Question

There is a unique positive real number $x$ such that the three numbers $\log_8(2x),$ $\log_4 x,$ and $\log_2 x,$ in that order, form a geometric progression with positive common ratio. The number $x$ can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Answer

Solution:

Let $t = \log_2 x.$ Then $\log_4 x = \frac{t}{2}$ and $\log_8(2x) = \frac{1 + t}{3}.$ In a geometric progression the middle term squared equals the product of the outer terms: $\left(\frac{t}{2}\right)^2 = \frac{1 + t}{3} \cdot t.$

Since $t = 0$ gives no valid ratio, divide by $t:$ $\frac{t}{4} = \frac{1 + t}{3},$ so $3t = 4 + 4t$ and $t = -4.$ Thus $x = 2^{-4} = \frac{1}{16},$ and the progression is $-1,$ $-2,$ $-4$ with common ratio $2,$ which is positive as required.

Therefore $m + n = 1 + 16 = 17.$

2020 AIME I Problem 2

Solution:

Problem 2 in Other Years