Introduction of logarithms
- Logarithms were introduced by the great Muslim mathematician Abu Muhammad Musa al-Khwarizmi.
- Abu Muhammad Musa al-Khwarizmi first gave an idea of logarithms.
- Logarithms developed In $17^{th}$ century by John Napier the concept of the logarithm further and developed/prepared tables for logarithms
- These tables utilized the base "$e$" The estimated value of the irrational number $e$ is $2.71828...$.
- The renowned mathematician "Euler" discovered the characteristics of the number "$e$". Therefore, this number is represented by the initial letter of his name.
- John Napier's writings caught the curiosity of Professor Henry Briggs.
- He created base-10 logarithm tables in the amount of $1631.
- Joost Burgi created the antilogarithm table in the year 1620 AD.
Importance of Logarithm
- Using logarithms makes complex and challenging computations easier.
Scientific Notation
- Scientific notation is a short way to write very large and very small numbers.
- Generally, approximate values of the given numbers are used to write them in scientific notation.
- In the problems of mathematics and other branches of science, we deal with very small or very large numbers.
For examples:
$1.$ $0.000000847$
$2.$ $56,784,234,079$
$3.$ The weight of the electron is
$4.$ The weight of earth
$6,000,000,000,000,000,000,000,000$ $kg$
- For the sake of convenience, we write such numbers, by using a special notation which is known as scientific notation.
- In this notation, the given number is expressed as the product of two factors.
- One of which is a number equal to or greater than $1$ but less than $10$, and the other is a power of $10$.
- If the given number is $n$: then in scientific notation $n=s\times 10^m$ where $\le s\lt 10$ and $m$ is an integer.
Example:
write the numbers in scientific notation
$i)$ $5373.458$
solution:
- The first factor is obtained by placing a decimal point after the first-hand non-zero digit.
$5.373458\times 1000$
- The second factor is a power of $10$ and the exponent is obtained by counting the number of digits that must be passed over to move from the original position of the decimal point to its new position.
$5.373458\times 10^3$
- If the decimal point moves from left to right then the exponent is negative.
$0.0004302$
- If the decimal point
$4300$
$ii)$ $89000000$
solution:
$8.9\times 1000000$
$8.9\times 10^7$ Ans.
$iii)$ $0.0007689$
solution:
$\frac{7.689}{10000}$
$\frac {7.689}{10^4}=7.689\times 10^{-4}$ Ans.
$iv)$ $0.00000015$
solution:
$\frac {1.5}{1000000}=\frac {1.5}{10^7}=1.5\times 10^{-7}$ Ans.
Definition of Logarithm
Logarithm form Exponential form
$(36)^{3/2}=216\Leftrightarrow \log_{36} 216=\frac {3}{2}$
$2^{-7}=\frac {1}{128}\Leftrightarrow \log_{2} \frac {1}{128}=-7$
$10^{-2}=0.01\Leftrightarrow \log_{10} 0.01=-2$
Write in exponential form
$\log_{5}25=2\Leftrightarrow 5^2=25$
$\log_{10}0.001=-3\Leftrightarrow 10^{-2}=0.001$
$\log_{2}\frac{1}{8}=-3\Leftrightarrow 2^{-3}=\frac {1}{8}$
In other words, a number cannot have two distinct $log$ to the same base.
for example $1^1=1$, $1^2=1$, $1^3=1$, $1^4=1$, etc $y$ may assume any of the values $1,2,3,4,...$
The logarithm of $1$ to any base is zero- If $y=1$, then $x=a^1\Rightarrow a=a^1$ Therefore $log_{a} a=1$
The logarithm of the base to itself is $1$
Solution:
$32^{-\frac{1}{5}}=x$
$\frac{1}{(2^5)^{1/5}}=x$
$x=\frac{1}{2}$ Ans
$ii)$ $log_{10} x=-4$
Solution:
$10^{-4}=x$
$x=\frac{1}{10^4}$
$x=\frac{1}{10000}$ Ans.
Problem: 2 Find the value of $a$ in the following
$i)$ $log_{a} 3=\frac{1}{2}$
Solution:
$a^{1/2}=3$
Squaring on both sides
$a=3$ Ans.
$ii)$ $log_{a}{\frac{1}{25}}=-\frac{2}{3}$
Solution:
$log_{a}{\frac{1}{25}}=-\frac{2}{3}$ $\rightarrow a^{-\frac{2}{3}}=\frac{1}{25}$
$a^{-\frac{2}{3}}=25^{-1}$
$a^{\frac{2}{3}}=25$
$a^\frac{1}{3}=5$
$a=125$ Ans.
