2log2(log2x)+log0 5(log2x)=1 Как решить?

2log2(log2x)+log0, 5(log2x)=1;
2log2(log2x)+log( 5 / 10)(log2x)=1;
2log2(log2x)+log( 1 / 2)(log2x)=1;
2log2(log2x)+log( 2 ^ ( - 1 ))(log2x)=1;
log2(log2x) ^ 2 +1 / ( - 1 ) * log2 (log2x)=1;
log2(log2x) ^ 2 - log2 (log2x)=1;
log2(log2x) ^ 2 - log2 (log2x)- 1 = 0;
Пусть log2(log2x) = а, тогда:
A ^ 2 a 1 = 0;
Отсюда, а1 = (1 - 5) /2;
А2 = (1 + 5) /2;
Тогда:
1 ) log2(log2x) = (1 - 5) /2;
log2x = ((1 - 5) /2) ^ 2;
log2x = ((1 - 25 + 5 ) /4 ) ;
log2x = ((6 - 25 ) /4 ) ;
log2x = ((3 - 5 ) /2 ) ;
x = ((3 - 5 ) /2 ) ^ 2;
x = ((9 - 65 + 5 ) /4 );
x = ((14 - 65 ) /4 );
x = ((7 - 35 ) /2 );
2 ) log2(log2x) = (1 +5) /2;
X = ((7 + 35 ) /2 );
Ответ: x = ((7 - 35 ) /2 );
х = ((7 + 35 ) /2 ).