Fundamentals of Logarithms Tutorial

Fundamentals of Logarithms Tutorial

Fundamentals of Logarithms

Key Definitions

  1. Logarithm: For $a > 0$, $a \neq 1$, and $x > 0$, the logarithm of $x$ to the base $a$ is denoted by $\log_a x$ and is defined as the unique real number $y$ such that $a^y = x$. That is, $$ \log_a x = y \iff a^y = x $$
  2. Common Logarithm: The logarithm with base 10 is called the common logarithm and is denoted by $\log x$ (without specifying the base). That is, $\log x = \log_{10} x$.
  3. Natural Logarithm: The logarithm with base $e$ (where $e \approx 2.71828\ldots$) is called the natural logarithm and is denoted by $\ln x$. That is, $\ln x = \log_e x$.
  4. Characteristic and Mantissa: For a positive real number, the logarithm can be expressed as $\log N = \text{characteristic} + \text{mantissa}$, where the characteristic is an integer and the mantissa is a non-negative real number less than 1.
  5. Antilogarithm: If $\log_a x = y$, then $x$ is called the antilogarithm of $y$ to the base $a$, written as $x = \text{antilog}_a y$.

Theory and Concepts

Relationship Between Exponents and Logarithms

Exponential and logarithmic functions are inverses of each other. For $a > 0$, $a \neq 1$:

$$ \log_a (a^x) = x \quad \text{for all } x \in \mathbb{R} $$ $$ a^{\log_a x} = x \quad \text{for all } x > 0 $$

Fundamental Laws of Logarithms

For $a > 0$, $a \neq 1$, $m, n > 0$, and $k \in \mathbb{R}$:

  1. Product Rule: $\log_a (mn) = \log_a m + \log_a n$
  2. Quotient Rule: $\log_a \left(\dfrac{m}{n}\right) = \log_a m – \log_a n$
  3. Power Rule: $\log_a (m^k) = k \log_a m$
  4. Base Change Rule: $\log_a m = \dfrac{\log_b m}{\log_b a}$ for any $b > 0$, $b \neq 1$
  5. $\log_a a = 1$ and $\log_a 1 = 0$
  6. $\log_a b = \dfrac{1}{\log_b a}$ (provided $b \neq 1$)
  7. $\log_{a^k} m = \dfrac{1}{k} \log_a m$ for $k \neq 0$

Important Properties

  1. $\log_a x = \log_a y \iff x = y$ (One-to-one property)
  2. If $a > 1$, then $\log_a x < \log_a y \iff x < y$
  3. If $0 < a < 1$, then $\log_a x < \log_a y \iff x > y$
  4. $\log_a x = \dfrac{\log x}{\log a}$ (Common logarithm form)
  5. $\log_a x = \dfrac{\ln x}{\ln a}$ (Natural logarithm form)

Graphical Representation

Class 12 Maths Solution

Solved Examples

Solved Example 1.1

Find the value of $\log_2 32$.

Solution:

Let $\log_2 32 = x$. Then by definition, $2^x = 32$. Since $32 = 2^5$, we have $2^x = 2^5$, which implies $x = 5$. Therefore, $\log_2 32 = 5$.

Solved Example 1.2

Find the value of $\log_5 \sqrt[3]{25}$.

Solution:

Let $x = \log_5 \sqrt[3]{25}$. Then $5^x = \sqrt[3]{25} = (25)^{1/3} = (5^2)^{1/3} = 5^{2/3}$. Since the bases are equal, we have $x = \dfrac{2}{3}$. Therefore, $\log_5 \sqrt[3]{25} = \dfrac{2}{3}$.

Solved Example 1.3

Express $\log_5 7$ in terms of $\log_{10} 5$ and $\log_{10} 7$.

Solution:

Using the change of base formula with base 10: $$ \log_5 7 = \frac{\log_{10} 7}{\log_{10} 5} $$ Thus, $\log_5 7 = \dfrac{\log 7}{\log 5}$ (where $\log$ denotes common logarithm).

Solved Example 1.4

Simplify: $\log_2 24 – \log_2 3$.

Solution:

Using the quotient rule: $$ \log_2 24 – \log_2 3 = \log_2 \left(\frac{24}{3}\right) = \log_2 8 $$ Since $8 = 2^3$, $\log_2 8 = 3$. Therefore, the simplified value is $3$.

Solved Example 1.5

If $\log_{10} 2 = 0.3010$ and $\log_{10} 3 = 0.4771$, find $\log_{10} 72$.

