Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The rule: "Characteristic of a number($\lt$ 1) with n zeros after decimal and before a significant digit = $\overline{( n + 1 )}$" This rule is mentioned in my module,Goggle lead me to this link.

When I am trying to verify with some random numbers like Log[10, 0.0004356] its giving me $-3.36091$ and Log[10, 0.725] gives me $-0.139662$. Giving $-3$ and $-0$ as characteristic respectivley, however shouldn't the characteristic be $-4$ and $-1$ respectively,consistent to the rule ? what exactly I am missing here ?

share|improve this question
    
Either the rule is wrong or whatever says 3 and 0 are wrong. –  Mark Hurd Nov 18 '10 at 13:04
    
@Mark:You can check out the proof here :math-help-ace.com/Logarithm-Tables.html –  Quixotic Nov 18 '10 at 15:07

2 Answers 2

up vote 3 down vote accepted

Read carefully: you take the characteristic as $-(n+1)$, but then you add a positive number (the mantissa); so the actual number you will get will be $-(n+1)+c$, with $c\gt 0$, and so you will get a number $\ell$ that satisfies $$-(n+1) \leq \ell \lt -n.$$

So in your example $0.0004356$, you start with a characteristic of $-4$, but then you add a positive number to get a number strictly larger than $-4$, but less than $-3$. With $0.725$, you start with $-1$, then add a positive number to get a number strictly larger than $-1$ but less than $0$.

Added: Basically, what you are doing is taking your number and multiplying by a sufficiently large power of $10$ so that you just get it to $1$ or over. If the first significant digit occurs in position $n+1$ after $n$ zeroes, then $10^{n+1}x$ will "shift" the decimal point $n+1$ spaces to the right, so you will get a number of the form $d_{n+1}.d_{n+2}d_{n+3}d_{n+4}\ldots$ (where your original number was $x = 0.0\ldots0d_{n+1}d_{n+2}d_{n+3}\ldots$, with $n$ zeroes after the decimal). You already know how to figure out these logarithms: they have characteristic $0$, and a certain mantissa $m$.

So what you get then is that $\log_{10}(10^{n+1}x) = 0.m$. But by the properties of the logarithm, you know that $$0.m = \log_{10}(10^{n+1}x) = \log_{10}(10^{n+1}) + \log_{10}(x) = (n+1) + \log_{10}(x).$$ Since $\log_{10}(x)$ is what you really want to know, that tells you that $$\log_{10}(x) = -(n+1)+0.m$$ So, we take $-(n+1)$ as the "characteristic", and then add the positive number $m$. So we get $-(n+1)\leq \log_{10}(x) \lt -n$.You get equality if and only if $m=0$, which happens if and only if $d_{n+2}=d_{n+3}=\cdots = 0$, which is exactly when $x=10^{-(n+1)}$. Otherwise, you get a number strictly between $-(n+1)$ and $-n$. But these numbers, when written in decimal, have $-n$ before the decimal, not $-(n+1)$. That's why you are getting $-3$ and $0$, and not $-4$ and $-1$, in your examples.

share|improve this answer

When I was an electrical engineer in the 1950s we used logarithms (or a slide rule) to calculate numerical answers, in the days before computers and hand held electronic calculators. $Log_{10}$ tables give the logarithms of numbers from 1 to 10, i.e. values between 0 and 1. Let's do the simple calculation of $0.345 \times 3 = 1.035$ using logarithms. We generally used 'four figure tables' to get three significant figures (3 s.f.) in our answers, and, surprisingly, this was accurate enough for most of our needs. We also had five figure tables, and, for really accurate work, we used "Chamber's six figure tables" to get 5 s.f.

Looking up $3.45$ in the tables gives us $0.5378$. Now $0.345$ is $3.45$ times $0.1$ so $log(0.345)$ is $log(3.45) + log(0.1)$ or $0.5378 -1 = -0.4622$, which is what you get if you look up $log(0.345)$ on a calculator. $Log(3)$ is $0.4771$, so $log(3\times0.345)$ is $0.4771-0.4622 = 0.0149$. This is the log of $1.0349$, or $1.035$ to 3 s.f. (which we would look up in "anti-log" tables).

In a long calculation it was awkward to have some additions and some subtractions in the process, so we note that we could have written the sum as $(3.45\times0.1)\times3$ and the log of the answer is $log(3.45)+log(0.1)+log(3)$ or $0.5378+(-1)+0.4771$. We did not have to subtract to find $log(0.345)$, instead we have replaced it by the simple addition of $-1$.

This was incorporated into the practice of using logarithms, by writing the log as two parts: a value between 1 and 10, which we could look up directly, called the 'Mantissa', and a multiplier, a power of ten, always a whole number and called the 'Characteristic'. So, since $0.345$ is $3.45\times10^{- 1}$, its logarithm could be written as $-1+0.5378$. A second short cut was to write this as $\overline{1}.5378$, the overline being a reminder to treat just this integer part as a negative number when adding up.

So, we would have written this sum as:

Number Log
$0.345\ \ \ \ \ \ \ \overline{1}.5378$
$3\ \ \ \ \ \ \ \ \ +\ \underline{0.4771}$
$\ \ \ \ \ \ \ \ \ \ \ = 0.0149\ \ Ans:\ 1.0349$

We would have left out the column headings, and the plus and equals signs in day-to-day calculations.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.