Math Problem Statement

Question 34 until 36

Solution

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

Question 34: Solve the equation 42x+164+16=04^{2x+1} - 64 + 16 = 0

  1. Simplify the equation: 42x+164+16=04^{2x+1} - 64 + 16 = 0 42x+148=04^{2x+1} - 48 = 0 42x+1=484^{2x+1} = 48

  2. Express 48 as a power of 4: 48=41×1248 = 4^1 \times 12 However, this form is not easily solvable directly. Let's convert the base 4 term: 42x+1=(22)2x+1=24x+24^{2x+1} = (2^2)^{2x+1} = 2^{4x+2} So, 24x+2=482^{4x+2} = 48

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

  4. Solve for xx: 4x+2=log(48)log(2)4x + 2 = \frac{\log(48)}{\log(2)} x=log(48)log(2)24x = \frac{\frac{\log(48)}{\log(2)} - 2}{4} Substitute the values: xlog(48)/log(2)24=5.585243.58540.896x \approx \frac{\log(48)/\log(2) - 2}{4} = \frac{5.585 - 2}{4} \approx \frac{3.585}{4} \approx 0.896 x0.896x \approx 0.896

    Therefore, the solution is approximately x0.896x \approx 0.896.

Question 35: Solve the inequality x1<x232x+5x - 1 < x^2 - 3 \leq 2x + 5

  1. Consider the inequality x1<x23x - 1 < x^2 - 3: x2x2>0x^2 - x - 2 > 0 Factorize: (x2)(x+1)>0(x - 2)(x + 1) > 0

    Critical points: x=2x = 2 and x=1x = -1.

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

  2. Consider the second part x232x+5x^2 - 3 \leq 2x + 5: x22x80x^2 - 2x - 8 \leq 0 Factorize: (x4)(x+2)0(x - 4)(x + 2) \leq 0

    Critical points: x=2x = -2 and x=4x = 4.

    This implies xx belongs to the interval [2,4][-2, 4].

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

Question 36: Solve the inequality 3x+11x1\frac{3}{x+1} \leq \frac{1}{x-1}

  1. Start by simplifying the inequality: 3x+11x10\frac{3}{x+1} - \frac{1}{x-1} \leq 0 Combine the fractions: 3(x1)(x+1)(x+1)(x1)0\frac{3(x-1) - (x+1)}{(x+1)(x-1)} \leq 0 Simplify the numerator: 3x3x1(x+1)(x1)0\frac{3x - 3 - x - 1}{(x+1)(x-1)} \leq 0 2x4(x+1)(x1)0\frac{2x - 4}{(x+1)(x-1)} \leq 0 Factorize the numerator: 2(x2)(x+1)(x1)0\frac{2(x-2)}{(x+1)(x-1)} \leq 0

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

    Test intervals around these critical points:

    • For x<1x < -1, the expression is positive.
    • For 1<x<1-1 < x < 1, the expression is negative.
    • For 1<x<21 < x < 2, the expression is positive.
    • For x>2x > 2, the expression is positive.
  3. The inequality holds when x[1,1)[2,)x \in [-1, 1) \cup [2, \infty).


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.

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Math Problem Analysis

Mathematical Concepts

Exponential equations
Inequalities
Logarithms

Formulas

Exponential properties
Logarithm rules

Theorems

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Suitable Grade Level

Grades 10-12