Question

The equation log_x² 16 + log_2x 64 = 3 has,

Options

• one irrational solution

• no prime solution

• two real solutions

• one integral solution

Answer 1

two real solutions

Explanation;

log_x² 16 + log_2x 64 = 3

∴ `(log_(2)16)/(log_(2)x^2) + (log_(2)64)/(log_2(2x))` = 3

∴ `(log_(2)2^4)/(2log_(2)x) + (log_(2)2^6)/(log_(2)2 + log_(2)x` = 3

∴ `(4log_(2)2)/(2log_(2)x) + (6log_(2)2)/(1 + log_(2)x)` = 3

∴ `2/"m"+6/(1+"m")` = 3 .....[m = log₂ x]

∴ 2(1 + m) + 6m = 3m(1 + m)

∴ 2 + 2m + 6m = 3m + 3m²

∴ 3m² – 5m – 2 = 0

∴ 3m² – 6m + 1m – 2 = 0

∴ 3m(m – 2) + 1(m – 2) = 0

∴ (m – 2)(3m + 1) = 0

∴ m – 2 = 0 or 3m + 1 = 0

∴ m = 2 or m = `-1/3`

∴ log₂ x = 2 or log₂ x = `-1/3`

∴ x = 4 or x = `2^(-1/3)`