Question

Positive real numbers $b \ne 1$ and $n$ satisfy the equations $\sqrt{\log_b n} = \log_b \sqrt{n}$ and $b \cdot \log_b n = \log_b (bn).$ The value of $n$ is $\frac{j}{k},$ where $j$ and $k$ are relatively prime positive integers. Find $j + k.$

Answer

Answer: 881

Concepts:logarithm substitution

Difficulty rating: 2100

Solution:

Let $x = \log_b n.$ The first equation says $\sqrt{x} = \log_b n^{1/2} = \frac{x}{2},$ so $x = \frac{x^2}{4},$ giving $x = 0$ or $x = 4.$ If $x = 0$ then $n = 1,$ and the second equation would read $0 = \log_b b = 1,$ impossible; so $x = 4.$

The second equation says $bx = \log_b b + \log_b n = 1 + x,$ so $4b = 5$ and $b = \frac{5}{4}.$ Then $n = b^4 = \left(\frac{5}{4}\right)^4 = \frac{625}{256},$ which is in lowest terms, so $j + k = 625 + 256 = 881.$

2023 AIME I Problem 2

Answer: 881

Difficulty rating: 2100

Solution:

Problem 2 in Other Years