# Logarithms

Why logarithms:

We know that $32 = {2^5}$.  ie., 32 can be represented as a power of 2.  But how do we represent 32 as some power of 10?
This is where logarithms comes.  32 when represented a power of 10 is equal to ${\log _{10}}32 = 1.5051$.  i.e., $32 = {10^{1.5051}}$

Properties of Logarithms:

1. ${\log _a}1 = 0$ because 0 is the power to which a must be raised to obtain 1.
2. ${\log _a}a = 1$ because 1 is the power to which a must be raised to obtain a.
3. ${\log _a}{a^N} = N$ because N is the power to which a must be raised to obtain ${a^N}$
4. ${a^{{{\log }_a}b}} = b$ because ${\log _a}b$ is the power to which a must be raised to obtain b.
5. ${a^{{{\log }_b}c}} = {c^{{{\log }_b}a}}$
6. ${\log _{{a^b}}}M = \displaystyle\frac{1}{b}{\log _a}M$
7. ${\log _b}M = \displaystyle\frac{{{{\log }_a}M}}{{{{\log }_a}b}}$
8. ${\log _a}b = \displaystyle\frac{1}{{{{\log }_b}a}}$
9. ${\log _a}b.{\log _b}c = {\log _a}c$

Laws of logarithm:
1. Product rule: ${\log _a}MN = {\log _a}M + {\log _a}N$
2. Division rule = ${\log _a}\displaystyle\frac{M}{N} = {\log _a}M - {\log _a}N$

Remember:
${\log _a}(b + c) \ne {\log _a}b + {\log _a}c$

Note: When no base is given we generally assume the base as 10.  These are called common logarithms
When "e" is a base then we call them as natural logarithms.

Characteristic of logarithm:
The characteristic of the logarithm of a number greater than one is one less than the number of digits in it.
Eg: characteristic of 98765 = 4 so total digits in the given number is 5.
We know that log 100 = $\log {10^2} = 2\log 10 = 2.00$.  As the characteristic of 100 is 2, then total digits in 100 are 3.
log 10 = $\log {10^1} = 1\log 10 = 1$ So total digits are 2 in 10.
log 99 = 1.9956 so total digits are 1 + 1 = 2.

Working Rule to find the number digits in ${a^b}$ format number:
1. Calculate the logarithm (you will get some positive number)
2. Adding one to the intezer part will give you the number of digits in that original number

Solved Example 1: (Important model)
How many digits are contained in the number ${2^{100}}$
Sol: $\log {2^{100}}$ = 100 x log 2 = 100 x 0.3010 = 30.10
Number of digits in ${2^{100}}$ are 30 + 1 = 31

To determine the characteristic of the logarithm of a decimal fraction: (Numbers between 0 to 1)
Look at this example:
Find the total zeroes after he decimal point of the expression $2^6$
We know that $2^6$ = $\dfrac{1}{{64}}$ = 0.015625
log $\dfrac{1}{{64}}$  = -1.806
Now when you calculate Antilog of -1.806 using calculator, you will get 0.0156.
But if you want to use antilog tables, you have to follow this procedure.
Now log $\dfrac{1}{{64}}$ = log $\dfrac{1}{{{2^6}}}$ = log ${2^{ - 6}}$ = -6 log 2.
We know that log 2 = 0.301
Now log $\dfrac{1}{{64}}$ = -6 × 0.301 = -1.806.
Important: Now if you look at the antilog table for 0.80 and 6, you will get wrong answer.  Why? Because -1.806 = -1 + (-0.806)
But mantissa is always positive.
-1.806 should be written as -2 + (1 - 0.806) = -2 + 0.194
Now when you look at the anti log table 0.194 gives you 1563.
So characteristic is 2.

That is the characteristic of the logarithm of a decimal fraction is one more than than the number of zeroes immediately after the decimal point  and is negative.

