Applied Mathematics for Class 11th & 12th (Concepts and Questions) | ||
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11th | Concepts | Questions |
12th | Concepts | Questions |
Content On This Page | ||
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Binary Numbers | Indices, Logarithm and Antilogarithm | Laws and Properties of Logarithms |
Simple Applications of Logarithm and Antilogarithm |
Chapter 1 Numbers and Quantification (Concepts)
Welcome to this foundational exploration of essential numerical concepts and quantitative techniques that form the bedrock of Applied Mathematics. Before delving into complex models and applications, it's crucial to possess a strong number sense and familiarity with specific tools designed for practical quantification and calculation. This introductory section revisits the structure of the real number system and then focuses specifically on powerful concepts like prime numbers, the fascinating world of modulo arithmetic, and the indispensable utility of logarithms. Understanding these elements is not merely an academic exercise; it equips you with the fundamental language and computational aids necessary to effectively analyze and solve problems encountered across diverse applied fields, from computer science and cryptography to finance and the physical sciences. Our journey here emphasizes practical understanding and application over abstract theory.
We begin with a brief review of the real number system, ensuring clarity on the distinction between rational numbers (those expressible as a ratio of integers, $\frac{p}{q}$) and irrational numbers (like $\sqrt{2}$ or $\pi$, whose decimal expansions are non-terminating and non-repeating). Building upon this base, we introduce the concept of prime numbers – integers greater than 1 divisible only by 1 and themselves. These numbers are the fundamental building blocks of integers through the Fundamental Theorem of Arithmetic (every integer greater than 1 is either prime or can be uniquely factorized into primes) and hold profound significance in areas such as number theory and modern cryptography, underpinning the security of many digital systems.
Next, we delve into the practical and intriguing system of modulo arithmetic, a system concerned with remainders after division. The core concept is congruence, expressed as $a \equiv b \pmod{m}$, which means that $a$ and $b$ leave the same remainder when divided by $m$ (or equivalently, $m$ divides $a-b$). This seemingly simple idea has far-reaching applications. Modulo arithmetic is fundamental for understanding:
- Cyclical patterns and repetitive processes (like days of the week or clock arithmetic).
- Error detection and correction codes (e.g., ISBN check digits).
- Scheduling algorithms and resource allocation.
- Hashing algorithms and various operations in computer science.
- Basic cryptographic techniques.
Finally, we introduce logarithms as a powerful computational tool and the inverse operation to exponentiation. The statement $y = \log_b x$ is equivalent to $x = b^y$, where $b$ is the base of the logarithm. We distinguish between common logarithms (base 10, often written as $\log x$) and natural logarithms (base $e$, where $e \approx 2.71828...$, written as $\ln x$ or $\log_e x$). The true power of logarithms lies in their properties, which allow the simplification of complex calculations:
- $\log(mn) = \log m + \log n$ (Product Rule: turns multiplication into addition)
- $\log(\frac{m}{n}) = \log m - \log n$ (Quotient Rule: turns division into subtraction)
- $\log(m^n) = n \log m$ (Power Rule: turns exponentiation into multiplication)
- Change of Base Formula: $\log_b a = \frac{\log_c a}{\log_c b}$
Historically crucial before calculators, these properties enabled efficient computation using logarithm tables. Their relevance persists in understanding phenomena measured on logarithmic scales, such as earthquake intensity (Richter scale), sound intensity (decibels, $dB$), and acidity levels (pH scale). We will explore how to conceptually use logarithms and their inverse operation, antilogarithms ($y = \text{antilog}_b x \iff x = \log_b y$), to simplify calculations and appreciate their role in modeling real-world scales. Mastering these quantification techniques – number properties, modulo arithmetic, and logarithms – provides an essential toolkit for tackling the quantitative challenges presented throughout Applied Mathematics.
Binary Numbers
The number system we commonly use in our daily lives is the Decimal Number System (also known as Base-10 system). This system is based on powers of 10 and uses ten distinct digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. The value or magnitude of a digit within a decimal number is determined by its position, or place value. Each position represents a power of 10, increasing from right to left, starting with $10^0$. For instance, in the decimal number $754$:
$754_{10} = 7 \times 10^2 + 5 \times 10^1 + 4 \times 10^0 = 7 \times 100 + 5 \times 10 + 4 \times 1 = 700 + 50 + 4 = 754$.
In contrast to the decimal system, the Binary Number System (also known as Base-2 system) is a positional numeral system that uses only two digits: 0 and 1. It is the fundamental language of computers and digital systems because it aligns perfectly with devices that have only two stable states, such as switches being ON (1) or OFF (0), or electrical signals representing High (1) or Low (0) voltage. This two-state system makes digital circuits simpler and more reliable.
Place Value in Binary Numbers
Just like in the decimal system, the value of each digit in a binary number is determined by its position. However, in the binary system, the place values are powers of 2, not 10. Starting from the rightmost digit, which is the Least Significant Bit (LSB), the positions correspond to powers of 2 increasing as we move leftwards. The place values are $2^0, 2^1, 2^2, 2^3, 2^4$, and so on.
For a binary number represented as $(b_n b_{n-1} ... b_1 b_0)_2$, where $b_i$ is either 0 or 1 (a binary digit, also called a bit), its equivalent value in the decimal system is calculated as the sum of each digit multiplied by its corresponding power of 2. This can be expressed by the following formula:
$(b_n b_{n-1} ... b_1 b_0)_2 = b_n \times 2^n + b_{n-1} \times 2^{n-1} + ... + b_1 \times 2^1 + b_0 \times 2^0$
... (1)
Here, $n$ represents the highest position index (starting from 0 on the right). Each term $b_i \times 2^i$ contributes to the total decimal value. If $b_i$ is 0, the term is 0. If $b_i$ is 1, the term is $2^i$.
Conversion from Binary to Decimal
To convert a binary number into its decimal equivalent, we directly apply the formula mentioned above. We identify the position of each bit (starting from 0 for the rightmost bit), calculate the corresponding power of 2 for each position, multiply each bit by its respective power of 2, and finally, sum up all these products.
