Log2^2 (4x) + log2^2 (2x) = 1 с объяснением (желательно)

И тут квадрат не над 2-кой, а над Log2

[log2 (4x) * log2 (4x) ] + [log2 (2x) * log2 (2x) ] = 1

[log2 (4) + log2 (x) ]^2 + [log2 (2) + log2 (x) ]^2=1

[2+log2 (x) ]^2 + [1+log2 (x) ]^2=1

4+4log2 (x) + [log2 (x) ]^2 + 1+2log2 (x) + [log2 (x) ]^2-1=0

2[log2 (x) ]^2 + 6log2 (x) + 4=0 |:2

[log2 (x) ]^2+3log2 (x) + 2=0

Замена: log2 (x) = t

t^2+3t+2=0

D=3^2-4*1*2=1

t1 = (-3-1) / 2=-2

t2 = (-3+1) / 2=-1

Обратная замена:

1) log2 (x) = - 2

log2 (x) = log2 (1/4)

x=1/4

2) log2 (x) = - 1

log2 (x) = log2 (1/2)

x=1/2

Ответ: 1/4; 1/2