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Question
The equation logx2 16 + log2x 64 = 3 has,
Options
one irrational solution
no prime solution
two real solutions
one integral solution
Solution
two real solutions
Explanation;
logx2 16 + log2x 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)`
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