Example 1. Convert the binary number $(11011)_2$ to its decimal equivalent.
Answer:
The given binary number is $(11011)_2$. We identify the digits from right to left and their corresponding place values (powers of 2):
Rightmost digit (position 0): 1 ($2^0 = 1$)
Digit at position 1: 1 ($2^1 = 2$)
Digit at position 2: 0 ($2^2 = 4$)
Digit at position 3: 1 ($2^3 = 8$)
Digit at position 4: 1 ($2^4 = 16$)
Now, we multiply each digit by its place value and sum the results:
$(11011)_2 = 1 \times 2^4 + 1 \times 2^3 + 0 \times 2^2 + 1 \times 2^1 + 1 \times 2^0$
... (i)
$= 1 \times 16 + 1 \times 8 + 0 \times 4 + 1 \times 2 + 1 \times 1$
... (ii)
$= 16 + 8 + 0 + 2 + 1$
... (iii)
$= 27$
... (iv)
Thus, the decimal equivalent of $(11011)_2$ is $27$. We write this as $27_{10}$.
Example 2. Convert the binary number $(1000110)_2$ to its decimal equivalent.
Answer:
The given binary number is $(1000110)_2$. We determine the place value for each digit from right to left:
Position 0: 0 ($2^0 = 1$)
Position 1: 1 ($2^1 = 2$)
Position 2: 1 ($2^2 = 4$)
Position 3: 0 ($2^3 = 8$)
Position 4: 0 ($2^4 = 16$)
Position 5: 0 ($2^5 = 32$)
Position 6: 1 ($2^6 = 64$)
Now, we calculate the weighted sum:
$(1000110)_2 = 1 \times 2^6 + 0 \times 2^5 + 0 \times 2^4 + 0 \times 2^3 + 1 \times 2^2 + 1 \times 2^1 + 0 \times 2^0$
$= 1 \times 64 + 0 \times 32 + 0 \times 16 + 0 \times 8 + 1 \times 4 + 1 \times 2 + 0 \times 1$
$= 64 + 0 + 0 + 0 + 4 + 2 + 0$
$= 70$
So, $(1000110)_2 = 70_{10}$.
Conversion from Decimal to Binary
To convert a decimal number to its binary equivalent, the most common method is the method of successive division by 2. In this method, we repeatedly divide the decimal number by 2 and record the remainder (which will always be either 0 or 1) at each step. We continue this process with the quotient until the quotient becomes 0. The binary representation is then formed by reading the remainders from bottom to top, where the last remainder is the Most Significant Bit (MSB) and the first remainder is the Least Significant Bit (LSB).
Example 3. Convert the decimal number $45$ to its binary equivalent.
Answer:
We use successive division by 2:
$\begin{array}{c|cc} & \text{Quotient} & \text{Remainder} \\2 & 45 \\ \hline 2 & 22 & 1 \;\uparrow \\ \hline 2 & 11 & 0 \;\uparrow \\ \hline 2 & 5 & 1 \;\uparrow \\ \hline 2 & 2 & 1\;\uparrow \\ \hline & 1 & 0 \;\uparrow \\ \hline & 0 & 1 \;\uparrow \\ & \xrightarrow{\text{Read Up}} & \end{array}$We stop when the quotient is 0. Reading the remainders from bottom to top (the direction of the upward arrows), we get the sequence of bits $101101$.
So, the binary equivalent of $45_{10}$ is $(101101)_2$.
Example 4. Convert the decimal number $156$ to its binary equivalent.
Answer:
We use successive division by 2:
$\begin{array}{c|cc} & \text{Quotient} & \text{Remainder} \\2 & 156 \\ \hline 2 & 78 & 0 \;\uparrow \\ \hline 2 & 39 & 0 \;\uparrow \\ \hline 2 & 19 & 1 \;\uparrow \\ \hline 2 & 9 & 1 \;\uparrow \\ \hline 2 & 4 & 1 \;\uparrow \\ \hline 2 & 2 & 0 \;\uparrow \\ \hline 2 & 1 & 0 \;\uparrow \\ \hline & 0 & 1 \;\uparrow \\ & \xrightarrow{\text{Read Up}} & \end{array}$Reading the remainders from bottom to top, we get $10011100$.
So, the binary equivalent of $156_{10}$ is $(10011100)_2$.
Indices, Logarithm and Antilogarithm
Indices (Exponents)
Indices, also known as exponents or powers, are a mathematical notation used to represent repeated multiplication of a number by itself. If $a$ is a real number and $n$ is a positive integer, then $a^n$ (read as "$a$ raised to the power of $n$") is defined as the product of $n$ factors, each equal to $a$:
$$a^n = \underbrace{a \times a \times a \times ... \times a}_{n \text{ times}}$$
... (2)
In the expression $a^n$, $a$ is called the base and $n$ is called the exponent or index. This notation provides a concise way to write very large or very small numbers involving repeated multiplication.
Laws of Indices (Recap)
For any non-zero real numbers $a$ and $b$, and rational exponents $m$ and $n$ (assuming the bases are such that the expressions are defined and real):
Product Rule: When multiplying exponential terms with the same base, add the exponents.
$a^m \times a^n = a^{m+n}$
Example: $2^3 \times 2^4 = 2^{3+4} = 2^7 = 128$. (Because $2^3 = 8$ and $2^4 = 16$, $8 \times 16 = 128$.)
Quotient Rule: When dividing exponential terms with the same base, subtract the exponent of the denominator from the exponent of the numerator.
$$\frac{a^m}{a^n} = a^{m-n}$$, for $a \neq 0$
Example: $\frac{5^6}{5^2} = 5^{6-2} = 5^4 = 625$. (Because $5^6 = 15625$ and $5^2 = 25$, $15625 \div 25 = 625$.)
Power Rule: When raising an exponential term to another power, multiply the exponents.
$(a^m)^n = a^{mn}$
Example: $(3^2)^3 = 3^{2 \times 3} = 3^6 = 729$. (Because $(3^2)^3 = 9^3 = 9 \times 9 \times 9 = 81 \times 9 = 729$.)
