Question

Answer the following:

Solve : `sqrt(log_2 x^4) + 4log_4 sqrt(2/x)` = 2

Answer 1

`sqrt(log_2 x^4) + 4log_4 sqrt(2/x)` = 2

∴ `sqrt(4log_2x) + (4log_2(sqrt(2/x)))/log_2 4` = 2

∴ `2sqrt(log_2x) + (4 xx 1/2log_2 (2/x))/(2log_2 2)` = 2

∴ `2sqrt(log_2x) + log_2 (2/x)` = 2   ...[∵ log₂2 = 1]

∴ `2sqrt(log_2x) + log_2 2 - log_2 x` = 2

∴ `2sqrt(log_2 x) + 1 - log_2 x` = 2

∴ `2sqrt(log_2 x)` = 1 + log₂x

Squaring both sides, we get,

4 log₂x = (1 + log₂x)²

∴ 4 log₂x = 1 + 2log₂x + (log₂x)²

∴ (log₂x)² – 2log₂x + 1 = 0

∴ (log₂x – 1)² = 0

∴ log₂x – 1 = 0

∴ log₂x = 1 = log2²

∴ x = 2.