values of <math>b</math> since <math>200-2+1 = 199</math>. | + | <math>\log_b a=0</math> has exactly <math>199</math> solutions with the restricted domain of <math>[2,200]</math> since this equation will always have a solution in the form of <math>(1, b)</math>, and there are <math>199</math> possible values of <math>b</math> since <math>200-2+1 = 199</math>. |
− | We proceed to solve the other factor, <math>(log_b a)2016-2017</math>. We add <math>2017</math> to both sides, and take the <math>2016th</math> root, this gives us <math>log_b a=\sqrt[2016]{2017}</math> <math>\sqrt[2016]{2017}</math> is a real number, and therefore <math>a=b^{\sqrt[2016]{2017}}</math> Again, there are <math>199</math> solutions, as | + | We proceed to solve the other factor, <math>(\log_b a)2016-2017</math>. We add <math>2017</math> to both sides, and take the <math>2016th</math> root, this gives us <math>\log_b a=\sqrt[2016]{2017}</math> <math>\sqrt[2016]{2017}</math> is a real number, and therefore <math>a=b^{\sqrt[2016]{2017}}</math> Again, there are <math>199 \cdot 2 = 398</math> solutions, as <math>b^{\sqrt[2016]{2017}}</math> must be a real number (It's a real number raised to a real number), and we also have to multiply by two because when we take the <math>2016th</math> root, we must also consider the negative root which is valid because the taking the reciprocal of <math>a</math> negates <math>log_ba</math>. |
− | <math>b^{\sqrt[2016]{2017}}</math> must be a real number (It's a real number raised to a real number). | ||
− | Therefore, | + | Therefore, the answer is <math>199 + 398 = \boxed{\textbf{E } 597}</math>. |
− | Note: | + | Note: the former solution was incorrect because when we take the <math>2016th</math> root, we must also consider the negative root which is valid because the taking the reciprocal of <math>a</math> negates <math>log_ba</math>. Therefore the answer is <math>199 \cdot 3</math> or <math>\textbf{E}</math>. |
==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 19:17, 8 February 2017
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 . Then, subtracting
from each side yields
. We then proceed to factor out the term
which results in
. Then, we set both factors equal to zero and solve.
has exactly
solutions with the restricted domain of
since this equation will always have a solution in the form of
, and there are
possible values of
since
.
We proceed to solve the other factor, . We add
to both sides, and take the
root, this gives us
is a real number, and therefore
Again, there are
solutions, as
must be a real number (It's a real number raised to a real number), and we also have to multiply by two because when we take the
root, we must also consider the negative root which is valid because the taking the reciprocal of
negates
.
Therefore, the answer is .
Note: the former solution was incorrect because when we take the root, we must also consider the negative root which is valid because the taking the reciprocal of
negates
. Therefore the answer is
or
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.