Power of a Product: When raising a product of bases to a power, raise each base to that power.
$(ab)^n = a^n b^n$
Example: $(2 \times 5)^3 = 2^3 \times 5^3 = 8 \times 125 = 1000$. (Because $(2 \times 5)^3 = 10^3 = 1000$.)
Power of a Quotient: When raising a quotient of bases to a power, raise the numerator and the denominator to that power.
$$(\frac{a}{b})^n = \frac{a^n}{b^n}$$, for $b \neq 0$
Example: $(\frac{4}{2})^3 = \frac{4^3}{2^3} = \frac{64}{8} = 8$. (Because $(\frac{4}{2})^3 = 2^3 = 8$.)
Zero Exponent: Any non-zero base raised to the power of zero is equal to 1. The expression $0^0$ is typically considered undefined in elementary algebra, though in some contexts (like calculus or set theory), it might be defined as 1.
$$a^0 = 1$$, for $a \neq 0$
Example: $7^0 = 1$, $(\frac{1}{2})^0 = 1$.
Negative Exponent: A base raised to a negative exponent is equal to the reciprocal of the base raised to the corresponding positive exponent.
$$a^{-n} = \frac{1}{a^n}$$, for $a \neq 0$
$$\frac{1}{a^{-n}} = a^n$$, for $a \neq 0$
Example: $4^{-2} = \frac{1}{4^2} = \frac{1}{16}$.
Fractional Exponent: A base raised to a fractional exponent $m/n$ represents the $n$-th root of the base raised to the power of $m$. If $m=1$, it represents the $n$-th root.
$$a^{m/n} = \sqrt[n]{a^m} = (\sqrt[n]{a})^m$$
$$a^{1/n} = \sqrt[n]{a}$$
Assuming $\sqrt[n]{a}$ is a real number. If $n$ is even, $a$ must be non-negative for the real root to exist. If $n$ is odd, $a$ can be any real number.
Examples: $25^{1/2} = \sqrt[2]{25} = \sqrt{25} = 5$. $8^{2/3} = (\sqrt[3]{8})^2 = (2)^2 = 4$.
Logarithm
Logarithms are fundamentally linked to indices as they are the inverse operation to exponentiation. While exponentiation answers "What is $a$ raised to the power of $x$?", logarithms answer "To what power must the base $a$ be raised to get the number $N$?".
Formally, the logarithm of a positive number $N$ with respect to a positive base $a$ (where $a \neq 1$) is the exponent $x$ such that $a$ raised to the power of $x$ equals $N$. This relationship can be written in two equivalent forms:
Exponential Form: $a^x = N$
Logarithmic Form: $\log_a N = x$
... (3)
These two forms are interchangeable and express the same mathematical relationship. In $\log_a N = x$:
- $a$ is the base of the logarithm. It must be a positive number and $a \neq 1$. The reason the base cannot be 1 is that $1^x = 1$ for any $x$, so you cannot uniquely determine $x$ if $N$ is not 1. If $N=1$, $\log_1 1$ could be any number, again not unique.
- $N$ is the argument or number for which the logarithm is being calculated. It must be a positive number ($N > 0$). This is because a positive base raised to any real power will always result in a positive number ($a^x > 0$).
- $x$ is the logarithm or the exponent. It is the value we are trying to find. The logarithm can be any real number (positive, negative, or zero).
Types of Logarithms
Although the base $a$ can be any positive real number other than 1, certain bases are used frequently and have special notation:
Common Logarithms: These are logarithms with base 10. They are often written as $\log N$ or $\text{log}_{10} N$. The base 10 is convenient because our decimal number system is based on powers of 10. Common logarithms are useful in fields like engineering, physics, and chemistry, especially for calculations involving scaling and magnitude (e.g., Richter scale for earthquakes, pH scale for acidity).
If $10^x = N$, then $\log_{10} N = x$ or $\log N = x$.
Example: $\log_{10} 100 = 2$ because $10^2 = 100$. $\log_{10} 0.1 = -1$ because $10^{-1} = 0.1 = \frac{1}{10}$.
Natural Logarithms: These are logarithms with base $e$, where $e$ is an important irrational mathematical constant approximately equal to $2.71828$. Natural logarithms are denoted by $\ln N$ or $\log_e N$. The base $e$ arises naturally in calculus and phenomena involving exponential growth or decay, such as compound interest, population growth, and radioactive decay.
If $e^x = N$, then $\log_e N = x$ or $\ln N = x$.
Example: $\ln e = 1$ because $e^1 = e$. $\ln 1 = 0$ because $e^0 = 1$.
Conversion between Exponential and Logarithmic Forms
Understanding the equivalence between $a^x = N$ and $\log_a N = x$ is crucial for working with logarithms. You can convert any equation from one form to the other.
Example 1. Convert the exponential equation $3^4 = 81$ into logarithmic form.
Answer:
The given equation is $3^4 = 81$. This is in the exponential form $a^x = N$, where $a=3$, $x=4$, and $N=81$.
The equivalent logarithmic form is $\log_a N = x$.
Substituting the values, we get:
$\log_3 81 = 4$
... (i)
This means "the power to which 3 must be raised to get 81 is 4".
Example 2. Convert the logarithmic equation $\log_5 125 = 3$ into exponential form.
Answer:
The given equation is $\log_5 125 = 3$. This is in the logarithmic form $\log_a N = x$, where $a=5$, $N=125$, and $x=3$.
The equivalent exponential form is $a^x = N$.
Substituting the values, we get:
$5^3 = 125$
... (i)
This confirms that $5 \times 5 \times 5 = 125$.
Example 3. Find the value of $\log_2 8$.
Answer:
We want to find the value of $\log_2 8$. Let this value be $x$.
$\log_2 8 = x$
... (i)
Converting this logarithmic form to exponential form ($a^x = N$), we have $a=2$, $N=8$, and the exponent is $x$.
$2^x = 8$
... (ii)
We need to find the power $x$ to which 2 must be raised to get 8. We know that $8 = 2 \times 2 \times 2 = 2^3$.
