Logarithm inequalities and applications questions

Logarithm inequalities and applications questions

Inequalities and Applications

Key Definitions

  1. Logarithmic Inequality: An inequality that contains a logarithm of a variable expression. For example, $\log_a x > b$ or $\log_a (x^2 – 3x) \le 2$.
  2. Domain of Logarithmic Function: For $f(x) = \log_a g(x)$, the domain is the set of all $x$ such that $g(x) > 0$, $a > 0$, and $a \neq 1$.
  3. Range of Logarithmic Function: For $f(x) = \log_a x$ with $a > 0$, $a \neq 1$, and domain $x > 0$, the range is all real numbers $\mathbb{R}$.
  4. Monotonicity Property:
    • If $a > 1$, then $\log_a x$ is strictly increasing: $x_1 < x_2 \iff \log_a x_1 < \log_a x_2$
    • If $0 < a < 1$, then $\log_a x$ is strictly decreasing: $x_1 < x_2 \iff \log_a x_1 > \log_a x_2$
  5. Critical Points in Inequalities: Points where the expression equals zero or becomes undefined. These partition the real line into intervals for sign analysis.

Theory and Concepts

Domain of Logarithmic Functions

For $f(x) = \log_a (h(x))$, the domain is determined by:

  1. $h(x) > 0$ (argument must be positive)
  2. $a > 0$ and $a \neq 1$ (base conditions)
Class 12 Maths Solution

Solving Logarithmic Inequalities

Step-by-step approach:

  1. Determine the domain of all logarithmic expressions.
  2. If necessary, combine logarithms using product/quotient rules.
  3. Express the inequality in the form $\log_a f(x) \le \log_a g(x)$ (or $\ge$, $<$, $>$).
  4. Apply the monotonicity property based on the base $a$:
    • If $a > 1$: $f(x) \le g(x)$ (inequality direction preserved)
    • If $0 < a < 1$: $f(x) \ge g(x)$ (inequality direction reversed)
  5. Solve the resulting algebraic inequality.
  6. Intersect the solution with the domain conditions.

Important Results

  1. For $a > 1$: $$ \log_a x > 0 \iff x > 1 $$ $$ \log_a x < 0 \iff 0 < x < 1 $$
  2. For $0 < a < 1$: $$ \log_a x > 0 \iff 0 < x < 1 $$ $$ \log_a x < 0 \iff x > 1 $$
  3. $\log_a x \ge b \implies \begin{cases} x \ge a^b & \text{if } a > 1 \\ 0 < x \le a^b & \text{if } 0 < a < 1 \end{cases}$
Class 12 Maths Solution

Solved Examples

Solved Example 5.1

Solve the inequality: $\log_{1/3} (x^2 – 5x + 7) > 0$.

Solution:

Domain condition: $x^2 – 5x + 7 > 0$

Discriminant: $25 – 28 = -3 < 0$, so the quadratic is always positive. Domain: $x \in \mathbb{R}$.

Since base $a = \frac{1}{3}$ satisfies $0 < a < 1$, the logarithmic function is decreasing.

$$ \log_{1/3} (x^2 – 5x + 7) > 0 = \log_{1/3} 1 $$

Since the function is decreasing:

$$ x^2 – 5x + 7 < 1 $$ $$ x^2 - 5x + 6 < 0 $$ $$ (x - 2)(x - 3) < 0 $$

Therefore, $2 < x < 3$.

The solution set is $x \in (2, 3)$.

Solved Example 5.2

Solve: $\log_2 (x^2 – 4x – 5) \le 3$.

Solution:

Domain: $x^2 – 4x – 5 > 0 \implies (x – 5)(x + 1) > 0 \implies x < -1 \text{ or } x > 5$.

Since base $a = 2 > 1$, the function is increasing:

$$ \log_2 (x^2 – 4x – 5) \le 3 = \log_2 8 $$ $$ x^2 – 4x – 5 \le 8 $$ $$ x^2 – 4x – 13 \le 0 $$

Roots: $x = \frac{4 \pm \sqrt{16 + 52}}{2} = \frac{4 \pm \sqrt{68}}{2} = \frac{4 \pm 2\sqrt{17}}{2} = 2 \pm \sqrt{17}$

So $2 – \sqrt{17} \le x \le 2 + \sqrt{17}$. Intersecting with domain: $x \in (-\infty, -1) \cup (5, \infty)$ gives:

$$ x \in [2 – \sqrt{17}, -1) \cup (5, 2 + \sqrt{17}] $$

Solved Example 5.3

Solve: $\log_x 2 > \log_x 3$.