Working Rule to find the number of zeroes in a decimal number:
1. Calculate the logarithm (you will get some negative number)
2. Subtract the decimal part from one and increase the integer part by 1 to make mantissa positive
3. Number of zeroes of that number = Integer part - 1

Solved Example 2: (Important model)
How many zeroes are there between the decimal point and the first significant digit in ${\left( {\dfrac{1}{2}} \right)^{1000}}$
log ${\left( {\dfrac{1}{2}} \right)^{1000}}$ = 1000 × log (1/2) = 1000 × -0.30102 = -301.02
But in logarithms the decimal point should be positive. (By using
- 301.02 = - 301 + -0.02 = -302 + (1 - 0.02) = $\mathop {302}\limits^{\_\_\_}$.98
So number of zeroes are 302 - 1 = 301

Solved Example 3:
$\displaystyle\frac{1}{{1 + {{\log }_a}bc}} + \displaystyle\frac{1}{{1 + {{\log }_b}ca}} + \displaystyle\frac{1}{{1 + {{\log }_c}ab}} =$
 a. 0 b. 3 c. 2 d. 1
Explanation:
$\displaystyle\frac{1}{{1 + {{\log }_a}bc}}$ + $\displaystyle\frac{1}{{1 + {{\log }_b}ca}}$ + $\displaystyle\frac{1}{{1 + {{\log }_c}ab}}$=
$\displaystyle\frac{1}{{{{\log }_a}a + {{\log }_a}bc}}$+ $\displaystyle\frac{1}{{{{\log }_b}b + {{\log }_b}ca}}$ + $\displaystyle\frac{1}{{{{\log }_c}c + {{\log }_c}ab}}$
= $\displaystyle\frac{1}{{{{\log }_a}abc}}$ + $\displaystyle\frac{1}{{{{\log }_b}abc}}$ + $\displaystyle\frac{1}{{{{\log }_c}abc}}$ =
= ${\log _{abc}}a + {\log _{abc}}b + {\log _{abc}}c = {\log _{abc}}abc = 1$

Solved Example 4:
The value of  ${(yz)^{\log y - \log z}} \times {(zx)^{\log y - \log x}} \times {(xy)^{\log x - \log y}}$

 a. 2 b. 1 c. 0 d. 3
Explanation:
Assume K = ${(yz)^{\log y - \log z}}$ × ${(zx)^{\log y - \log x}}$ × ${(xy)^{\log x - \log y}}$
Taking log on both sides
Log K = log (${(yz)^{\log y - \log z}}$ × ${(zx)^{\log y - \log x}}$ × ${(xy)^{\log x - \log y}}$)
= $\log {(yz)^{\log y - \log z}}$ + $\log {(zx)^{\log y - \log x}}$ + $\log {(xy)^{\log x - \log y}}$
= $(\log y - \log z)\log (yz)$ + $(\log z - \log x)\log (zx)$ + $(\log x - \log y)\log (xy)$
$(\log y - \log z)(\log y + \log z)$ + $(\log z - \log x)(\log z + \log x)$ + $(\log x - \log y)(\log x + \log y)$ =0
log K = 0$\Rightarrow$ K = 1

Solved Example 5:
$\log \left( {\dfrac{{x + y}}{3}} \right)$ = $\displaystyle\frac{1}{2}$ (logx + logy)  then ($\dfrac{x}{y} + \dfrac{y}{x}$) is
 a. 5 b. 7 c. 9 d. 0
Explanation:
As the answer does not have any log, first we try to remove log from the given equation by simplifying it.
log$\left( {\displaystyle\displaystyle\frac{{x + y}}{3}} \right)$ =
$\displaystyle\frac{1}{2}$(logx + logy)
$\Rightarrow 2 log \left( {\displaystyle\frac{{x + y}}{3}} \right)$= log xy
$\Rightarrow$ $\log {\left( {\dfrac{{x + y}}{3}} \right)^2} = \log xy$
$\Rightarrow $${x^2} + {y^2} + 2xy = 9xy \Rightarrow$${x^2} + {y^2} = 7xy$
$\Rightarrow$ $\displaystyle\frac{x}{y} + \displaystyle\frac{y}{x} = 7$