So, $2^x = 2^3$
... (iii)
Since the bases are equal and positive (and not equal to 1), the exponents must be equal:
$x = 3$
... (iv)
Therefore, $\log_2 8 = 3$.
Example 4. Find the value of $\log_{10} 0.001$.
Answer:
Let $\log_{10} 0.001 = x$.
$\log_{10} 0.001 = x$
... (i)
Converting to exponential form:
$10^x = 0.001$
... (ii)
We need to express $0.001$ as a power of 10. $0.001 = \frac{1}{1000} = \frac{1}{10^3}$.
Using the rule for negative exponents ($a^{-n} = \frac{1}{a^n}$), we have $\frac{1}{10^3} = 10^{-3}$.
So, $10^x = 10^{-3}$
... (iii)
Equating the exponents (since the bases are equal):
$x = -3$
... (iv)
Therefore, $\log_{10} 0.001 = -3$.
Antilogarithm
The antilogarithm is the inverse operation of the logarithm. Finding the antilogarithm of a number $x$ with respect to a base $a$ means finding the original number $N$ whose logarithm base $a$ is $x$.
In simpler terms, if you know the logarithm of a number ($\log_a N = x$), the antilogarithm operation helps you find the number ($N$) itself.
By definition, if $\log_a N = x$, then the exponential form is $N = a^x$. Therefore, the antilogarithm of $x$ to the base $a$ is simply $a^x$.
If $\log_a N = x$, then $N = \text{antilog}_a(x) = a^x$
... (4)
Commonly encountered antilogarithms are with base 10 and base $e$:
- For common logarithms (base 10): If $\log_{10} N = x$, then $N = \text{antilog}_{10}(x) = 10^x$. This is also called the inverse logarithm or the base-10 exponential function.
- For natural logarithms (base $e$): If $\log_e N = x$, then $N = \text{antilog}_e(x) = e^x$. This is also known as the exponential function.
Example 5. If $\log_{10} N = 2$, find $N$.
Answer:
We are given $\log_{10} N = 2$. This means the logarithm of $N$ to the base 10 is 2. We need to find the number $N$. This is an antilogarithm problem.
Using the definition $\log_a N = x \iff N = a^x$, here $a=10$, $x=2$, and we need to find $N$.
$N = \text{antilog}_{10}(2) = 10^2$
... (i)
$N = 100$
... (ii)
So, the number whose base-10 logarithm is 2 is 100.
Example 6. Find the antilogarithm of $-1.5$ to the base 10.
Answer:
We need to find the number $N$ such that $\log_{10} N = -1.5$. This is $\text{antilog}_{10}(-1.5)$.
Using the definition $N = 10^x$, with $x = -1.5$:
$N = 10^{-1.5}$
... (i)
We can write the exponent $-1.5$ as $-\frac{3}{2}$.
$N = 10^{-3/2}$
... (ii)
Using the laws of indices ($a^{-n} = 1/a^n$ and $a^{m/n} = \sqrt[n]{a^m}$):
$$N = \frac{1}{10^{3/2}} = \frac{1}{\sqrt{10^3}} = \frac{1}{\sqrt{1000}}$$$$
... (iii)
Alternatively, $10^{-1.5} = 10^{-(1+0.5)} = 10^{-1} \times 10^{-0.5} = \frac{1}{10} \times \frac{1}{10^{0.5}} = \frac{1}{10 \sqrt{10}}$.
Calculating the numerical value (often requires a calculator or log tables which we will discuss later): $\sqrt{1000} \approx 31.62$. $N \approx \frac{1}{31.62} \approx 0.03162$.
So, $\text{antilog}_{10}(-1.5) \approx 0.03162$.
Laws and Properties of Logarithms
Logarithms are powerful tools for simplifying calculations, especially those involving multiplication, division, and exponentiation. Their utility comes from a set of fundamental laws and properties that are directly derived from the corresponding laws of indices (exponents). Understanding these laws is essential for solving problems involving logarithms.
Fundamental Laws of Logarithms
Let $a$ be a positive real number such that $a \neq 1$. This is the base of the logarithm. Let $M$ and $N$ be positive real numbers. These are the arguments of the logarithm. Let $p$ be any real number (can be positive, negative, or zero, rational or irrational). The following laws hold true:
1. Product Rule: The logarithm of a product is the sum of the logarithms.
$$\log_a (MN) = \log_a M + \log_a N$$
This law states that the logarithm of the product of two positive numbers is equal to the sum of their individual logarithms, provided they have the same base.
Derivation:
Let $\log_a M = x$ and $\log_a N = y$.
By the definition of logarithm, converting these to exponential form gives:
$a^x = M$
... (i)
$a^y = N$
... (ii)
Now, let's consider the product $MN$. Substituting the exponential forms from equations (i) and (ii):
$$MN = a^x \times a^y$$
(Substituting from (i) and (ii))
According to the product rule of indices ($a^m \times a^n = a^{m+n}$), we can combine the terms on the right side:
$$MN = a^{x+y}$$
... (iii)
Equation (iii) is in exponential form. Converting this back into logarithmic form using the definition $a^z = P \iff \log_a P = z$, where $a$ is the base, $P$ is the number ($MN$), and $z$ is the exponent ($x+y$):
$$\log_a (MN) = x+y$$
... (iv)
Finally, substitute back the original values of $x$ and $y$ (which were $\log_a M$ and $\log_a N$ respectively) into equation (iv):
$$\log_a (MN) = \log_a M + \log_a N$$
[From (iv)]
This completes the derivation of the product rule for logarithms.
2. Quotient Rule: The logarithm of a quotient is the difference of the logarithms.
$$\log_a \left(\frac{M}{N}\right) = \log_a M - \log_a N$$
This law states that the logarithm of the quotient of two positive numbers is equal to the difference between the logarithm of the numerator and the logarithm of the denominator, provided they have the same base.
Derivation:
Let $\log_a M = x$ and $\log_a N = y$.
