2023 AIME I Problem 2

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

Set

x = \log_b n

and rewrite both equations in terms of

x

and

b.

The first equation forces

\sqrt{x} = \frac{x}{2};

discard the root that breaks the second equation, which reads

bx = 1 + x.

Let

x = \log_b n.

The first equation says

\sqrt{x} = \log_b n^{\frac{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.

Problem 2 in other years