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Break it on down......



Reading from the table for figures close to A:
\log_{10} (18790) = 4.27393    and     \log_{10} (18800) = 4.27416
Now if we linearly interpolate between these two figures, for greater accuracy, we obtain the approximation
\log_{10} (A) = 4.274005
Reading from the table for figures close to B:
\log_{10} (54770) = 4.73854    and     \log_{10} (54780) = 4.73862
Now if we linearly interpolate between these two figures, for greater accuracy, we obtain the approximation
\log_{10} (B) = 4.738605
Next, we use the property of logarithms mentioned earlier to estimate the logarithm of AB:
\log_{10} (AB) = \log_{10} (A) + \log_{10} (B) = 4.274005 + 4.738605 = 9.01261
The process of adding logarithms is very easy, and this is the point of the method. We’ve taken a relatively complicated problem (multiplying two numbers that have many digits) and converted it to a much easier problem (adding two numbers that have many digits). Now we have to convert the result back into the realm of the initial problem.

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