In exponential form, these are:
$a^x = M$
... (i)
$a^y = N$
... (ii)
Consider the quotient $\frac{M}{N}$. Substitute the exponential forms from equations (i) and (ii):
$$\frac{M}{N} = \frac{a^x}{a^y}$$
(Substituting from (i) and (ii))
Using the quotient rule of indices ($\frac{a^m}{a^n} = a^{m-n}$), we simplify the right side:
$$\frac{M}{N} = a^{x-y}$$
... (iii)
Equation (iii) is in exponential form. Converting this back to logarithmic form ($\log_a P = z$, where $P = \frac{M}{N}$ and $z = x-y$):
$$\log_a \left(\frac{M}{N}\right) = x-y$$
... (iv)
Finally, substitute back the original values of $x$ and $y$ ($\log_a M$ and $\log_a N$) into equation (iv):
$$\log_a \left(\frac{M}{N}\right) = \log_a M - \log_a N$$
[From (iv)]
This completes the derivation of the quotient rule for logarithms.
3. Power Rule: The logarithm of a number raised to a power is the power multiplied by the logarithm of the number.
$$\log_a (M^p) = p \log_a M$$
This law states that when the argument of a logarithm is raised to a power, that power can be brought down as a multiplier of the logarithm of the base number.
Derivation:
Let $\log_a M = x$.
In exponential form, this is:
$a^x = M$
... (i)
Now consider the term $M^p$. Substitute the expression for $M$ from equation (i):
$$M^p = (a^x)^p$$
(Substituting from (i))
Using the power rule of indices ($(a^m)^n = a^{mn}$), we multiply the exponents:
$$M^p = a^{xp}$$
... (ii)
Equation (ii) is in exponential form. Convert this back to logarithmic form ($\log_a P = z$, where $P = M^p$ and $z = xp$):
$$\log_a (M^p) = xp$$
... (iii)
Finally, substitute back the original value of $x$ ($\log_a M$) into equation (iii):
$$\log_a (M^p) = p \log_a M$$
[Rearranging the right side of (iii)]
This completes the derivation of the power rule for logarithms.
4. Change of Base Formula: Changing the base of a logarithm.
$$\log_b a = \frac{\log_c a}{\log_c b}$$
This is a very useful formula that allows us to convert a logarithm from any base $b$ to another base $c$ (where $c>0, c \neq 1$). This is particularly important when using calculators or software that only provide logarithms for specific bases, like base 10 (log) or base $e$ (ln).
Derivation:
Let the logarithm we want to evaluate be $x$.
$\log_b a = x$
... (i)
Convert this logarithmic equation to exponential form ($b^x = a$):
$$b^x = a$$
... (ii)
Now, take the logarithm of both sides of equation (ii) with respect to a new, arbitrary base $c$ (where $c>0, c \neq 1$):
$$\log_c (b^x) = \log_c a$$
... (iii)
Apply the power rule of logarithms ($\log_c (M^p) = p \log_c M$) to the left side of equation (iii), treating $b$ as $M$ and $x$ as $p$:
$$x \log_c b = \log_c a$$
... (iv)
We want to find $x$. Since $b \neq 1$ and $c \neq 1$, $\log_c b$ will be a non-zero finite value (unless $b=1/c$). We can divide both sides of equation (iv) by $\log_c b$ (assuming $\log_c b \neq 0$, which implies $b \neq 1$):
$$x = \frac{\log_c a}{\log_c b}$$
... (v)
Substitute back the original expression for $x$ from equation (i):
$$\log_b a = \frac{\log_c a}{\log_c b}$$
[From (v)]
This completes the derivation of the change of base formula.
Other Important Properties
Besides the fundamental laws, there are several other properties that directly follow from the definition of logarithms and the laws of indices. For any valid base $a$ ($a>0, a \neq 1$) and positive number $N$:
Logarithm of the Base: The logarithm of the base itself is always 1.
$$\log_a a = 1$$
Explanation: This is true because, by definition, $\log_a a = x$ means $a^x = a$. Since $a = a^1$, we have $a^x = a^1$, which implies $x=1$.
Logarithm of One: The logarithm of 1 to any valid base is always 0.
$$\log_a 1 = 0$$
Explanation: This is true because, by definition, $\log_a 1 = x$ means $a^x = 1$. From the laws of indices, we know that any non-zero number raised to the power of 0 is 1 ($a^0 = 1$). Therefore, $a^x = a^0$, which implies $x=0$.
Inverse Property 1: Raising the base to the power of the logarithm of a number (with the same base) results in the number itself.
$$a^{\log_a N} = N$$
Explanation: This property highlights the inverse relationship between exponentiation and logarithm. If $\log_a N = x$, then by definition $a^x = N$. Substituting $x = \log_a N$ back into the exponential form gives $a^{\log_a N} = N$.
Inverse Property 2: The logarithm of the base raised to a power is the power itself.
$$\log_a (a^x) = x$$
Explanation: This property also stems from the inverse relationship. Let $a^x = P$. By definition, $\log_a P = x$. Substituting $P=a^x$ back into the logarithmic form gives $\log_a (a^x) = x$. Alternatively, this can be seen as a special case of the power rule: $\log_a (a^x) = x \log_a a$. Since $\log_a a = 1$, we get $x \times 1 = x$.
Logarithm with a Power in the Base: If the base of the logarithm is a power of some other number.
$$\log_{a^k} N = \frac{1}{k} \log_a N$$
Derivation: We can derive this using the change of base formula. Let the new base be $a$.
$$\log_{a^k} N = \frac{\log_a N}{\log_a (a^k)}$$
[Using change of base with base $a$]
Apply the power rule to the denominator: $\log_a (a^k) = k \log_a a = k \times 1 = k$.
$$= \frac{\log_a N}{k}$$
$$= \frac{1}{k} \log_a N$$
This property is useful when the base is a power, e.g., $\log_{16} 8 = \log_{2^4} 2^3$. Using the formula, $\log_{2^4} 2^3 = \frac{1}{4} \log_2 2^3 = \frac{1}{4} \times 3 = \frac{3}{4}$. Alternatively, using change of base: $\log_{16} 8 = \frac{\log_2 8}{\log_2 16} = \frac{3}{4}$.
