Summer Problem Solving Marathon Solution #43

Let log₈(log₂ x) = y. Then log₂ x = 8^y = 2^3y, and x = 2^23y.

We then also have log₂(log₈ x) = y, so log₈ = 2^y, and x = 8^2y = 2^3(2y).

Since log₈(log₂ x) = log₂(log₈ x), we have 2^23y = 2^3(2y). Thus 3^3y = 3(2^y), and 2^2y = 3, and 2y = log₂ 3, and y = (log₂3)/2.

Since x = 2^3(2y), log₂ x = 3(2^y), and (log₂ x)² = 9(2^2y).

We now substitute (log₂3)/2 for y in this expression. This gives 9(2^log₂3) = 9 x 3 = 27.