Given log 2 = 0.3010, log 3 = 0.4771. Find the value of a) log 12 b) log 5.

a) log 12 = log3*4= log3 + log4=log3 + log 2^2=log3 + 2log2= 0.4771 +2*0.3010=1.0791b) log5= log10/2= log10 - log2=1 - 0.3110= 0.6989

[tex]a)\ \ \ log12=log(4\cdot3)=log4+log3=log2^2+log3=2log2+log3\approx\\\\\approx2\cdot0.3010+0.4771=0.602+0.4771=1.0791\\\\2)\ \ \ log5=log \frac{10}{2} =log10-log2\approx1-0.3010=0.699[/tex]