Reciprocal Property: The logarithm of a number to a certain base is the reciprocal of the logarithm of the base to that number.
$$\log_a b = \frac{1}{\log_b a}$$
Derivation: This can also be derived using the change of base formula. Use base $c=a$.
$$\log_a b = \frac{\log_a b}{\log_a a} = \frac{\log_a b}{1}$$
[Using change of base formula]
Now, use base $c=b$ in the change of base formula for $\log_a b$:
$$\log_a b = \frac{\log_b b}{\log_b a}$$
[Using change of base with base $b$]
Since $\log_b b = 1$ (logarithm of the base property), we get:
$$= \frac{1}{\log_b a}$$
This property shows how to swap the base and the argument of a logarithm.
Examples Applying Logarithm Laws
Example 1. Evaluate $\log_3 9 + \log_3 27$.
Answer:
We can use the Product Rule: $\log_a M + \log_a N = \log_a (MN)$. Here $a=3$, $M=9$, $N=27$.
$\log_3 9 + \log_3 27 = \log_3 (9 \times 27)$
$= \log_3 (243)$
Now, we need to find the power to which the base 3 must be raised to get 243. We can list powers of 3:
$3^1=3$
$3^2=9$
$3^3=27$
$3^4=81$
$3^5=243$
So, $3^5 = 243$. By the definition of logarithm, $\log_3 243 = 5$.
$= 5$
... (i)
Alternatively, we could evaluate each logarithm separately first:
For $\log_3 9$: $3^x = 9 \implies 3^x = 3^2 \implies x = 2$. So, $\log_3 9 = 2$.
$\log_3 9 = \log_3 3^2 = 2$
[Using $\log_a (a^x) = x$]
For $\log_3 27$: $3^x = 27 \implies 3^x = 3^3 \implies x = 3$. So, $\log_3 27 = 3$.
$\log_3 27 = \log_3 3^3 = 3$
[Using $\log_a (a^x) = x$]
Adding the results:
$\log_3 9 + \log_3 27 = 2 + 3 = 5$
... (ii)
Both methods yield the same result, $5$.
Example 2. Simplify $\log_2 (\frac{32}{4})$.
Answer:
We can use the Quotient Rule: $\log_a (\frac{M}{N}) = \log_a M - \log_a N$. Here $a=2$, $M=32$, $N=4$.
$\log_2 \left(\frac{32}{4}\right) = \log_2 32 - \log_2 4$
Evaluate each logarithm:
For $\log_2 32$: $2^x = 32$. Since $2^5 = 32$, $\log_2 32 = 5$.
For $\log_2 4$: $2^x = 4$. Since $2^2 = 4$, $\log_2 4 = 2$.
$= 5 - 2$
$= 3$
... (i)
Alternatively, we can simplify the fraction inside the logarithm first:
$\log_2 \left(\frac{32}{4}\right) = \log_2 (8)$
Now, evaluate $\log_2 8$: $2^x = 8$. Since $2^3 = 8$, $\log_2 8 = 3$.
$= 3$
... (ii)
Both methods give the same result, $3$.
Example 3. Evaluate $\log_5 125^2$.
Answer:
We can use the Power Rule: $\log_a (M^p) = p \log_a M$. Here $a=5$, $M=125$, $p=2$.
$\log_5 125^2 = 2 \log_5 125$
Now, evaluate $\log_5 125$: $5^x = 125$. Since $5^3 = 125$, $\log_5 125 = 3$.
$= 2 \times 3$
$= 6$
... (i)
Alternatively, we could calculate $125^2$ first: $125^2 = 15625$. Then find $\log_5 15625$.
We need to find the power to which 5 must be raised to get 15625. $5^1=5, 5^2=25, 5^3=125, 5^4=625, 5^5=3125, 5^6=15625$. So, $5^6 = 15625$. By the definition of logarithm, $\log_5 15625 = 6$.
$\log_5 125^2 = \log_5 15625 = 6$
... (ii)
Both methods give the result, $6$.
Example 4. Evaluate $\log_4 64$ using the change of base formula with base 2.
Answer:
The change of base formula is $\log_b a = \frac{\log_c a}{\log_c b}$. Here, we want to evaluate $\log_4 64$, so $a=64$ and $b=4$. We are asked to use base $c=2$.
$$\log_4 64 = \frac{\log_2 64}{\log_2 4}$$
Evaluate the logarithms in the numerator and the denominator:
For $\log_2 64$: $2^x = 64$. Since $2^6 = 64$, $\log_2 64 = 6$.
For $\log_2 4$: $2^x = 4$. Since $2^2 = 4$, $\log_2 4 = 2$.
$$= \frac{6}{2}$$
$= 3$
... (i)
To verify the result using the definition directly: $\log_4 64 = x$ means $4^x = 64$. Since $4^3 = 4 \times 4 \times 4 = 16 \times 4 = 64$, $x=3$. The result matches.
Example 5. Express $\log \left(\frac{x^3 y^2}{\sqrt{z}}\right)$ in terms of $\log x$, $\log y$, and $\log z$. Assume a base of 10 for $\log$.
Answer:
We need to expand the given logarithmic expression using the laws of logarithms.
$$\log \left(\frac{x^3 y^2}{\sqrt{z}}\right)$$
First, apply the Quotient Rule: $\log \left(\frac{M}{N}\right) = \log M - \log N$. Here $M = x^3 y^2$ and $N = \sqrt{z}$.
$$= \log (x^3 y^2) - \log (\sqrt{z})$$
[Quotient Rule]
Next, apply the Product Rule to the first term: $\log (MN) = \log M + \log N$. Here $M = x^3$ and $N = y^2$.
$$= (\log x^3 + \log y^2) - \log (\sqrt{z})$$
[Product Rule]
Rewrite the square root using a fractional exponent: $\sqrt{z} = z^{1/2}$.
$$= \log x^3 + \log y^2 - \log (z^{1/2})$$
[Substituting $\sqrt{z} = z^{1/2}$]
Now, apply the Power Rule to each term: $\log (M^p) = p \log M$.
$$= 3 \log x + 2 \log y - \frac{1}{2} \log z$$
[Power Rule]
This is the expanded form of the expression in terms of $\log x$, $\log y$, and $\log z$.
