Comprehensive guide and practice questions for CBSE Class 11 Applied Mathematics curriculum (2026-27).
1.3 Logarithm and Antilogarithm: Foundations & Inverses
Every positive real number can be expressed as a power of a suitable base. For any positive real number $a$, there exists a real number $x$ such that:
$$10^x = b \iff x = \log_{10} b \quad \text{and} \quad b = 10^{\log_{10} b}$$
More generally, for any base $b > 0, b \ne 1$, and positive number $N > 0$:
$$b^x = N \iff x = \log_b N \quad (N = b^{\log_b N})$$
Thus, logarithms are the inverse operation of indices/exponentials.
Systems of Logarithms:
1. Natural Logarithm (Napierian Logarithm): Calculated with base $e \approx 2.71828$. Usually denoted as $\ln x$ or $\log_e x$.
2. Common Logarithm (Briggsian Logarithm): Calculated with base 10. Written simply as $\log x$ or $\log_{10} x$.
3. Interconversion Identity: $\log_{10} N = 0.4343 \log_e N$ and $\log_e N = 2.303 \log_{10} N$.
Algebraic Laws & Properties of Logarithms
1. Logarithm of a Product:
$$\log_b(MN) = \log_b M + \log_b N$$
Follows from $b^{u+v} = b^u \cdot b^v$. E.g. $\log_2(4 \times 8) = \log_2 4 + \log_2 8 = 2 + 3 = 5$.
2. Logarithm of a Quotient:
$$\log_b\left(\frac{M}{N}\right) = \log_b M - \log_b N$$
Follows from $b^{u-v} = b^u / b^v$. E.g. $\log_3(81/9) = \log_3 81 - \log_3 9 = 4 - 2 = 2$.
3. Logarithm of a Power:
$$\log_b(a^n) = n \log_b a$$
Example: $\log_2(3^{100}) = 100 \log_2 3$.
4. Base Change Formula:
$$\log_b N = \frac{\log_a N}{\log_a b} = (\log_a N)(\log_b a)$$
Also: $\log_b a = \frac{1}{\log_a b}$.
1.3.3 Logarithmic Tables: Characteristic and Mantissa
The logarithm of any positive real number consists of two components:
Characteristic: The integral part of the logarithm. It can be positive, zero, or negative.
Rule: Write number in scientific notation $N = m \times 10^p$ where $1 \le m < 10$. The power $p$ is the characteristic.
• $1297.3 = 1.2973 \times 10^3 \implies \text{char} = 3$
• $129.73 = 1.2973 \times 10^2 \implies \text{char} = 2$
• $1.2973 = 1.2973 \times 10^0 \implies \text{char} = 0$
• $0.12973 = 1.2973 \times 10^{-1} \implies \text{char} = -1 = \bar{1}$
• $0.0012973 = 1.2973 \times 10^{-3} \implies \text{char} = -3 = \bar{3}$
Mantissa: The fractional (decimal) part. The mantissa is always positive ($0 \le m < 1$) and is read directly from logarithmic tables.
Negative Logarithms: If $\log_{10} N = -1.2572$, we must add and subtract 1 to ensure a positive mantissa:
$$-1.2572 = -(1 + 0.2572) = -1 - 0.2572 = (-1 - 1) + (1 - 0.2572) = -2 + 0.7428 = \bar{2}.7428$$
Here, characteristic is $\bar{2}$ (-2) and mantissa is $+0.7428$.
Example 7 (CBSE Support Material): Express as a single logarithm: $2\log\left(\frac{9}{14}\right) + \log\left(\frac{35}{24}\right) - \log\left(\frac{15}{16}\right)$.
Show Step-by-Step Solution
Example 14: Evaluate $\sqrt{320}$ using logarithms.
Show Step-by-Step Solution
Let $x = \sqrt{320} = (320)^{1/2}$.
Taking log on both sides: $\log x = \frac{1}{2}\log 320$.
Since $320 = 3.20 \times 10^2$, $\text{characteristic} = 2$. From log tables, mantissa for 32 under 0 is $0.5051$.
$\log 320 = 2.5051$.
$\log x = \frac{1}{2}(2.5051) = 1.2525$.
$x = \text{Antilog}(1.2525)$. Characteristic = 1 (decimal after $1+1=2$ digits).
Antilog table for 0.25 under 2 and mean diff 5 gives $1788$. Answer: $x = \mathbf{17.88}$.
CHECK YOUR PROGRESS 1.3 & 1.4
Solve these authentic textbook problems covering logarithmic expansions, table evaluations, and real-life scientific models.
Problem 1Proofs & Expansions
Prove the following identities using the fundamental laws of logarithms:
(i) $\log 15 = \log 3 + \log 5$
(ii) $\log 500 = 2\log 2 + 3\log 5$
(iii) $\log\left(\frac{x^2}{yz}\right) + \log\left(\frac{y^2}{zx}\right) + \log\left(\frac{z^2}{xy}\right) = 0$
Apply quotient law followed by product and power laws:
$$\log_a(a^2 b^3) - \log_a(c^4 d^5) = (\log_a a^2 + \log_a b^3) - (\log_a c^4 + \log_a d^5)$$
$$= 2\log_a a + 3\log_a b - 4\log_a c - 5\log_a d$$
Since $\log_a a = 1$: Answer: $\mathbf{2 + 3\log_a b - 4\log_a c - 5\log_a d}$.
