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**

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.**