Simple Applications of Logarithm and Antilogarithm
While the fundamental definition of logarithms relates them directly to exponents, their true power lies in their ability to simplify complex mathematical operations and model real-world phenomena. Historically, before the widespread availability of calculators and computers, logarithms were indispensable for performing tedious arithmetic calculations quickly and accurately. Today, they remain crucial in solving certain types of equations and representing data across vast scales in various scientific and practical applications.
Simplifying Calculations (Historical Context)
The invention of logarithms by John Napier and Joost Bürgi in the early 17th century revolutionized computation. They created extensive tables of logarithms (Napier's used a base related to $e$, while common logarithms, base 10, were later developed). By converting numbers to their logarithms, complicated operations could be replaced by simpler ones, as demonstrated by the laws of logarithms:
- Multiplication becomes addition ($\log_a (MN) = \log_a M + \log_a N$).
- Division becomes subtraction ($\log_a (\frac{M}{N}) = \log_a M - \log_a N$).
- Raising to a power becomes multiplication ($\log_a (M^p) = p \log_a M$).
- Finding a root becomes division ($\log_a (\sqrt[n]{M}) = \log_a (M^{1/n}) = \frac{1}{n} \log_a M$).
To perform a calculation like $P = M \times N$ using log tables (typically base 10):
1. Find the logarithm of each number from the log table: Find $\log_{10} M$ and $\log_{10} N$.
2. Use the Product Rule: $\log_{10} P = \log_{10} M + \log_{10} N$. Add the logarithms found in step 1.
3. Find the number whose logarithm is $\log_{10} P$. This is done using antilogarithm tables (which are essentially inverse log tables). $P = \text{antilog}_{10}(\log_{10} P)$.
Similarly, for division $Q = \frac{M}{N}$:
1. Find $\log_{10} M$ and $\log_{10} N$ from log tables.
2. Use the Quotient Rule: $\log_{10} Q = \log_{10} M - \log_{10} N$. Subtract $\log_{10} N$ from $\log_{10} M$.
3. Find the antilogarithm of the result: $Q = \text{antilog}_{10}(\log_{10} Q)$.
For finding powers $R = M^p$:
1. Find $\log_{10} M$ from log tables.
2. Use the Power Rule: $\log_{10} R = p \times \log_{10} M$. Multiply the logarithm by the power $p$.
3. Find the antilogarithm of the result: $R = \text{antilog}_{10}(\log_{10} R)$.
Characteristic and Mantissa (for Base 10 Logarithms)
When using log tables for base 10 logarithms ($\log_{10}$), it is helpful to understand the structure of the logarithm of a positive number. Any positive real number $N$ can be expressed in scientific notation (or standard form) as $N = m \times 10^c$, where $1 \leq m < 10$ and $c$ is an integer.
Taking the base-10 logarithm of both sides:
$$\log_{10} N = \log_{10} (m \times 10^c)$$
Using the Product Rule of logarithms ($\log_a (MN) = \log_a M + \log_a N$):
$$= \log_{10} m + \log_{10} 10^c$$
[Product Rule]
Using the Power Rule of logarithms ($\log_a (M^p) = p \log_a M$) on the second term:
$$= \log_{10} m + c \log_{10} 10$$
[Power Rule]
Since $\log_{10} 10 = 1$:
$$= \log_{10} m + c \times 1$$
[$\log_{10} 10 = 1$]
Rearranging the terms, we get:
$$\log_{10} N = c + \log_{10} m$$
... (5)
In this equation:
- $c$ is an integer and is called the characteristic of $\log_{10} N$. It determines the magnitude (number of digits) of the original number $N$.
- $\log_{10} m$ is the decimal part and is called the mantissa of $\log_{10} N$. Since $1 \leq m < 10$, taking $\log_{10}$ gives $\log_{10} 1 \leq \log_{10} m < \log_{10} 10$, which means $0 \leq \log_{10} m < 1$. The mantissa is always a non-negative decimal fraction less than 1. Log tables provide the mantissa for a sequence of digits, ignoring the decimal point.
The characteristic $c$ can be determined by the position of the decimal point in $N$:
- If $N \geq 1$, the characteristic is one less than the number of digits in the integer part of $N$. For example, if $N=123.45$, the integer part has 3 digits, so $c = 3-1=2$. $\log_{10} 123.45 \approx 2.0915$.
- If $0 < N < 1$, the characteristic is negative and its magnitude is one more than the number of zeros immediately after the decimal point. The negative characteristic is usually written with a bar above the digit, e.g., $\bar{1}, \bar{2}$. For example, if $N=0.0123$, there is one zero after the decimal point, so $c = -(1+1) = -2$, written as $\bar{2}$. $\log_{10} 0.0123 \approx \bar{2}.0899 = -2 + 0.0899 = -1.9101$. The mantissa ($0.0899$) is always positive.
Antilog tables work by using the mantissa to find the significant digits of the number $N$ and the characteristic to place the decimal point.
Solving Exponential and Logarithmic Equations
One of the most significant modern applications of logarithms is in solving equations where the unknown variable appears in the exponent (exponential equations) or as the argument of a logarithm (logarithmic equations). The key is to use the inverse relationship between exponentiation and logarithms, and the laws of logarithms to isolate the variable.
Example 1. Solve for $x$: $2^x = 10$.
Answer:
This is an exponential equation where the variable $x$ is in the exponent. To solve for $x$, we can take the logarithm of both sides of the equation. We can use any valid base for the logarithm, but base 10 (common log) or base $e$ (natural log) are convenient as their values are available on calculators.
