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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}yRule (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:
NN in
standard form
\log NCharacteristicMantissalog 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.
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