Question 144344
A.  substituting __ 2=a2^b


substituting __ 16=a4^b


dividing __ 16/2=(a4^b)/(a2^b) __ 8=(4^b)/(2^b) __ 8=(4/2)^b __ 8=2^b


__ taking logarithm __ log(8)=b(log(2)) __ log(8)/log(2)=b __ 3=b


substituting __ 2=a(2^3) __ 2=8a __ 1/4=a



B.  substituting __ 3=a3^b


substituting __ 12=a6^b


dividing __ 12/3=(a6^b)/(a3^b) __ 4=(6^b)/(3^b) __ 4=(6/3)^b __ 4=2^b


__ taking logarithm __ log(4)=b(log(2)) __ log(4)/log(2)=b __ 2=b


substituting __ 3=a(3^2) __ 3=9a __ 1/3=a