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> | |
+ | ==Video Solution & More by MegaMath== | ||
+ | https://www.youtube.com/watch?v=jxY7BBe-4gU | ||
− | + | ==Solution 1== | |
− | |||
Denote <math>x = \log_b n</math>. | Denote <math>x = \log_b n</math>. | ||
Hence, the system of equations given in the problem can be rewritten as | Hence, the system of equations given in the problem can be rewritten as | ||
<cmath> | <cmath> | ||
\begin{align*} | \begin{align*} | ||
− | \sqrt{x} & = \frac{1}{2} x | + | \sqrt{x} & = \frac{1}{2} x , \\ |
bx & = 1 + x . | bx & = 1 + x . | ||
\end{align*} | \end{align*} | ||
</cmath> | </cmath> | ||
+ | Solving the system gives <math>x = 4</math> and <math>b = \frac{5}{4}</math>. | ||
+ | Therefore, | ||
+ | <cmath>n = b^x = \frac{625}{256}.</cmath> | ||
+ | Therefore, the answer is <math>625 + 256 = \boxed{881}</math>. | ||
− | + | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | |
− | + | ||
− | < | + | ==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> |
− | n | + | |
− | + | ~idk12345678 | |
− | \ | ||
− | </ | ||
− | |||
− | + | ==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}} | ||
+ | [[Category:Intermediate Algebra Problems]] |
Revision as of 13:10, 4 April 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
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.