Solution:

Domain: $x > 0$, $x \neq 1$.

The inequality is $\log_x 2 – \log_x 3 > 0 \implies \log_x \left(\frac{2}{3}\right) > 0$.

Consider two cases:

Case 1: $x > 1$ (base > 1, function increasing)

$$ \log_x \left(\frac{2}{3}\right) > 0 \implies \frac{2}{3} > x^0 = 1 \implies \frac{2}{3} > 1 \quad \text{(False)} $$

No solution in this case.

Case 2: $0 < x < 1$ (base between 0 and 1, function decreasing)

$$ \log_x \left(\frac{2}{3}\right) > 0 \implies \frac{2}{3} < x^0 = 1 \implies \frac{2}{3} < 1 \quad \text{(True for all $x$ in this domain)} $$

Therefore, the solution is $0 < x < 1$.

Solved Example 5.4

Find the domain of $f(x) = \log_{|x-1|} (x^2 – 5x + 6)$.

Solution:

Domain conditions:

  1. Base $|x-1| > 0$ and $|x-1| \neq 1 \implies x \neq 1$ and $|x-1| \neq 1 \implies x \neq 0, 2$
  2. Argument $x^2 – 5x + 6 > 0 \implies (x – 2)(x – 3) > 0 \implies x < 2 \text{ or } x > 3$

Combining: $x < 2$ and $x \neq 0, 1, 2$ gives $x \in (-\infty, 0) \cup (0, 1) \cup (1, 2)$. Also $x > 3$ gives $x \in (3, \infty)$.

Therefore, domain $= (-\infty, 0) \cup (0, 1) \cup (1, 2) \cup (3, \infty)$.

Solved Example 5.5

Solve: $\log_{1/2} (x^2 – 3x + 2) \ge -1$.

Solution:

Domain: $x^2 – 3x + 2 > 0 \implies (x – 1)(x – 2) > 0 \implies x < 1 \text{ or } x > 2$.

Base $a = \frac{1}{2} \in (0, 1)$, so function is decreasing:

$$ \log_{1/2} (x^2 – 3x + 2) \ge -1 = \log_{1/2} \left(\frac{1}{2}\right)^{-1} = \log_{1/2} 2 $$

Since decreasing:

$$ x^2 – 3x + 2 \le 2 $$ $$ x^2 – 3x \le 0 $$ $$ x(x – 3) \le 0 \implies 0 \le x \le 3 $$

Intersect with domain: $x \in [0, 1) \cup (2, 3]$.

Solved Example 5.6

Find the range of $f(x) = \log_2 (x^2 – 2x + 3)$.

Solution:

First, find the minimum value of $g(x) = x^2 – 2x + 3$.

$$ g(x) = (x – 1)^2 + 2 \ge 2 $$

So $g(x) \in [2, \infty)$. Since $f(x) = \log_2 g(x)$ is increasing for $g(x) > 0$:

$$ f(x) \ge \log_2 2 = 1 $$

Also, as $x \to \pm\infty$, $g(x) \to \infty$, so $f(x) \to \infty$.

Therefore, range $= [1, \infty)$.

Solved Example 5.7

Solve: $\log_{0.2} (x^2 – 2x – 3) < \log_{0.2} (x + 5)$.

Solution:

Domain: $x^2 – 2x – 3 > 0 \implies (x – 3)(x + 1) > 0 \implies x < -1 \text{ or } x > 3$

Also $x + 5 > 0 \implies x > -5$

Combined domain: $x \in (-5, -1) \cup (3, \infty)$. Base $a = 0.2 = \frac{1}{5} \in (0, 1)$, so function is decreasing:

$$ x^2 – 2x – 3 > x + 5 $$ $$ x^2 – 3x – 8 > 0 $$

Roots: $x = \frac{3 \pm \sqrt{9 + 32}}{2} = \frac{3 \pm \sqrt{41}}{2}$

So $x < \frac{3 - \sqrt{41}}{2}$ or $x > \frac{3 + \sqrt{41}}{2}$. Now $\frac{3 – \sqrt{41}}{2} \approx -1.7015$ and $\frac{3 + \sqrt{41}}{2} \approx 4.7015$.

