Difference between revisions of "2023 AIME I Problems/Problem 2"
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==Solution2== | ==Solution2== | ||
| − | Denote b=x^n | + | Denote b=n^x. Hence, the system of equations given in the problem can be rewritten as |
| + | sqrt(x)=x/2, b*x=1+x | ||
| + | Thus, x=x^2/4, x=4. So, n=b^4 | ||
| + | Then, 4x=1+x. So, x=1/3 | ||
==Video Solution by TheBeautyofMath== | ==Video Solution by TheBeautyofMath== | ||
Revision as of 08:35, 30 September 2023
Problem
Positive real numbers 
and 
satisfy the equations ![\[\sqrt{\log_b n} = \log_b \sqrt{n} \qquad \text{and} \qquad b \cdot \log_b n = \log_b (bn).\]](http://latex.artofproblemsolving.com/e/1/8/e188ed1d68418cfaed35d91cc7b0f333a84f6b1b.png)
The value of is 
where 
and 
are relatively prime positive integers. Find 
Solution
Denote 
.
Hence, the system of equations given in the problem can be rewritten as
Solving the system gives 
and 
.
Therefore,
![\[n = b^x = \frac{625}{256}.\]](http://latex.artofproblemsolving.com/f/d/a/fda6b134d08ca7b814f36204c2f6a165af9ea24b.png)
Therefore, the answer is 
.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution2
Denote b=n^x. Hence, the system of equations given in the problem can be rewritten as sqrt(x)=x/2, b*x=1+x Thus, x=x^2/4, x=4. So, n=b^4 Then, 4x=1+x. So, x=1/3
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. 
