Question

The equation log_x² 16 + log_2x64 = 3 has,

(A) one irrational solution

(B) no prime solution

(C) two real solutions

(D) one integral solution

Answer 1

(A), (B), (C), (D)

log _x²  16 + log_2x64 = 3

\(\therefore\frac{log16}{logx^2}+\frac{log64}{log2x}=3\)

∴ 4 log 2 [log x + log 2] + (6 log 2) (2 log x)

= 3 (2 log x) (log 2 + log x)

Let log 2 = a, log x = t. Then

∴ 4at + 4a² + 12at = 6at + 6t²

∴ 6t²– 10at – 4a² = 0

∴ 3t² – 5at – 2a² = 0

∴ (3t + a) (t – 2a) = 0

∴ t = \(-\frac13\)a, 2a

∴ log x = log(2)^-1/3, log(2²)

∴ x = 2^-1/3, 4

∴ x = \(\frac{1}{\sqrt[3]2},4\)