Difference between revisions of "2023 AIME I Problems/Problem 2"
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==Problem== | ==Problem== | ||
Positive real numbers <math>b \not= 1</math> and <math>n</math> satisfy the equations <cmath>\sqrt{\log_b n} = \log_b \sqrt{n} \qquad \text{and} \qquad b \cdot \log_b n = \log_b (bn).</cmath> The value of <math>n</math> is <math>\frac{j}{k},</math> where <math>j</math> and <math>k</math> are relatively prime positive integers. Find <math>j+k.</math> | Positive real numbers <math>b \not= 1</math> and <math>n</math> satisfy the equations <cmath>\sqrt{\log_b n} = \log_b \sqrt{n} \qquad \text{and} \qquad b \cdot \log_b n = \log_b (bn).</cmath> The value of <math>n</math> is <math>\frac{j}{k},</math> where <math>j</math> and <math>k</math> are relatively prime positive integers. Find <math>j+k.</math> | ||
+ | ==Video Solution & More by MegaMath== | ||
+ | https://www.youtube.com/watch?v=jxY7BBe-4gU | ||
==Solution 1== | ==Solution 1== | ||
Line 11: | Line 13: | ||
\end{align*} | \end{align*} | ||
</cmath> | </cmath> | ||
− | + | Solving the system gives <math>x = 4</math> and <math>b = \frac{5}{4}</math>. | |
Therefore, | Therefore, | ||
<cmath>n = b^x = \frac{625}{256}.</cmath> | <cmath>n = b^x = \frac{625}{256}.</cmath> | ||
Line 18: | Line 20: | ||
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
− | ==Solution 2 ( | + | ==Solution 2== |
− | + | We can use the property that <math>\log(xy) = \log(x) + \log(y)</math> on the first equation. We get <math>b \log_b(n) = 1 + \log_b(n)</math>. Then, subtracting <math>\log_b(n)</math> from both sides, we get <math>(b-1) \log_b(n) = 1</math>, therefore <math>\log_b(n) = \frac{1}{b-1}</math>. Substituting that into our first equation, we get <math>\frac{1}{2b-2} = \sqrt{\frac{1}{b-1}}</math>. Squaring, reciprocating, and simplifying both sides, we get the quadratic <math>4b^2 - 9b + 5 = 0</math>. Solving for <math>b</math>, we get <math>\frac{5}{4}</math> and <math>1</math>. Since the problem said that <math>b \neq 1</math>, <math>b = \frac{5}{4}</math>. To solve for <math>n</math>, we can use the property that <math>\log_b(n) = \frac{1}{b-1}</math>. <math>\log_\frac{5}{4}(n) = 4</math>, so <math>n = \frac{5^4}{4^4} = \frac{625}{256}</math>. Adding these together, we get <math>\boxed{881}</math> | |
− | + | ~idk12345678 | |
− | + | ==Solution 3 (quick)== | |
+ | |||
+ | We can let <math>n=b^{4x^2}</math>. Then, in the first equation, the LHS becomes <math>2x</math> and the RHS becomes <math>2x^2</math>. Therefore, <math>x</math> must be <math>1</math> (<math>x</math> can't be <math>0</math>). So now we know <math>n=b^4</math>. So we can plug this into the second equation to get <math>n=b^4</math>. This gives <math>b\cdot4=5</math>, so <math>b= \frac{5}{4}</math> and <math>n= b^4=\frac{625}{256}</math>. Adding the numerator and denominator gives <math>\boxed{881}</math>. | ||
+ | |||
+ | Honestly this problem is kinda misplaced. | ||
+ | |||
+ | ~yrock | ||
+ | |||
+ | ==Video Solution by TheBeautyofMath== | ||
+ | https://youtu.be/U96XHH23zhA | ||
+ | |||
+ | ~IceMatrix | ||
==See also== | ==See also== | ||
{{AIME box|year=2023|num-b=1|num-a=3|n=I}} | {{AIME box|year=2023|num-b=1|num-a=3|n=I}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | [[Category:Intermediate Algebra Problems]] |
Latest revision as of 19:46, 23 May 2024
Contents
Problem
Positive real numbers and
satisfy the equations
The value of
is
where
and
are relatively prime positive integers. Find
Video Solution & More by MegaMath
https://www.youtube.com/watch?v=jxY7BBe-4gU
Solution 1
Denote .
Hence, the system of equations given in the problem can be rewritten as
Solving the system gives
and
.
Therefore,
Therefore, the answer is
.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 2
We can use the property that on the first equation. We get
. Then, subtracting
from both sides, we get
, therefore
. Substituting that into our first equation, we get
. Squaring, reciprocating, and simplifying both sides, we get the quadratic
. Solving for
, we get
and
. Since the problem said that
,
. To solve for
, we can use the property that
.
, so
. Adding these together, we get
~idk12345678
Solution 3 (quick)
We can let . Then, in the first equation, the LHS becomes
and the RHS becomes
. Therefore,
must be
(
can't be
). So now we know
. So we can plug this into the second equation to get
. This gives
, so
and
. Adding the numerator and denominator gives
.
Honestly this problem is kinda misplaced.
~yrock
Video Solution by TheBeautyofMath
~IceMatrix
See also
2023 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.