Write down the logarithm of $2^{32}$ and use it to state the number of digits in the numeral for $2^{32}$ (given $\log_{10} 2 = 0.30103$).
Show Step-by-Step Solution
Step 1: $\log_{10}(2^{32}) = 32 \times \log_{10} 2 = 32 \times 0.30103 = 9.63296$. Step 2: The characteristic of $\log_{10}(2^{32})$ is 9. Step 3: If the characteristic is $n$, the number of digits before the decimal is $n + 1$.
Number of digits $= 9 + 1 = \mathbf{10\text{ digits}}$.
Exercise 1.3 Revision Worksheet
Printable practice drill on logarithms, antilogarithms, characteristic-mantissa operations, and exponential equations.
State with reasons why $\log_5(0)$ is undefined in real numbers.
Show Model Solution
Answer: By definition, $\log_b N = x \iff b^x = N$. Since $5^x > 0$ for all real $x$, there is no real exponent $x$ such that $5^x = 0$. Hence $\log_5(0)$ is Not Defined.
Q21 Mark
Determine the characteristic of: (i) $\log_{10}(1297.3)$ and (ii) $\log_{10}(0.0012973)$.
Section C: High-Yield Applied Word Problems2 Questions
Q63 Marks
An earthquake registers a wave amplitude 2020 times the baseline amplitude $A_0$. Calculate the Richter magnitude $R = \log_{10}\left(\frac{A}{A_0}\right)$ (given $\log_{10} 2.02 \approx 0.3054$).
A bacterial colony grows according to $b(t) = 200(\sqrt{2})^{t/20}$. Determine the number of hours $t$ until the population reaches 500 bacteria (given $\log 2 = 0.3010, \log 2.5 = 0.3979$).
Problem: Given $\log_{10} x = -c.m$ (where $c$ is integer and $m$ is fractional part). Step 1: Write as $-(c + 0.m) = -c - 0.m$. Step 2: Add and subtract 1: $(-c - 1) + (1 - 0.m)$. Step 3: The new characteristic is $-(c+1) = \overline{c+1}$ (written with a bar), and the new positive mantissa is $(1 - 0.m)$. Example: $-2.0258 = -2 - 0.0258 = -3 + 0.9742 = \mathbf{\bar{3}.9742}$.
Examination Pitfalls
False Log Distributive Law
$\log(A+B) \ne \log A + \log B$ and $\frac{\log A}{\log B} \ne \log A - \log B$.
Decimal Placement in Antilog
If characteristic is $n \ge 0$, place decimal after $n+1$ digits. If characteristic is $\bar{n}$, prefix $n-1$ zeros after the decimal point!
Time: 45 Minutes Max Marks: 20 CBSE Board Blueprint
Solution: $\log(3+4) = \log 7$, whereas $\log(3 \times 4) = \log 12$. Logarithms do not distribute over addition.
Question 2MCQ • 1 Mark
What is the value of $\log_5(0)$?
(A) Not Defined
(B) 5
(C) 0
(D) 1
Solution: Logarithm of zero is undefined for any real base because $5^x > 0$ for all real $x$.
Question 3MCQ • 1 Mark
What is the characteristic of $\log_{10}(0.0012973)$?
(A) 3
(B) -2
(C) -4
(D) $\bar{3}$ (or -3)
Solution: $0.0012973 = 1.2973 \times 10^{-3}$. Therefore, the characteristic is $-3$ or $\bar{3}$.
Question 4Assertion-Reason • 1 Mark
Assertion (A): The mantissa of the logarithm of any positive real number is always non-negative. Reason (R): When a logarithm is negative, such as $-1.2572$, it is transformed into $\bar{2}.7428$ by adding and subtracting 1 to keep the fractional part strictly positive.
(A) Both A and R are true, and R is the correct explanation of A.
(B) Both A and R are true, but R is not the correct explanation of A.
(C) A is true, but R is false.
(D) A is false, but R is true.
Solution: Both A and R are true, and R explains why mantissas in mathematical log tables are always constrained to be positive.
Section B: Short Answer Type I (Q5 – Q6)2 Questions • 2 Marks Each
Question 52 Marks
Find the number of digits in $2^{32}$ given that $\log_{10} 2 = 0.30103$.
Section D: Long Answer Type (Q9)1 Question • 4 Marks
Question 94 Marks
Evaluate $\sqrt{320}$ using logarithmic and antilogarithmic tables given $\log 320 = 2.5051$ and $\text{antilog}(1.2525) = 17.88$.
Show Marking Scheme & Solution (4 Marks)
Step 1: Setup (1 Mark): Let $x = (320)^{1/2} \implies \log x = \frac{1}{2}\log 320$. Step 2: Log value (1 Mark): $\log x = \frac{1}{2}(2.5051) = 1.2525$. Step 3: Antilog setup (1 Mark): $x = \text{Antilog}(1.2525)$. Characteristic is 1, mantissa is 0.2525. Step 4: Answer (1 Mark): Insert decimal after $1+1=2$ places: $x = \mathbf{17.88}$.
Section E: Case-Based / Source-Based Problem (Q10)1 Case Study • 4 Marks
Question 10: Case Study4 Marks Total
Case Background: The magnitude of an earthquake on the Richter scale is defined by $R = \log_{10}\left(\frac{A}{A_0}\right)$, where $A$ is the measured wave amplitude and $A_0$ is the standard reference amplitude.
(i) [1 Mark]: An earthquake records amplitude $A = 10,000 A_0$. Determine its Richter scale magnitude.