2020 AIME I Problem 2

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

Set

t = \log_2 x;

the three terms become

\frac{1+t}{3},

\frac{t}{2},

and

t.

The middle term squared equals the product of its neighbors; solve for

t,

discarding

t = 0.

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.

Problem 2 in other years