Solve for x the equation log2(x)=log2(4)+3logx(2)
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We have to solve log2(x)=log2(4)+3logx(2) for x
log2(x) = log2(4)+3logx(2)
=> log2(x) - log2(4) = 3logx(2)
=> log2(x) - 2 = 3logx(2)
=> log2(x) - 2 = logx(8)
=> log2(x) - 2 = log2(8)/log2(x)
let log2(x) = y
=> y - 2 = 3/y
=> y^2 - 2y - 3 = 0
=> y^2 - 3y + y - 3 = 0
=> y(y - 3) + 1(y - 3) = 0
=> (y + 1)(y - 3) = 0
=> y = -1 and y = 3
log2(x) = -1 => x = 1/2
log2(x) = 3 => x = 8
The solutions of the equation are x = 1/2 and x = 8
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We'll write log2 (4) = log2 (2^2)
We'll apply the power property:
log2 (2^2) = 2*log2 (2)
But log2 (2) = 1 => log2 (4) = 2
We'll write log(x) 8 = 1/log (8) x
log(2) x =log(8) x*log(2) 8
log(2) 8= log(2) (2^3)
We'll use the power property of logarithms:
log(2) (2^3) = 3log(2) (2) = 3
log(2) x =3*log(8) x => log(8) x=log(2) x/3
The equation will become:
log(2) x - 2 - 3/log(2) x = 0
[log(2) x]^2 - 2log(2) x - 3 = 0
Let log(2) x be t:
t^2 - 2t - 3 = 0
t1 = [2+sqrt(4 + 12)]/2
t1 = (2+4)/2
t1 = 3
t2 = -1
log(2) x = 3 => x = 2^3 => x = 8
log(2) x = -1 => x = 2^-1 => x = 1/2
The requested solutions of the equation are: x = 1/2 and x = 8.
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