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Here are the solutions for questions 34 to 36 from the image provided:

### Question 34: Solve the equation \( 4^{2x+1} - 64 + 16 = 0 \)

1. Simplify the equation:
   \[
   4^{2x+1} - 64 + 16 = 0
   \]
   \[
   4^{2x+1} - 48 = 0
   \]
   \[
   4^{2x+1} = 48
   \]

2. Express 48 as a power of 4:
   \[
   48 = 4^1 \times 12
   \]
   However, this form is not easily solvable directly. Let's convert the base 4 term:
   \[
   4^{2x+1} = (2^2)^{2x+1} = 2^{4x+2}
   \]
   So,
   \[
   2^{4x+2} = 48
   \]

3. Taking the logarithm of both sides:
   \[
   (4x+2) \log(2) = \log(48)
   \]

4. Solve for \( x \):
   \[
   4x + 2 = \frac{\log(48)}{\log(2)}
   \]
   \[
   x = \frac{\frac{\log(48)}{\log(2)} - 2}{4}
   \]
   Substitute the values:
   \[
   x \approx \frac{\log(48)/\log(2) - 2}{4} = \frac{5.585 - 2}{4} \approx \frac{3.585}{4} \approx 0.896
   \]
   \[
   x \approx 0.896
   \]

   Therefore, the solution is approximately \( x \approx 0.896 \).

### Question 35: Solve the inequality \( x - 1 < x^2 - 3 \leq 2x + 5 \)

1. Consider the inequality \( x - 1 < x^2 - 3 \):
   \[
   x^2 - x - 2 > 0
   \]
   Factorize:
   \[
   (x - 2)(x + 1) > 0
   \]

   Critical points: \( x = 2 \) and \( x = -1 \).

   This implies \( x \) belongs to the intervals \( (-\infty, -1) \cup (2, \infty) \).

2. Consider the second part \( x^2 - 3 \leq 2x + 5 \):
   \[
   x^2 - 2x - 8 \leq 0
   \]
   Factorize:
   \[
   (x - 4)(x + 2) \leq 0
   \]

   Critical points: \( x = -2 \) and \( x = 4 \).

   This implies \( x \) belongs to the interval \( [-2, 4] \).

3. The solution for the inequality combines the results:
   \[
   x \in (2, 4]
   \]

### Question 36: Solve the inequality \( \frac{3}{x+1} \leq \frac{1}{x-1} \)

1. Start by simplifying the inequality:
   \[
   \frac{3}{x+1} - \frac{1}{x-1} \leq 0
   \]
   Combine the fractions:
   \[
   \frac{3(x-1) - (x+1)}{(x+1)(x-1)} \leq 0
   \]
   Simplify the numerator:
   \[
   \frac{3x - 3 - x - 1}{(x+1)(x-1)} \leq 0
   \]
   \[
   \frac{2x - 4}{(x+1)(x-1)} \leq 0
   \]
   Factorize the numerator:
   \[
   \frac{2(x-2)}{(x+1)(x-1)} \leq 0
   \]

2. Determine the critical points:
   \( x = 2, -1, 1 \).

   Test intervals around these critical points:
   - For \( x < -1 \), the expression is positive.
   - For \( -1 < x < 1 \), the expression is negative.
   - For \( 1 < x < 2 \), the expression is positive.
   - For \( x > 2 \), the expression is positive.

3. The inequality holds when \( x \in [-1, 1) \cup [2, \infty) \).

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Let me know if you need further explanations or details on any step. Would you like to go over these or explore some additional problems?

### 8 Related Questions:
1. How do you solve exponential equations involving different bases?
2. What are the techniques for solving quadratic inequalities?
3. Can inequalities involving rational expressions be solved graphically?
4. What is the difference between strict and non-strict inequalities?
5. How can logarithmic and exponential functions be combined in equations?
6. What methods are useful for factoring cubic polynomials?
7. How do you determine intervals of increase and decrease for a function?
8. What are common mistakes to avoid when solving inequalities?

### Tip:
When solving inequalities, always consider the critical points where the expression changes sign, and don't forget to test intervals around these points to determine where the inequality holds true.

Explore how to solve exponential equations and inequalities, including logarithmic conversions. Detailed step-by-step solutions provided for problems involving exponential functions and logarithms. Suitable for students in Grades 10-12.

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