Problem: 3 Find the value of $y$ in the following
$(i)$ $log_{5}{\frac{1}{5}}=y$
Solution:
$5^y=\frac{1}{5}$
$5^y=5^{-1}$
∴ $log_{a}{\frac{1}{a}}=-1$
$y=-1$ Ans.
$(ii)$ $log_{55} 55=y$
Solution:
$55^y=55$
$y=1$ Ans.
Problem: 4 Find the logarithms
$i)$ $1728$ to base $2\sqrt {3}$
Solution:
Let $log_{2\sqrt {3}} 1728=x$
$(2\sqrt 3)^x=1728$
$(2.3^{1/2})^x=2^6.3^3$
$(2.3^{1/2})^x=2^6.3^{2/2.3}$
$(2.3^{1/2})^x=(2.3^{1/2})^6$
$x=6$ Ans.
$ii)$ $125$ to base $5\sqrt 5$
Solution:
let $log_{5} {\sqrt{5}}125=x$
$(5\sqrt{5})^x=125$
$(5.5^{1/2})^x=5^3$
$(5^{1+1/2})^x=5^3$
$(5\sqrt{5})^x=125$
$(5.5^{1/2})^x=5^3$
$(5^{1+1/2})^x=5^3$
$(5^{3/2})^x=5^{3/2\times 2}$
$(5^3/2)^x=(5^{3/2})^2$
$x=2$ Ans.
Laws of Logarithm
We shall state and prove three laws of logarithms that enable us to simplify numerical calculations.
1. First laws:
For real numbers $m$, $n$, and $a$ $\gt 0$, $\neq 1$.
$log_{a} mn=log_{a} m+log_{a} n$
Proof:
Let $log_{a} m=x$ and $log_{a} n=y$
Exponential Form
$a^x=m$ and $a^y=n$
Multiplying both
Law of indices
$a^x.a^y=mn$
$a^{a+b}=mn$
Logarithm form
$log_{a} {mn}=log_{a} m+log_{a} n$ proved.
The logarithm of the product of two or more two numbers is the sum of their logarithms.
$log_{a} (xyz)=log_{a} x+log_{a} y+log_{a} z$
2. Second Law:
For real numbers $m$, $n$, $a$ and $a\gt 0$, $a\neq 1$
$log_{a}{\frac{m}{n}}=log_{a}{m}-log_{a}{n}$
Proof:
Let $log_{a}{m}=x$ and $log_{a}{n}=y$
$a^x=m$ and $a^y=n$
Dividing $1^{st}$ by $2^{nd}$
Exponential form
$\frac{a^x}{a^y}=\frac{m}{n}$
Index Law
$a^{x-y}=\frac{m}{n}$
Logarithmic form
$log_{a}\frac{m}{n}=x-y$
$log_{a}{\frac{m}{n}}=log_{a} m-log_{a} n$
3. Third Law
For real numbers $m$, $n$, $a$ and $a\gt 0$, $a\neq 1$
$log_{a} m^n=n log_{a} m$
Proof:
Let $log_{a} m=x$
$a^x=m$
$(a^x)^n=m$
$\left( a^x \right)^n=m^n$
$a^{nx}=m^n$
$(a)^{nx}=\left( m^n \right)$
$log_{a}m^n=nx$
$log_{a}m^n=n\text{ }log_{a}m$ proved.
- The logarithm of a number raised to a power $n$ is the product of the exponent $n$ and the logarithm.
Change of Base in Logarithms
- The logarithm's change of base rule is as follows:
$log_{a}n=\frac{log_{b}\text{ }n}{log_{b}\text{ }a}$
Proof:
Let $log_{a}{n}=x$
$a^x=n$
Taking $log$ on both sides to the base $b$
we have
$log_{b}{a^x}=log_{b}{n}$
$x log_{b}{a}=log_{b}{n}$
$x=\frac{log_{b}n}{log_{b}a}$ Proved.
$2.$ Special case:
$log_{a}{b}.log_{b}{a}=1$
Proof:
let $log_{a}{b}=x$
$a^x=b$
Taking $log$ on both sides to the base $b$
$log_{b}{a^x}=log_{b}{b}$
[Since $log_{b}{b}=1$]
$x\text{ } log_{b}{a}=1$
$log_{a}{b}.log_{b}{a}=1$
3. Third Law
For real numbers $m$, $n$, $a$ and $a\gt 0$, $a\neq 1$
$log_{a}{m^n}=nlog_{a}{m}$
Proof:
let $log_{a}{m}=x$
$a^x=m$
$\left( a^x \right)^n=m^n$
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