Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2023 AIME I Problems/Problem 2"

m (→‎Solution 2 (extremely similar to above))
m (→‎Solution 2 (extremely similar to above))
Line 19: Line 19:
  
 
==Solution 2 (extremely similar to above)==
 
==Solution 2 (extremely similar to above)==
−
First, take the first equation and convert <math>\log_b\sqrt{n}</math> to <math>\log_b n^{\cfrac12}=\dfrac12\log_b n</math>. Square both sides to get <math>\log_b n=1/4 (\log_b n)^2</math>. Because a logarithm cannot be equal to <math>0</math>, <math>log_b n=4</math>.
+
First, take the first equation and convert <math>\log_b\sqrt{n}</math> to <math>\log_b n^{\cfrac12}=\dfrac12\log_b n</math>. Square both sides to get <math>\log_b n=1/4 (\log_b n)^2</math>. Because a logarithm cannot be equal to <math>0</math>, <math>\log_b n=4</math>.
  
−
By another logarithm rule, <math>log_b(bn)=log_b b+log_b n=1+4=5</math>. Therefore, <math>4b=5</math>, and <math>b=\dfrac54</math>. Since <math>b^4=n</math>, we have <math>n=\dfrac{625}{256}</math>, and <math>a+b=\boxed{881}</math>.
+
By another logarithm rule, <math>\log_b(bn)=\log_b b+\log_b n=1+4=5</math>. Therefore, <math>4b=5</math>, and <math>b=\dfrac54</math>. Since <math>b^4=n</math>, we have <math>n=\dfrac{625}{256}</math>, and <math>a+b=\boxed{881}</math>.
  
 
~wuwang2002 (feel free to remove if this is too similar to the above)
 
~wuwang2002 (feel free to remove if this is too similar to the above)

Revision as of 18:55, 8 February 2023

Problem

Positive real numbers $b \not= 1$ and $n$ satisfy the equations \[\sqrt{\log_b n} = \log_b \sqrt{n} \qquad \text{and} \qquad b \cdot \log_b n = \log_b (bn).\] The value of $n$ is $\frac{j}{k},$ where $j$ and $k$ are relatively prime positive integers. Find $j+k.$

Solution 1

Denote $x = \log_b n$. Hence, the system of equations given in the problem can be rewritten as \begin{align*} \sqrt{x} & = \frac{1}{2} x , \\ bx & = 1 + x . \end{align*} Thus, $x = 4$ and $b = \frac{5}{4}$. Therefore, \[n = b^x = \frac{625}{256}.\] Therefore, the answer is $625 + 256 = \boxed{881}$.

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution 2 (extremely similar to above)

First, take the first equation and convert $\log_b\sqrt{n}$ to $\log_b n^{\cfrac12}=\dfrac12\log_b n$. Square both sides to get $\log_b n=1/4 (\log_b n)^2$. Because a logarithm cannot be equal to $0$, $\log_b n=4$.

By another logarithm rule, $\log_b(bn)=\log_b b+\log_b n=1+4=5$. Therefore, $4b=5$, and $b=\dfrac54$. Since $b^4=n$, we have $n=\dfrac{625}{256}$, and $a+b=\boxed{881}$.

~wuwang2002 (feel free to remove if this is too similar to the above)

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AIME_I_Problems/Problem_2&oldid=189118"