Question

помогите плизз
a)log6(12)+log6(3)=
б)log49(84)-log49(12)=
в)
$\frac{lg9}{lg3} =$
г)log2log2(2)
д)2^1-log2(7)

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Answer 1

a)log6(12)+log6(3) = log6(36) = 2

б)log49(84)-log49(12) = log49(7) = 1/2

в)lg9/lg3 = log3(9) = 2

г)log2log2(2) = log2(1) = 0

д)2^1-log2(7) = 2/2^log2(7) = 2/7