Difference between revisions of "2023 AIME I Problems/Problem 2"
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| Line 7: | Line 7: | ||
<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> | ||
| − | |||
Thus, <math>x = 4</math> and <math>b = \frac{5}{4}</math>. | Thus, <math>x = 4</math> and <math>b = \frac{5}{4}</math>. | ||
Therefore, | Therefore, | ||
| − | <cmath> | + | <cmath>n = b^x = \frac{625}{256}.</cmath> |
| − | |||
| − | n | ||
| − | |||
| − | |||
| − | </cmath> | ||
| − | |||
Therefore, the answer is <math>625 + 256 = \boxed{881}</math>. | Therefore, the answer is <math>625 + 256 = \boxed{881}</math>. | ||
Revision as of 13:35, 8 February 2023
Contents
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 1
Denote 
.
Hence, the system of equations given in the problem can be rewritten as
Thus, 
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)
Solution 2
Solution by someone else
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 | ||
