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##log(5*8) = 2log(2)+1 ~~ 1.60206##

If ##x y > 0## then:

##log(xy) = log(x)+log(y)##

(This follows from ##10^(a+b) = 10^a * 10^b## and the definition of common logarithms)

##color(white)()##

In general for any valid base ##b## (i.e. ##b > 0## and ##b != 1##) we also have:

##log_b b = 1##

(This follows from ##b^1 = b## and the definition of logarithms in general)

So in particular for common (i.e. base ##10##) logarithms we have:

##log 10 = 1##

##color(white)()##

It is also useful to know that:

##log 2 ~~ 0.30103##

This is a very good approximation. The actual value is nearer ##0.301029995664##

##color(white)()##

Putting this all together we find:

##log(5*8) = log(2*2*10)##

##color(white)(log(5*8)) = log(2)+log(2)+log(10)##

##color(white)(log(5*8)) = 2log(2)+1##

##color(white)(log(5*8)) ~~ 2*0.30103+1 = 1.60206##