The equation logx2 16 + log2x64 = 3 has,(A) one irrational solution(B)

1.	The equation logx2 16 + log2x64 = 3 has,(A) one irrational solution(B) no prime solution(C) two real solutions(D) one integral solution
Answer» (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 = \(\frac{1}{\sqrt[3]2},4\)

The equation logx2 16 + log2x64 = 3 has,(A) one irrational solution(B) no prime solution(C) two real solutions(D) one integral solution

Answer»

(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 = \(\frac{1}{\sqrt[3]2},4\)