2(log2)2 - 7log2x + log28 = 0
let log2x = x
In index form 2x = 8 therefor 2x = 23
x = 3
Let log2x = y
2y2 - 7y + 3 = 0
P = 6; S = -7; N = -1, -6
(2y2 - y) - (6y + 3) = 0
y(2y - 1) - 3(2y - 1) = 0
(y - 3)(2y - 1) = 0
y - 3 = 0
so log2x x = 3; 23 = x
y = 3
x = 8
also 2y - 1 = 0
2y = 1 therefor y = $\frac{1}{2}$
So log2x = $2^{\frac{1}{2}}$ = x
x = $\sqrt2$
johnmulu answered the question on March 7, 2017 at 06:27
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