Math Problem Statement

Please solve this

Please solve the logarithmic inequalities provided in the image.

Solution

Let's solve each problem one by one from the image provided. I'll focus on each inequality and logarithmic equation.


Problem 492

log4(4x4)log2xwith the restriction x>1\log_4 (4x - 4) \leq \log_2 x \quad \text{with the restriction } x > 1

  1. Rewrite log4(4x4)\log_4 (4x - 4) as log2(4x4)2\frac{\log_2 (4x - 4)}{2}: log2(4x4)2log2x\frac{\log_2 (4x - 4)}{2} \leq \log_2 x

  2. Multiply both sides by 2: log2(4x4)2log2x\log_2 (4x - 4) \leq 2 \log_2 x

  3. Rewrite 2log2x2 \log_2 x as log2x2\log_2 x^2: log2(4x4)log2x2\log_2 (4x - 4) \leq \log_2 x^2

  4. Since the logarithmic function is increasing, we can remove the logs: 4x4x24x - 4 \leq x^2

  5. Rearrange this inequality: x24x+40x^2 - 4x + 4 \geq 0

  6. Factor: (x2)20(x - 2)^2 \geq 0

Since (x2)20(x - 2)^2 \geq 0 is true for all xx, the solution under the constraint x>1x > 1 is: x>1x > 1


Problem 493

(log1/4x)2+52log1/4x>32with 0<x<12 or x>64\left( \log_{1/4} x \right)^2 + \frac{5}{2} \log_{1/4} x > \frac{3}{2} \quad \text{with } 0 < x < \frac{1}{2} \text{ or } x > 64

  1. Let y=log1/4xy = \log_{1/4} x, then the equation becomes: y2+52y>32y^2 + \frac{5}{2} y > \frac{3}{2}

  2. Rearrange this as a quadratic inequality: y2+52y32>0y^2 + \frac{5}{2} y - \frac{3}{2} > 0

  3. Solve the quadratic inequality for yy using the quadratic formula: y=52±(52)2+4322y = \frac{-\frac{5}{2} \pm \sqrt{\left(\frac{5}{2}\right)^2 + 4 \cdot \frac{3}{2}}}{2}

Solving this will give values of yy, which you can then substitute back for xx.


Problem 494

1logx3logx<2with 110<x<1 or x>103\frac{1}{\log x} - 3 \log x < 2 \quad \text{with } \frac{1}{10} < x < 1 \text{ or } x > \sqrt[3]{10}

This is a complex inequality involving both logx\log x and 1logx\frac{1}{\log x}. We can let y=logxy = \log x and rewrite it as: 1y3y<2\frac{1}{y} - 3y < 2 Then solve for yy and back-substitute for xx.


Problem 495

log(3x)22log(4+x)<0with x>12 and x3\log (3 - x)^2 - 2 \log (4 + x) < 0 \quad \text{with } x > -\frac{1}{2} \text{ and } x \neq 3

  1. Simplify using logarithmic properties: 2log(3x)2log(4+x)<02 \log (3 - x) - 2 \log (4 + x) < 0

  2. Factor out the 2: 2(log(3x)log(4+x))<02 (\log (3 - x) - \log (4 + x)) < 0

  3. Simplify further: log3x4+x<0\log \frac{3 - x}{4 + x} < 0

Then solve for xx from this inequality.


Problem 496

log1/3(2x+8)log1/3(6x1)with x12\log_{1/3} (2x + 8) \geq \log_{1/3} (6x - 1) \quad \text{with } x \geq \frac{1}{2}

Since log1/3\log_{1/3} is a decreasing function, we can reverse the inequality: 2x+86x12x + 8 \leq 6x - 1 Then solve for xx.


Let me know if you'd like detailed steps for each solution or if there are specific parts you need more clarification on!


Questions for Further Exploration

  1. What happens if you have an inequality with mixed bases in logarithms?
  2. How do properties of logarithms change when the base is between 0 and 1?
  3. How do you handle equations where logarithmic expressions are squared?
  4. What additional techniques help with inequalities involving fractional logarithmic expressions?
  5. Can logarithmic inequalities have solutions that need domain restrictions?

Tip

For inequalities with logarithmic terms, consider converting all terms to the same base or variable if possible, as this can greatly simplify the problem.

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Math Problem Analysis

Mathematical Concepts

Logarithmic Inequalities
Logarithmic Properties
Algebraic Manipulations
Quadratic Inequalities

Formulas

log_a (x^m) = m * log_a(x)
log_a (xy) = log_a(x) + log_a(y)
Change of base formula: log_b(x) = log_c(x) / log_c(b)

Theorems

Properties of Logarithms
Rules of Inequalities (e.g., reversing inequality when multiplying by a negative)

Suitable Grade Level

Grades 10-12