Chapter 1 • Unit I

Exercise 1.3: Logarithm and Antilogarithm

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:

$$10^x = b \iff x = \log10 b \quad and \quad b = 10\log10 b$$

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 $e ≈ 2.71828$. Usually denoted as $\ln x$ or $\log_e x$.
2. Common Logarithm (Briggsian Logarithm): Calculated with base 10. Written simply as $\log x$ or $\log10 x$.
3. Interconversion Identity: $\log10 N = 0.4343 \log_e N$ and $\log_e N = 2.303 \log10 N$.

Algebraic Laws & Properties of Logarithms

1. Logarithm of a Product: $$\log_b(MN) = \log_b M + \log_b N$$ Follows from $bu+v = b^u · b^v$. E.g. $\log_2(4 × 8) = \log_2 4 + \log_2 8 = 2 + 3 = 5$.
2. Logarithm of a Quotient: $$\log_b(MN) = \log_b M - \log_b N$$ Follows from $bu-v = b^u / b^v$. 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: $\log_2(3100) = 100 \log_2 3$.
4. Base Change Formula: $$\log_b N = \log_a N\log_a b = (\log_a N)(\log_b a)$$ Also: $\log_b a = 1\log_a b$.

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 $N = m × 10^p$ where $1 ≤ m < 10$. The power $p$ is the characteristic.
    • $1297.3 = 1.2973 × 10^3 \implies char = 3$
    • $129.73 = 1.2973 × 10^2 \implies char = 2$
    • $1.2973 = 1.2973 × 10^0 \implies char = 0$
    • $0.12973 = 1.2973 × 10-1 \implies char = -1 = \bar{1}$
    • $0.0012973 = 1.2973 × 10-3 \implies char = -3 = \bar{3}$
  • Mantissa: The fractional (decimal) part. The mantissa is always positive ($0 ≤ m < 1$) and is read directly from logarithmic tables.
  • Negative Logarithms: If $\log10 N = -1.2572$, 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: $2\log(914) + \log(3524) - \log(1516)$.
Show Step-by-Step Solution
Apply power rule: $2\log(914) = \log(914)^2 = \log(81196)$
Now apply product and quotient rules:
$$= \log[(81196) × (3524)(1516)] = \log[81196 × 3524 × 1615]$$ Simplify fractions: $8124 = 278$; $3515 = 73$; $7196 = 128$; $27 × 168 × 3 × 28 = 914$.
Result: $\mathbf{\log(914)}$.
Example 14: Evaluate $√320$ using logarithms.
Show Step-by-Step Solution
Let $x = √320 = (320)1/2$.
Taking log on both sides: $\log x = 12\log 320$.
Since $320 = 3.20 × 10^2$, $characteristic = 2$. From log tables, mantissa for 32 under 0 is $0.5051$.
$\log 320 = 2.5051$.
$\log x = 12(2.5051) = 1.2525$.
$x = Antilog(1.2525)$. Characteristic = 1 (decimal after $1+1=2$ digits).
Antilog table for 0.25 under 2 and mean diff 5 gives $1788$.
Answer: $x = \mathbf{17.88}$.
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