Solved Example 6:
The value of $7{\log _a}\displaystyle\frac{{16}}{{15}} + 5{\log _a}\displaystyle\frac{{25}}{{24}} + 3{\log _a}\displaystyle\frac{{81}}{{80}}$ is
 a. ${\log _a}5$ b. ${\log _a}3$ c. ${\log _a}2$ d. ${\log _a}0$
Explanation:
$7{\log _a}\displaystyle\displaystyle\frac{{16}}{{15}}$ + $5{\log _a}\displaystyle\frac{{25}}{{24}}$ + $3{\log _a}\displaystyle\frac{{81}}{{80}}$ = $7{\log _a}\left[ {\displaystyle\frac{{{3^4}}}{{3 \times 5}}} \right]$ + $5{\log _a}\left[ {\displaystyle\frac{{{5^2}}}{{3 \times {2^3}}}} \right]$+ $3{\log _a}\left[ {\displaystyle\frac{{{3^4}}}{{5 \times {2^4}}}} \right]$

= $7\left[ {4{{\log }_a}2 - {{\log }_a}3 - {{\log }_a}5} \right]$ + $5\left[ {2{{\log }_a}5 - {{\log }_a}3 - 3{{\log }_a}2} \right]$+ $3\left[ {4{{\log }_a}3 - {{\log }_a}5 - 4{{\log }_a}2} \right]$
= ${\log _a}2$

Solved Example 7:
If  $x = {\log _{2a}}a, y = {\log _{3a}}2a, z = {\log _{4a}}3a$ then $\left( {x - 2} \right)yz$ =
 a. -1 b. -2 c.  1 d. 2
Explanation:
(x-2)yz = xyz - 2yz
Substituting values from the given values from above,
${\log _{2a}}a.{\log _{3a}}2a.{\log _{4a}}3a$ - $2{\log _{3a}}2a.{\log _{4a}}3a$ = ${\log _{4a}}a - 2{\log _{4a}}2a$
$\Rightarrow {\log _{4a}}a - {\log _{4a}}{(2a)^2} = {\log _{4a}}\left[ {a/{{(2a)}^2}} \right]$   (Division Rule)
$\Rightarrow lo{g_{4a}}\left[ {1/4a} \right]$ = ${\log _{4a}}4{a^{ - 1}}$ =  - 1

Solved Example 8:
If $a = {\log _4}5$ and $b = {\log _5}6$ then ${\log _2}3$ = ?
a. 1-2ab
b. 1+2ab
c. 2ab - 1
d. (a-b) / (a+b)
Explanation:
To solve questions like these, first observe the question asked. The given question contains 2 and 3.  If we remove 5 from the given values of a and b, we are left with 4 and 6.
$ab = {\log _4}5.{\log _5}6 = {\log _4}6 = \displaystyle\frac{1}{2}{\log _2}6$
By product rule, ${\log _2}6 = {\log _2}(3 \times 2) = {\log _2}3 + {\log _2}2$
$\displaystyle\frac{1}{2}({\log _2}3 + {\log _2}2) = \displaystyle\frac{1}{2}({\log _2}3 + 1)$
$\Rightarrow ab = \displaystyle\frac{1}{2}({\log _2}3 + 1)$
$\Rightarrow 2ab = {\log _2}3 + 1$
$\Rightarrow {\log _2}3 = 2ab - 1$

Solved Example 9:
If $a = {\log _{105}}7,b = {\log _7}5$ then ${\log _{35}}105$ =
 a. ab b. (b+1)a c. 1/ab d. 1/a(b+1)
$ab = {\log _{105}}7.{\log _7}5 = {\log _{105}}5$
Now ${\log _{35}}105 = \displaystyle\frac{1}{{{{\log }_{105}}35}}$ = $\displaystyle\frac{1}{{{{\log }_{105}}5 + {{\log }_{105}}7}}$ = $\displaystyle\frac{1}{{ab + a}}$ = $\displaystyle\frac{1}{{a(b + 1)}}$