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

Sagot :

Pulkitidn
a) log 12 
= log3*4
= log3 + log4
=log3 + log 2^2
=log3 + 2log2
= 0.4771 +2*0.3010
=1.0791

b) log5
= log10/2
= log10 - log2
=1 - 0.3110
= 0.6989
Kate200468idn
[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]
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