Solving Exponential Equations worksheet with math problems for students to solve.

To solve the given problem, we need to find all values of \( x \) that satisfy each of the following inequalities. Let's go through each inequality step by step.

---

(a) \( 2x - 3 < 7 \)

1. Start with the inequality:
\[
2x - 3 < 7
\]

2. Add 3 to both sides to isolate the term with \( x \):
\[
2x - 3 + 3 < 7 + 3
\]
\[
2x < 10
\]

3. Divide both sides by 2 to solve for \( x \):
\[
\frac{2x}{2} < \frac{10}{2}
\]
\[
x < 5
\]

Thus, the solution is:
\[
\boxed{x < 5}
\]

---

(b) \( x^2 - 4 \geq 0 \)

1. Start with the inequality:
\[
x^2 - 4 \geq 0
\]

2. Factor the left-hand side:
\[
(x - 2)(x + 2) \geq 0
\]

3. Determine the critical points by setting each factor to zero:
\[
x - 2 = 0 \quad \Rightarrow \quad x = 2
\]
\[
x + 2 = 0 \quad \Rightarrow \quad x = -2
\]

4. These critical points divide the real number line into three intervals: \( (-\infty, -2) \), \( (-2, 2) \), and \( (2, \infty) \). We test a point in each interval to determine where the inequality holds.

- For \( x \in (-\infty, -2) \), choose \( x = -3 \):
\[
(x - 2)(x + 2) = (-3 - 2)(-3 + 2) = (-5)(-1) = 5 \quad (\text{positive})
\]

- For \( x \in (-2, 2) \), choose \( x = 0 \):
\[
(x - 2)(x + 2) = (0 - 2)(0 + 2) = (-2)(2) = -4 \quad (\text{negative})
\]

- For \( x \in (2, \infty) \), choose \( x = 3 \):
\[
(x - 2)(x + 2) = (3 - 2)(3 + 2) = (1)(5) = 5 \quad (\text{positive})
\]

5. The inequality \( (x - 2)(x + 2) \geq 0 \) holds when the product is positive or zero. From the test points, the solution includes the intervals \( (-\infty, -2] \) and \( [2, \infty) \).

Thus, the solution is:
\[
\boxed{x \leq -2 \text{ or } x \geq 2}
\]

---

(c) \( |x - 1| > 3 \)

1. Start with the inequality:
\[
|x - 1| > 3
\]

2. The absolute value inequality \( |x - 1| > 3 \) can be split into two separate inequalities:
\[
x - 1 > 3 \quad \text{or} \quad x - 1 < -3
\]

3. Solve each inequality separately:
- For \( x - 1 > 3 \):
\[
x > 4
\]

- For \( x - 1 < -3 \):
\[
x < -2
\]

4. Combine the solutions:
\[
x < -2 \quad \text{or} \quad x > 4
\]

Thus, the solution is:
\[
\boxed{x < -2 \text{ or } x > 4}
\]

---

(d) \( x^2 - 9 < 0 \)

1. Start with the inequality:
\[
x^2 - 9 < 0
\]

2. Factor the left-hand side:
\[
(x - 3)(x + 3) < 0
\]

3. Determine the critical points by setting each factor to zero:
\[
x - 3 = 0 \quad \Rightarrow \quad x = 3
\]
\[
x + 3 = 0 \quad \Rightarrow \quad x = -3
\]

4. These critical points divide the real number line into three intervals: \( (-\infty, -3) \), \( (-3, 3) \), and \( (3, \infty) \). We test a point in each interval to determine where the inequality holds.

- For \( x \in (-\infty, -3) \), choose \( x = -4 \):
\[
(x - 3)(x + 3) = (-4 - 3)(-4 + 3) = (-7)(-1) = 7 \quad (\text{positive})
\]

- For \( x \in (-3, 3) \), choose \( x = 0 \):
\[
(x - 3)(x + 3) = (0 - 3)(0 + 3) = (-3)(3) = -9 \quad (\text{negative})
\]

- For \( x \in (3, \infty) \), choose \( x = 4 \):
\[
(x - 3)(x + 3) = (4 - 3)(4 + 3) = (1)(7) = 7 \quad (\text{positive})
\]

5. The inequality \( (x - 3)(x + 3) < 0 \) holds when the product is negative. From the test points, the solution includes the interval \( (-3, 3) \).

Thus, the solution is:
\[
\boxed{-3 < x < 3}
\]

---

(e) \( x^2 - 6x + 8 \leq 0 \)

1. Start with the inequality:
\[
x^2 - 6x + 8 \leq 0
\]

2. Factor the quadratic expression:
\[
x^2 - 6x + 8 = (x - 2)(x - 4)
\]

3. Determine the critical points by setting each factor to zero:
\[
x - 2 = 0 \quad \Rightarrow \quad x = 2
\]
\[
x - 4 = 0 \quad \Rightarrow \quad x = 4
\]

4. These critical points divide the real number line into three intervals: \( (-\infty, 2) \), \( (2, 4) \), and \( (4, \infty) \). We test a point in each interval to determine where the inequality holds.

- For \( x \in (-\infty, 2) \), choose \( x = 0 \):
\[
(x - 2)(x - 4) = (0 - 2)(0 - 4) = (-2)(-4) = 8 \quad (\text{positive})
\]

- For \( x \in (2, 4) \), choose \( x = 3 \):
\[
(x - 2)(x - 4) = (3 - 2)(3 - 4) = (1)(-1) = -1 \quad (\text{negative})
\]

- For \( x \in (4, \infty) \), choose \( x = 5 \):
\[
(x - 2)(x - 4) = (5 - 2)(5 - 4) = (3)(1) = 3 \quad (\text{positive})
\]

5. The inequality \( (x - 2)(x - 4) \leq 0 \) holds when the product is negative or zero. From the test points, the solution includes the interval \( [2, 4] \).

Thus, the solution is:
\[
\boxed{2 \leq x \leq 4}
\]

---

Final Answers:

1. \( 2x - 3 < 7 \): \(\boxed{x < 5}\)
2. \( x^2 - 4 \geq 0 \): \(\boxed{x \leq -2 \text{ or } x \geq 2}\)
3. \( |x - 1| > 3 \): \(\boxed{x < -2 \text{ or } x > 4}\)
4. \( x^2 - 9 < 0 \): \(\boxed{-3 < x < 3}\)
5. \( x^2 - 6x + 8 \leq 0 \): \(\boxed{2 \leq x \leq 4}\)