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 - 2√5 + 5 ) /4 ) ;log2x = ((6 - 2√5 ) /4 ) ;log2x = ((3 - √5 ) /2 ) ;x = ((3 - √5 ) /2 ) ^ 2;x = ((9 - 6√5 + 5 ) /4 );x = ((14 - 6√5 ) /4 );x = ((7 - 3√5 ) /2 );2 ) log2(log2x) = (1 +√5) /2;X = ((7 + 3√5 ) /2 );Ответ: x = ((7 - 3√5 ) /2 );х = ((7 + 3√5 ) /2 ).