(log6 2+log6 3+ 2^log2 4)log5 7
(log₆ 2+log₆ 3+ 2^log₂4)log₅ 7 =(log₆ (2· 3)+ 4)log₅ 7= (log₆ 6+ 4)log₅ 7= ( 1+ 4)log₅ 7= 5log₅ 7
или , если ошибка в условии
(log₆ 2+log₆ 3+ 2^log₂4)^log₅ 7 =(log₆ (2· 3)+ 4)^log₅ 7= (log₆ 6+ 4)^log₅ 7= ( 1+ 4)^log₅ 7=
5^log₅ 7 = 7
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(log6 2 + log6 3 + 2^log2 4)log5 7 = (log6 (2*3) + 2^log2 2^2)log5 7=
=(log6 6 + 2^2)log5 7=(1 + 4)log5 7 = 5log5 7
Ответ. 5log5 7