Laws and Properties of Logarithms

Laws and Properties of Logarithms

Laws and Properties of Logarithms

Key Definitions

  1. Product Rule: For any base $a > 0$, $a \neq 1$, and positive real numbers $m$ and $n$: $$ \log_a(mn) = \log_a m + \log_a n $$ The logarithm of a product equals the sum of the logarithms of the individual factors.
  2. Quotient Rule: For any base $a > 0$, $a \neq 1$, and positive real numbers $m$ and $n$: $$ \log_a\left(\frac{m}{n}\right) = \log_a m – \log_a n $$ The logarithm of a quotient equals the difference of the logarithms.
  3. Power Rule: For any base $a > 0$, $a \neq 1$, $m > 0$, and any real number $k$: $$ \log_a(m^k) = k \log_a m $$ The logarithm of a power equals the exponent times the logarithm of the base.
  4. Base Power Rule: For $a > 0$, $a \neq 1$, $b > 0$, $b \neq 1$, and any real number $k \neq 0$: $$ \log_{a^k} b = \frac{1}{k} \log_a b $$
  5. Reciprocal Rule: For $a > 0$, $a \neq 1$, $b > 0$, $b \neq 1$: $$ \log_a b = \frac{1}{\log_b a} $$

Theory and Concepts

Fundamental Laws of Logarithms

The laws of logarithms are derived directly from the laws of exponents. Let us examine each law in detail:

Product Rule:

$$ \begin{align*} \text{Let } \log_a m &= p \Rightarrow a^p = m \\ \log_a n &= q \Rightarrow a^q = n \\ \text{Then } mn &= a^p \cdot a^q = a^{p+q} \\ \Rightarrow \log_a(mn) &= p + q = \log_a m + \log_a n \end{align*} $$

Quotient Rule:

$$ \begin{align*} \text{Since } \frac{m}{n} &= \frac{a^p}{a^q} = a^{p-q} \\ \Rightarrow \log_a\left(\frac{m}{n}\right) &= p – q = \log_a m – \log_a n \end{align*} $$

Power Rule:

$$ \begin{align*} \text{Since } m^k &= (a^p)^k = a^{pk} \\ \Rightarrow \log_a(m^k) &= pk = k \log_a m \end{align*} $$

Change of Base Formula

For any positive bases $a$, $b$ ($a \neq 1$, $b \neq 1$) and $x > 0$:

$$ \log_a x = \frac{\log_b x}{\log_b a} $$

This formula allows conversion between different bases. A special case:

$$ \log_a b \cdot \log_b c \cdot \log_c a = 1 $$

Important Properties and Identities

  1. $\log_a 1 = 0$ (since $a^0 = 1$)
  2. $\log_a a = 1$ (since $a^1 = a$)
  3. $\log_a a^k = k$
  4. $a^{\log_a x} = x$ (for $x > 0$)
  5. $\log_a (x^2) = 2\log_a |x|$ (when domain allows)
  6. $\log_{a^m} b^n = \frac{n}{m} \log_a b$ (for $a > 0$, $a \neq 1$, $b > 0$)

Visual Representation of Logarithmic Laws

Class 12 Maths Solution

Solved Examples

Solved Example 2.1

Simplify: $\log_5 125 + \log_5 25 – \log_5 5$.

Solution:

$$ \begin{align*} \log_5 125 + \log_5 25 – \log_5 5 &= \log_5(125 \times 25) – \log_5 5 \quad \text{(Product rule)}\\ &= \log_5(3125) – \log_5 5\\ &= \log_5\left(\frac{3125}{5}\right) \quad \text{(Quotient rule)}\\ &= \log_5 625\\ &= \log_5(5^4) = 4 \end{align*} $$

Alternative approach using power rule directly:

$$ \log_5 125 = \log_5 5^3 = 3,\quad \log_5 25 = \log_5 5^2 = 2,\quad \log_5 5 = 1 $$ $$ 3 + 2 – 1 = 4 $$

Solved Example 2.2

Prove that $\log_{x^2} y^3 = \frac{3}{2} \log_x y$.

Solution:

Using the base power rule and power rule:

$$ \log_{x^2} y^3 = \frac{1}{2} \log_x y^3 \quad \text{(since $\log_{a^k} b = \frac{1}{k} \log_a b$)} $$ $$ = \frac{1}{2} \cdot 3 \log_x y \quad \text{(Power rule)} $$ $$ = \frac{3}{2} \log_x y $$

Solved Example 2.3

If $\log_{12} 18 = a$ and $\log_{24} 54 = b$, prove that $ab + 5(a – b) = 1$.

Solution:

First, express everything in terms of $\log_2 3$:

$$ a = \log_{12} 18 = \frac{\log_2 18}{\log_2 12} = \frac{\log_2(2 \times 3^2)}{\log_2(2^2 \times 3)} = \frac{1 + 2\log_2 3}{2 + \log_2 3} $$ $$ b = \log_{24} 54 = \frac{\log_2 54}{\log_2 24} = \frac{\log_2(2 \times 3^3)}{\log_2(2^3 \times 3)} = \frac{1 + 3\log_2 3}{3 + \log_2 3} $$

Let $t = \log_2 3$. Then:

$$ a = \frac{1 + 2t}{2 + t}, \quad b = \frac{1 + 3t}{3 + t} $$

Now compute $ab + 5(a – b)$:

$$ \begin{align*} ab &= \frac{(1+2t)(1+3t)}{(2+t)(3+t)} = \frac{1 + 5t + 6t^2}{(2+t)(3+t)}\\ a – b &= \frac{1+2t}{2+t} – \frac{1+3t}{3+t} = \frac{(1+2t)(3+t) – (1+3t)(2+t)}{(2+t)(3+t)}\\ &= \frac{(3 + 7t + 2t^2) – (2 + 7t + 3t^2)}{(2+t)(3+t)} = \frac{1 – t^2}{(2+t)(3+t)} \end{align*} $$

Therefore:

$$ ab + 5(a-b) = \frac{1 + 5t + 6t^2 + 5 – 5t^2}{(2+t)(3+t)} = \frac{6 + 5t + t^2}{(2+t)(3+t)} $$ $$ = \frac{(t+2)(t+3)}{(t+2)(t+3)} = 1 $$

Solved Example 2.4

If $\log_a x = 2$, $\log_b x = 3$, $\log_c x = 6$, find $\log_{abc} x$.

