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

(Solution 1)
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>
\begin{align*}
 
n & = b^x \
 
& = \frac{625}{256} .
 
\end{align*}
 
</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

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

Solution by someone else

See also

2023 AIME I (ProblemsAnswer KeyResources)
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