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The image shows two inequalities in Ukrainian, with the instruction "Розв'яжіть нерівності" which means "Solve the inequalities."

The given inequalities are:

a) \( 2^{3x-1} \leq 4 \)

b) \( 0.7^{2x+1} > 0.49 \)

Let's solve each inequality step by step.

### Part (a) \( 2^{3x-1} \leq 4 \)

1. **Rewrite 4 as a power of 2**:
   \[
   4 = 2^2
   \]
   So the inequality becomes:
   \[
   2^{3x-1} \leq 2^2
   \]

2. **Since the bases are the same, equate the exponents**:
   \[
   3x - 1 \leq 2
   \]

3. **Solve for \(x\)**:
   \[
   3x \leq 2 + 1
   \]
   \[
   3x \leq 3
   \]
   \[
   x \leq 1
   \]

Thus, the solution to part (a) is \( x \leq 1 \).

---

### Part (b) \( 0.7^{2x+1} > 0.49 \)

1. **Rewrite 0.49 as a power of 0.7**:
   \[
   0.49 = 0.7^2
   \]
   So the inequality becomes:
   \[
   0.7^{2x+1} > 0.7^2
   \]

2. **Since the base \(0.7\) is less than 1, reverse the inequality when equating exponents**:
   \[
   2x + 1 < 2
   \]

3. **Solve for \(x\)**:
   \[
   2x < 2 - 1
   \]
   \[
   2x < 1
   \]
   \[
   x < \frac{1}{2}
   \]

Thus, the solution to part (b) is \( x < \frac{1}{2} \).

---

### Final Answer:
- (a) \( x \leq 1 \)
- (b) \( x < \frac{1}{2} \)

Would you like more details or have any questions? Here are some related problems for practice:

1. Solve \( 3^{2x+1} > 9 \).
2. Solve \( 5^{x+1} \geq 25 \).
3. Solve \( 4^{x-2} < 16 \).
4. Solve \( 0.5^{3x} \leq 0.25 \).
5. Solve \( 0.8^{x+1} > 0.64 \).

**Tip:** When solving exponential inequalities, always pay attention to the base. If the base is less than 1, the inequality direction will reverse when you take the logarithm or equate the exponents.

Solve the inequalities: a) 2^{3x-1} ≤ 4; b) 0.7^{2x+1} > 0.49

Step-by-step solution to solve exponential inequalities: 2^{3x-1} ≤ 4 and 0.7^{2x+1} > 0.49. Learn how to handle powers and exponents in inequalities, using key algebraic principles. Suitable for high school students in Grades 9-12.

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