Intersect with domain: $x \in (-5, -1.7015) \cup (4.7015, \infty)$.

Solved Example 5.8

Find the domain of $f(x) = \sqrt{\log_{0.5} (x^2 – 2x)}$.

Solution:

For the square root to be defined:

$$ \log_{0.5} (x^2 – 2x) \ge 0 $$

Domain of logarithm: $x^2 – 2x > 0 \implies x(x – 2) > 0 \implies x < 0 \text{ or } x > 2$.

Base $a = 0.5 \in (0, 1)$, so function is decreasing:

$$ \log_{0.5} (x^2 – 2x) \ge 0 = \log_{0.5} 1 $$ $$ x^2 – 2x \le 1 $$ $$ x^2 – 2x – 1 \le 0 $$

Roots: $x = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}$

So $1 – \sqrt{2} \le x \le 1 + \sqrt{2}$. Now $1 – \sqrt{2} \approx -0.414$ and $1 + \sqrt{2} \approx 2.414$.

Intersect with $x < 0 \text{ or } x > 2$: $x \in [1 – \sqrt{2}, 0) \cup (2, 1 + \sqrt{2}]$.

Solved Example 5.9

Find the range of $f(x) = \log_{1/2} (x^2 + 2x + 3)$.

Solution:

$g(x) = x^2 + 2x + 3 = (x + 1)^2 + 2 \ge 2$.

Since base $a = \frac{1}{2} \in (0, 1)$, $f(x) = \log_{1/2} g(x)$ is decreasing.

As $g(x)$ increases from $2$ to $\infty$, $f(x)$ decreases from $\log_{1/2} 2$ to $-\infty$.

$$ \log_{1/2} 2 = \frac{\log_2 2}{\log_2 (1/2)} = \frac{1}{-1} = -1 $$

Therefore, range $= (-\infty, -1]$.

Solved Example 5.10

Solve: $\log_3 (2x – 1) \cdot \log_2 (3x – 2) \ge 0$.

Solution:

Domain: $2x – 1 > 0 \implies x > \frac{1}{2}$ and $3x – 2 > 0 \implies x > \frac{2}{3}$. So domain: $x > \frac{2}{3}$.

Let $A = \log_3 (2x – 1)$ and $B = \log_2 (3x – 2)$. We need $A \cdot B \ge 0$, i.e., $A$ and $B$ have the same sign or at least one is zero.

Case 1: $A = 0 \implies 2x – 1 = 1 \implies x = 1$ (valid)

Case 2: $B = 0 \implies 3x – 2 = 1 \implies x = 1$ (valid)

Case 3: $A > 0$ and $B > 0$

$A > 0 \implies 2x – 1 > 1 \implies x > 1$

$B > 0 \implies 3x – 2 > 1 \implies x > 1$

So $x > 1$ satisfies.

Case 4: $A < 0$ and $B < 0$

$A < 0 \implies 2x - 1 < 1 \implies x < 1$

$B < 0 \implies 3x - 2 < 1 \implies x < 1$

With domain $x > \frac{2}{3}$, we get $\frac{2}{3} < x < 1$.

Therefore, solution set: $x \in \left(\frac{2}{3}, 1\right] \cup [1, \infty) = \left(\frac{2}{3}, \infty\right)$.

Concept Application Exercise 5.1

  1. Solve: $\log_{1/2} (x^2 – 4x + 3) \le 0$.
  2. Find the domain of $f(x) = \log_{x-2} (x^2 – 3x + 2)$.
  3. Solve: $\log_2 (x – 1) + \log_2 (x – 2) < 1$.
  4. Find the range of $f(x) = \log_{1/3} (x^2 – 4x + 5)$.
  5. Solve: $\log_x 5 > \log_x 7$.
  6. Find the domain of $f(x) = \frac{1}{\sqrt{\log_{0.5} (x^2 – 4x + 3)}}$.
  7. Solve: $\log_{1/3} (x^2 – 5x + 7) < \log_{1/3} (2x - 1)$.
  8. Find the range of $f(x) = \log_2 (4 – x^2)$.
  9. Solve: $\log_2 (x + 3) \cdot \log_{1/2} (x – 1) \le 0$.
  10. Find the domain of $f(x) = \sqrt{\log_{0.2} (x^2 – 4x + 3)}$.