How do I find the value of log (0.3274) if the mantissa of the log (3274) is 0.5150?

Answer 1

NOTE: log is log base 10 below

Given log(3274) has mantissa 0.5150 (just a nit: more accurate to say it is 0.5151), we can write:

log(3274) = log(10^K * 0.3274) = K + 0.5150 where K is a nonnegative integer (we can easily plug into a calculator to see K=3 but let's assume it is unknown).

First note that the value 0.3274 is between 0.1 and 1, which is the same as saying 10^-1 < 0.3274 < 10⁰

This means x = log(0.3274) will be a negative number, in the range -1 < x < 0.

Using log(a/b) = log(a) - log(b)

x = log(0.3274) = log(3274/10000) = log(3274) - log(10000) = (K + 0.5150 ) - 4 = (K-4) + 0.5150

Now

-1 < x < 0

Substitue for x the far RHS of the last equation:

-1 < K - 4 + 0.5150 < 0

3 < K + 0.5150 < 4

So K = 3 ==> K-4+0.5150 = 3-4+0.5150 = -0.4850 (more accurately, -0.4849)

We could have also just looked at the mantissa of K + 0.5150 - M, which must have mantissa 0.4850 if K and M are such that M > K (i.e. M is big enough to make K-M negative, which we reasoned out above).