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Question
Answer the following:
Solve : `sqrt(log_2 x^4) + 4log_4 sqrt(2/x)` = 2
Solution
`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.
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