Suppose that someone wants to calculate approximately the product of 101,123,958,959,055 and 342,234,234,234,236 without using a computer. Since these numbers are so long, completely carrying out multiplication by hand would take a long time because each digit in one number must be multiplied by every digit in the other number. I understand how a logarithm together with a logarithm table can help approximate this product, but why is this better than approximating it by the product of 101,000,000,000,000 and 342,000,000,000,000 (or some other number of the digits replaced by zeros)?
Motivation of the Logarithm
logarithmsmath-historymotivation
Best Answer
Logarithms were not motivated by the desire to get exact solutions to problems such as multiplying two numbers with $15$ significant digits each.
With a decent table of logarithms you might have a precision of five digits, so indeed the answer you would get by using logarithms to compute $101{,}123{,}958{,}959{,}055 \times 342{,}234{,}234{,}234{,}236$ would be no more accurate than if you simply multiplied $101{,}120{,}000{,}000{,}000 \times 342{,}230{,}000{,}000{,}000.$ It might be possible to go to more digits with a better table and sophisticated interpolation techniques, but not many more digits.
Adding two five digit numbers is still easier to do than multiplying two five-digit numbers if the only tools at your disposal are pencil and paper.
This still might not have been enough motivation to purchase a table of logarithms if you only needed to do such a multiplication once a year. But there are problems that require you to perform many multiplications to get just one answer. Therefore there was an incentive to develop and publish tables of logarithms and promote their use.
Meanwhile, logarithms also turned out to have some interesting mathematical properties, for example providing a solution for the integral of $\frac1x$ (the only power of $x$ whose integral is not another power of $x$). I think the main reason we study logarithms today is because of things like that, not because we want to get the product of two $n$-digit numbers.