Solution:

\begin{align*} \log_{10} 72 &= \log_{10} (2^3 \times 3^2) \\ &= \log_{10} (2^3) + \log_{10} (3^2) \quad \text{(Product rule)} \\ &= 3\log_{10} 2 + 2\log_{10} 3 \quad \text{(Power rule)} \\ &= 3(0.3010) + 2(0.4771) \\ &= 0.9030 + 0.9542 \\ &= 1.8572 \end{align*}

Solved Example 1.6

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

Solution:

\begin{align*} \log_2 (x^2 – 4) &= 3 \\ \Rightarrow x^2 – 4 &= 2^3 \quad \text{(Definition of logarithm)} \\ \Rightarrow x^2 – 4 &= 8 \\ \Rightarrow x^2 &= 12 \\ \Rightarrow x &= \pm 2\sqrt{3} \end{align*} We must check the domain: $x^2 – 4 > 0 \Rightarrow x^2 > 4 \Rightarrow |x| > 2$. Both $2\sqrt{3} \approx 3.46$ and $-2\sqrt{3} \approx -3.46$ satisfy $|x| > 2$. Hence, both solutions are valid.

Solved Example 1.7

Prove that $\log_a b \cdot \log_b c \cdot \log_c a = 1$.

Solution:

Using the change of base formula: $$ \log_a b = \frac{\log b}{\log a}, \quad \log_b c = \frac{\log c}{\log b}, \quad \log_c a = \frac{\log a}{\log c} $$ Multiplying: $$ \log_a b \cdot \log_b c \cdot \log_c a = \frac{\log b}{\log a} \cdot \frac{\log c}{\log b} \cdot \frac{\log a}{\log c} = 1 $$

Solved Example 1.8

If $\log_2 x + \log_4 x + \log_8 x = \dfrac{11}{6}$, find $x$.

Solution:

Using the property $\log_{a^k} x = \frac{1}{k} \log_a x$: \begin{align*} \log_4 x &= \log_{2^2} x = \frac{1}{2} \log_2 x \\ \log_8 x &= \log_{2^3} x = \frac{1}{3} \log_2 x \end{align*} Therefore, $$ \log_2 x + \frac{1}{2} \log_2 x + \frac{1}{3} \log_2 x = \frac{11}{6} $$ $$ \left(1 + \frac{1}{2} + \frac{1}{3}\right) \log_2 x = \frac{11}{6} $$ $$ \frac{6 + 3 + 2}{6} \log_2 x = \frac{11}{6} $$ $$ \frac{11}{6} \log_2 x = \frac{11}{6} $$ $$ \log_2 x = 1 \Rightarrow x = 2 $$

Solved Example 1.9

Solve the inequality: $\log_{1/2} (x^2 – 5x + 7) > 0$.

Solution:

For base $a = \frac{1}{2}$ (which lies between 0 and 1), the logarithmic function is decreasing. Therefore, $$ \log_{1/2} (x^2 – 5x + 7) > 0 = \log_{1/2} 1 $$ Since the function is decreasing: $$ x^2 – 5x + 7 < 1 $$ $$ x^2 - 5x + 6 < 0 $$ $$ (x - 2)(x - 3) < 0 $$ $$ 2 < x < 3 $$ Also, domain condition: $x^2 - 5x + 7 > 0$. The discriminant $= 25 – 28 = -3 < 0$, so this quadratic is always positive. Hence, the solution is $x \in (2, 3)$.

Solved Example 1.10

Find the value of $\log_3 4 \cdot \log_4 5 \cdot \log_5 6 \cdot \log_6 7 \cdot \log_7 8 \cdot \log_8 9$.

Solution:

Using the change of base formula or the property $\log_a b \cdot \log_b c = \log_a c$: \begin{align*} &\log_3 4 \cdot \log_4 5 = \log_3 5 \\ &\log_3 5 \cdot \log_5 6 = \log_3 6 \\ &\log_3 6 \cdot \log_6 7 = \log_3 7 \\ &\log_3 7 \cdot \log_7 8 = \log_3 8 \\ &\log_3 8 \cdot \log_8 9 = \log_3 9 = \log_3 (3^2) = 2 \end{align*} Therefore, the product equals $2$.

Concept Application Exercise 1.1

  1. Find the value of $\log_{27} 81$.
  2. If $\log_{10} 2 = 0.3010$ and $\log_{10} 3 = 0.4771$, find the value of $\log_{10} (4.5)$.
  3. Simplify: $\log_5 75 – \log_5 3$.
  4. Solve for $x$: $\log_3 (x + 2) + \log_3 (x – 2) = 2$.
  5. Find the value of $\log_2 3 \cdot \log_3 4 \cdot \log_4 5 \cdot \log_5 6 \cdot \log_6 7 \cdot \log_7 8$.
  6. If $\log_2 (x^2 – 3x + 2) = 1$, find the values of $x$.
  7. Solve the inequality: $\log_{1/3} (x^2 – 4x + 3) < -1$.
  8. Find the value of $x$ satisfying: $\log_2 x + \log_4 x + \log_{16} x = \frac{21}{4}$.
  9. If $\log_a b = 2$ and $\log_b c = 3$, find the value of $\log_c a$.
  10. Prove that: $\log_a x = \log_b x \cdot \log_a b$.