Solution:

Given $\log_a x = 2 \Rightarrow x = a^2 \Rightarrow a = x^{1/2}$

Similarly, $b = x^{1/3}$, $c = x^{1/6}$

Therefore:

$$ abc = x^{1/2} \cdot x^{1/3} \cdot x^{1/6} = x^{(1/2 + 1/3 + 1/6)} = x^{1} $$

Thus $abc = x$, so $\log_{abc} x = \log_x x = 1$.

Solved Example 2.5

Simplify: $\log_2 3 \cdot \log_3 4 \cdot \log_4 5 \cdot \log_5 6 \cdot \log_6 7 \cdot \log_7 8$.

Solution:

Using the property $\log_a b \cdot \log_b c = \log_a c$:

$$ \begin{align*} &\log_2 3 \cdot \log_3 4 = \log_2 4 = 2\\ &\log_4 5 \cdot \log_5 6 = \log_4 6\\ &\log_6 7 \cdot \log_7 8 = \log_6 8 \end{align*} $$

So the product becomes: $2 \cdot \log_4 6 \cdot \log_6 8 = 2 \cdot \log_4 8 = 2 \cdot \log_{2^2} 2^3 = 2 \cdot \frac{3}{2} = 3$

Solved Example 2.6

Solve for $x$: $\log_2 x + \log_4 x + \log_8 x = 11$.

Solution:

Convert all logs to base 2:

$$ \log_4 x = \frac{\log_2 x}{\log_2 4} = \frac{\log_2 x}{2}, \quad \log_8 x = \frac{\log_2 x}{3} $$

Let $y = \log_2 x$. Then:

$$ y + \frac{y}{2} + \frac{y}{3} = 11 $$ $$ \frac{6y + 3y + 2y}{6} = 11 \Rightarrow \frac{11y}{6} = 11 \Rightarrow y = 6 $$

Therefore, $\log_2 x = 6 \Rightarrow x = 2^6 = 64$.

Solved Example 2.7

Find the value of $\log_{\sqrt{3}} 27$.

Solution:

$$ \log_{\sqrt{3}} 27 = \log_{3^{1/2}} 3^3 = \frac{3}{1/2} \log_3 3 = 6 \times 1 = 6 $$

Solved Example 2.8

If $\log_x y = \log_y z = \log_z x$, prove that $x = y = z$.

Solution:

Let $\log_x y = \log_y z = \log_z x = k$.

Then:

$$ y = x^k, \quad z = y^k, \quad x = z^k $$

Substituting:

$$ x = (y^k)^k = y^{k^2} = (x^k)^{k^2} = x^{k^3} $$

Since $x > 0$, $x \neq 1$, we have $k^3 = 1 \Rightarrow k = 1$ (since $k$ is real).

Therefore, $\log_x y = 1 \Rightarrow y = x$, $\log_y z = 1 \Rightarrow z = y$, hence $x = y = z$.

Solved Example 2.9

Solve: $\log_2 (x^2 + 4x – 3) – \log_2 (x – 1) = 2$.

Solution:

Using quotient rule:

$$ \log_2 \left( \frac{x^2 + 4x – 3}{x – 1} \right) = 2 $$

Converting to exponential form:

$$ \frac{x^2 + 4x – 3}{x – 1} = 2^2 = 4 $$ $$ x^2 + 4x – 3 = 4x – 4 $$ $$ x^2 + 4x – 3 – 4x + 4 = 0 $$ $$ x^2 + 1 = 0 \Rightarrow x^2 = -1 $$

No real solution exists.

Solved Example 2.10

Find the value of $\log_{27} \left( \frac{1}{81} \right)$.

Solution:

$$ \log_{27} \left( \frac{1}{81} \right) = \log_{3^3} (3^{-4}) = \frac{-4}{3} \log_3 3 = -\frac{4}{3} $$

Concept Application Exercise 2.1

  1. Simplify: $\log_7 49 + \log_7 343 – \log_7 7$.
  2. Find the value of $\log_{25} 125 \cdot \log_{125} 625$.
  3. If $\log_a x = 3$ and $\log_a y = 5$, find the value of $\log_a (x^2 y^3)$.
  4. Solve for $x$: $\log_3 (x^2 – 3x – 5) = \log_3 (7 – 2x)$.
  5. Prove that $\log_a b \cdot \log_b c \cdot \log_c a = 1$.
  6. Find the value of $\log_{0.2} 125$.
  7. If $\log_2 3 = a$ and $\log_3 5 = b$, express $\log_{30} 8$ in terms of $a$ and $b$.
  8. Solve: $\log_5 (x^2 – 4x + 3) – \log_5 (x – 1) = 1$.
  9. If $\log_x 2 = \log_4 x$, find the value of $x$.
  10. Simplify: $\log_2 3 \cdot \log_3 4 \cdot \log_4 5 \cdots \log_{n-1} n$.