Question

(log6 2+log6 3+ 2^log2 4)log5 7

Answer 1

(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

Answer 2

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

 

 

(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