2023 AIME I Problems/Problem 2

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)

2023 AIME I (Problems • Answer Key • Resources)
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2023 AIME I Problems/Problem 2

Contents

Problem

Solution 1

Solution 2 (extremely similar to above)

See also