If x > 10 and log₈x + logₓ8 = 10/3, then x equals

Let t = log₈x, so logₓ8 = 1/t. The equation becomes t + 1/t = 10/3.

Multiply by 3t: 3t² − 10t + 3 = 0 → (3t − 1)(t − 3) = 0 → t = 1/3 or t = 3.

t = log₈x = 3 gives x = 8³ = 512; t = 1/3 gives x = 8^(1/3) = 2.

Since x > 10, only x = 512 satisfies the condition.

Tip: log_a b and log_b a are always reciprocals of each other, so an equation in both reduces to t + 1/t = k.