Let's use the common logarithm (base 10):
$\log_{10} (2^x) = \log_{10} 10$
Now, apply the Power Rule of logarithms ($\log_a M^p = p \log_a M$) to the left side, treating $M=2$ and $p=x$. On the right side, use the property $\log_{10} 10 = 1$:
$$x \log_{10} 2 = 1$$
[Power Rule; $\log_{10} 10 = 1$]
Now, isolate $x$ by dividing both sides by $\log_{10} 2$ (since $2^x=10$, $x$ must be non-zero, so $\log_{10} 2 \neq 0$):
$$x = \frac{1}{\log_{10} 2}$$
... (i)
Using the approximate value $\log_{10} 2 \approx 0.3010$ (which can be found using a calculator or log tables):
$$x \approx \frac{1}{0.3010}$$
Performing the division:
$$x \approx 3.322$$
... (ii)
So, the solution is approximately $x \approx 3.322$. This means that 2 raised to the power of approximately 3.322 is equal to 10.
Example 2. Solve for $x$: $\log_2 (x+1) = 3$.
Answer:
This is a logarithmic equation. The simplest way to solve it is to convert it into its equivalent exponential form. The definition of logarithm states that $\log_a N = x$ is equivalent to $a^x = N$.
In the given equation, the base is $a=2$, the argument is $N=(x+1)$, and the logarithm is $x=3$ (the value on the right side of the equation).
Converting $\log_2 (x+1) = 3$ to exponential form:
$$x+1 = 2^3$$
[Converting to exponential form]
Calculate the value of $2^3$:
$$x+1 = 8$$
... (i)
Now, solve for $x$ by subtracting 1 from both sides:
$$x = 8 - 1$$
$$x = 7$$
... (ii)
It's important to check if the solution is valid by plugging it back into the original logarithmic equation. The argument of a logarithm must be positive.
Original equation: $\log_2 (x+1) = 3$
Substitute $x=7$: $\log_2 (7+1) = \log_2 8$.
We know that $2^3 = 8$, so $\log_2 8 = 3$. The equation holds true, and the argument $(x+1 = 8)$ is positive. Thus, $x=7$ is a valid solution.
Applications in Various Fields
Logarithms are widely used in many scientific, engineering, and financial disciplines, often because they help in handling very large or very small numbers by compressing the scale, or because the phenomena themselves follow logarithmic or exponential patterns.
Science and Engineering
pH Scale: This scale measures the acidity or alkalinity of an aqueous solution. It is defined as the negative of the base-10 logarithm of the molar concentration of hydrogen ions ($[H^+]$) in moles per litre.
$$pH = -\log_{10} [H^+]$$
A difference of 1 on the pH scale corresponds to a tenfold difference in hydrogen ion concentration. For example, a solution with pH 3 is ten times more acidic than a solution with pH 4.
Decibel (dB) Scale: Used in acoustics, electronics, and communication to measure ratios of power or intensity. It is a logarithmic unit that expresses the ratio of a value to a reference value.
$$L_{I} (\text{in dB}) = 10 \log_{10} \left(\frac{I}{I_0}\right)$$
Where $I$ is the intensity of the sound/signal and $I_0$ is a reference intensity (e.g., threshold of human hearing). A 10 dB increase represents a tenfold increase in intensity.
Richter Scale: Measures the magnitude of earthquakes. Developed by Charles Richter, this scale assigns a single number to quantify the energy released by an earthquake. It is a base-10 logarithmic scale.
$$M = \log_{10} A - \log_{10} A_0 = \log_{10} \left(\frac{A}{A_0}\right)$$
Where $M$ is the magnitude, $A$ is the amplitude of the seismic waves measured by a seismograph, and $A_0$ is the amplitude of a reference small earthquake. A magnitude 7 earthquake is ten times larger in amplitude (and roughly $10^{1.5} \approx 32$ times more energetic) than a magnitude 6 earthquake.
Star Brightness (Magnitude Scale): In astronomy, the apparent magnitude of a celestial body is a logarithmic measure of its brightness as seen by an observer on Earth. A difference of 1 magnitude corresponds to a brightness ratio of approximately 2.512 ($10^{0.4}$).
Finance and Growth Models
Compound Interest: Logarithms are used to solve problems involving compound interest, particularly when you need to find the time period or the interest rate. The formula for the future value $A$ of an investment with principal $P$ compounded annually at an interest rate $r$ for $n$ years is:
$$A = P(1 + r)^n$$
... (6)
To find $n$, you can use logarithms.
Exponential Growth and Decay: Phenomena like population growth, radioactive decay, and the charging/discharging of capacitors are often modeled by exponential functions of the form $N(t) = N_0 e^{kt}$ or $N(t) = N_0 a^t$. Logarithms (especially natural logarithms for base $e$ models) are used to solve for time $t$, initial amount $N_0$, or the rate constant $k$.
Example 3. How many years will it take for an investment to double if it earns 5% interest compounded annually?
Answer:
Let the initial principal amount be $P$. The annual interest rate is $r = 5\% = 0.05$. The number of years is $n$, which we need to find. The amount after $n$ years is $A$. The formula for compound interest compounded annually is $A = P(1 + r)^n$.
We are given that the investment doubles, so the amount after $n$ years is $A = 2P$.
Substitute $A=2P$ and $r=0.05$ into the compound interest formula:
$$2P = P(1 + 0.05)^n$$
$$2P = P(1.05)^n$$
Assuming $P \neq 0$, we can divide both sides by $P$:
$$2 = (1.05)^n$$
... (i)
This is an exponential equation. To solve for the exponent $n$, take the logarithm of both sides. Using the common logarithm (base 10):
$$\log_{10} 2 = \log_{10} (1.05)^n$$
Apply the Power Rule of logarithms ($\log_a M^p = p \log_a M$) to the right side:
$$\log_{10} 2 = n \log_{10} 1.05$$
[Power Rule]
Now, isolate $n$ by dividing both sides by $\log_{10} 1.05$:
$$n = \frac{\log_{10} 2}{\log_{10} 1.05}$$
... (ii)
Using a calculator or log tables to find the approximate values of the logarithms: $\log_{10} 2 \approx 0.3010$ and $\log_{10} 1.05 \approx 0.0212$.
$$n \approx \frac{0.3010}{0.0212}$$
Performing the division:
$$n \approx 14.198$$
... (iii)
So, it will take approximately 14.2 years for the investment to double at a 5% annual compound interest rate.