## Saturday, 9 August 2014

### chapter 2 Use of Logs in Calculations

Let us first discuss the general approach for using logs in calculation. Suppose that we have to evaluate the value of $N$ given by
 $N = \dfrac{{a \times b \times {c^{{e_1}}}}}{{x \times {y^{{e_2}}}}}$
One way is to use calculators! But you’ve only your pen and paper and what is called log-tables (we’ll soon discuss these) with you.
Taking the log of both sides to some base $B$, we have
 ${\log _B}N = {\log _B}\left( {\dfrac{{a \times b \times {c^{{e_1}}}}}{{x \times {y^{{e_2}}}}}} \right)$ $= {\log _B}\left( {a \times b \times {c^{{e_1}}}} \right) - {\log _B}\left( {x \times {y^{{e_2}}}} \right)$ Rule $(5)$ $= \left\{ {{{\log }_B}a + {{\log }_B}b + {{\log }_B}{c^{{e_1}}}} \right\} - \left\{ {{{\log }_B}x + {{\log }_B}{y^{{e_2}}}} \right\}$ Rule $(4)$ $= {\log _B}a + {\log _B}b + {e_1}{\log _B}c - {\log _B}x - {e_2}{\log _B}y$ Rule $(6)$
Thus, if we have the values of the various logs on the right, we can evaluate the value of ${\log _b}N$. Suppose this comes out to be $Z$. We then have,
 ${\log _B}N = Z$
Now we simply take the antilog of both sides, which is nothing but raising both sides to the power $B$, so that we have
 $N = {B^Z}$
Thus, we’ve succeded in evaluating N using only the operations of additions and subtractions (and simple multiplications). We’ve also taken logs and antilogs in the process – but for that we’ve log tables with us.
Before actually doing some calculations, let us understand how to use log tables. The first point is to note is that our default base that we use in $10$, i.e., $B = 10$, so you’ll find log tables generally in base $10$. This is because $10$ is a very easy base for our minds to grasp, and calculations in base $10$ are the easiest (we’ve used base $10$ ever since we learnt to count).
Suppose that we have to evaluate the log of $347$ to base $10$. From now on, this will be written simply as log $347$, where the base $10$ should be understood to be present. Note that we can write $347$ as
 $347 = 3.47 \times {10^2}$
Taking the log of both sides, we have
 $\log \,\,347\,\, = \log \left( {3.47} \right) + \log \left( {{{10}^2}} \right)$ $= 2 + \log \left( {3.47} \right)$
Thus, evaluating $\log 347$ is equivalent to evaluating $\log 3.47$ – the difference is simply of $2$. Consider another example:
 ${\rm{6478}}{\rm{.25 = 6}}{\rm{.47825 \times 1}}{{\rm{0}}^{\rm{3}}}$ $\Rightarrow\,\,\,\, \log \left( {6478.25} \right) = \,\,\,3 + \log \left( {6.47825} \right)$
In this manner, taking the log of any number can be reduced to finding the log of a corresponding number between $1$ and $10$. As another example,
 ${\rm{0}}{\rm{.000134 = 1}}{\rm{.34 \times 1}}{{\rm{0}}^{{\rm{ - 4}}}}$ $\Rightarrow \,\,\,\, \log \left( {0.000134} \right) = \,\,\, - 4 + \log \left( {1.34} \right)$
The integral part of any log is called its characteristic, while the non-integral part is its mantissa. Note that the mantissa will always lie between $0$ and $1$. Why? Because the mantissa is the log of some number $x$ between $1$ and $10$, and if
 $1 \le x < 10$ then $0 \le \log x < 1$
Thus, the mantissa lies in $[0,1)$.
Let us consider some more examples:
 $N$ $N$ instandard form $\log N$ Characteristic Mantissa log written as ${\rm{47}}.{\rm{234}}$ ${\rm{4}}.{\rm{7234 }} \times {\rm{ 1}}{0^{\rm{1}}}$ ${\rm{1}} + {\rm{ log }}\left( {{\rm{4}}.{\rm{7234}}} \right)$ $1$ ${\rm{log }}\left( {{\rm{4}}.{\rm{7234}}} \right)$ $1$. mantissa $0.000812$ ${\rm{8}}.{\rm{12 }} \times {\rm{ 1}}{0^{-{\rm{4}}}}$ $-{\rm{4}} + {\rm{log }}\left( {{\rm{8}}.{\rm{12}}} \right)$ $-{\rm{ 4}}$ ${\rm{log }}\left( {{\rm{8}}.{\rm{12}}} \right)$ $\mathop 4\limits^ -$. mantissa $96871112$ ${\rm{9}}.{\rm{6871112 }} \times {\rm{ 1}}{0^{\rm{7}}}$ ${\rm{7}} + {\rm{log }}\left( {{\rm{9}}.{\rm{6871112}}} \right)$ $7$ ${\rm{log}}\left( {{\rm{9}}.{\rm{6871112}}} \right)$ $7$. mantissa ${\rm{5}}.{\rm{24 }} \times {\rm{ 1}}{0^{{\rm{81}}}}$ ${\rm{5}}.{\rm{24 }} \times {\rm{ 1}}{0^{{\rm{81}}}}$ ${\rm{81 }} + {\rm{ log 5}}.{\rm{24}}$ $81$ ${\rm{log }}\left( {{\rm{5}}.{\rm{24}}} \right)$ $81$.mantissa[/latex]
Thus, the log of any number $N$ will be in the form
 $\log N = \left( {{\rm{Charecteristic}}} \right) \cdot \left( {{\rm{Mantissa}}} \right)$
Such that
 $N = \rm{antilog (Mantissa)}\, \times {10^{\rm{Characteristic}}}$
Where antilog (Mantissa) is simply ${10^{\rm{Mantissa}}}$
Note that the Mantissa is always positive (Why?); but characteristics will be negative for numbers less than $1.0$.
An alert reader should by now have realized that we need log tables only for values between $1$ and $10$.
Let us use log tables to find $\log (45.67)$. We have
 $45.67 = 4.567 \times {10^1}$ $\Rightarrow\,\,\,\, \log (45.67) = 1 + \log (4.567)$
Thus, we need to find $\log 4.567$. Look up a standard log table. In the left column, locate “$45$”. If the number that we were looking for were “$4500$”, the mantissa would have been $0.6532$. But now scan the topmost row, and find “$60$”. Thus, corresponding to “$4560$”, the mantissa would’ve been $0.6590$.
But we actually want the mantissa corresponding to “$4567$” – thus, add the number from the appropriate column in the right hand set of columns to the mantissa of the first three digits. Thus, the mantissa we are looking for is $0.6590 + 7 = 0.6597$.
 $\Rightarrow\,\,\,\, \log (4.567) = 0.6597$ $\Rightarrow\,\,\,\, \log (4.567) = 1.6597$
On the other hand, suppose we had been given that the $\log$ of $N$ is $1.6597$, i.e.,
 $\log N = 1.6597$
We need to find $N$. Now we’ll have to use antilog tables. The approach is the same: have a look at this table:
Thus
 $anti\log (1.6597) = 45.67$
Now that you know how to use log and antilog tables, let us perform some calculations using logs.