Math Problem Statement

Help me solve this problem

Help me solve this problem: ln(x^2 - 4e^2) < 1 + ln(3x)

Solution

The given inequality is:

ln(x24e2)<1+ln(3x)\ln(x^2 - 4e^2) < 1 + \ln(3x)

Steps to solve the inequality:

  1. Isolate the logarithmic terms: Subtract ln(3x)\ln(3x) from both sides to keep all logarithmic expressions on one side: ln(x24e2)ln(3x)<1\ln(x^2 - 4e^2) - \ln(3x) < 1

  2. Use logarithmic properties: Apply the property ln(a)ln(b)=ln(ab)\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right): ln(x24e23x)<1\ln\left(\frac{x^2 - 4e^2}{3x}\right) < 1

  3. Exponentiate both sides: To eliminate the natural logarithm, apply the exponential function (eln()=e^{\ln(\cdot)} = \cdot) to both sides: x24e23x<e1\frac{x^2 - 4e^2}{3x} < e^1 Simplifying: x24e23x<e\frac{x^2 - 4e^2}{3x} < e

  4. Solve the resulting inequality: Multiply both sides by 3x3x (assuming x>0x > 0 to keep 3x>03x > 0, since logarithms are only defined for positive arguments): x24e2<3xex^2 - 4e^2 < 3xe

  5. Rearrange terms: Move all terms to one side of the inequality: x23xe4e2<0x^2 - 3xe - 4e^2 < 0

  6. Solve the quadratic inequality: Treat this as a quadratic in xx: x23xe4e2=0x^2 - 3xe - 4e^2 = 0 Solve the quadratic equation using the quadratic formula: x=(3e)±(3e)24(1)(4e2)2(1)x = \frac{-(-3e) \pm \sqrt{(-3e)^2 - 4(1)(-4e^2)}}{2(1)} Simplifying: x=3e±9e2+16e22x = \frac{3e \pm \sqrt{9e^2 + 16e^2}}{2} x=3e±25e22x = \frac{3e \pm \sqrt{25e^2}}{2} x=3e±5e2x = \frac{3e \pm 5e}{2}

    The two solutions are: x1=3e+5e2=4e,x2=3e5e2=ex_1 = \frac{3e + 5e}{2} = 4e, \quad x_2 = \frac{3e - 5e}{2} = -e

  7. Determine the solution interval: The quadratic inequality x23xe4e2<0x^2 - 3xe - 4e^2 < 0 is satisfied between the roots x=ex = -e and x=4ex = 4e. However, we must also consider the domain of the original logarithmic expressions:

    • x24e2>0    x>2ex^2 - 4e^2 > 0 \implies x > 2e or x<2ex < -2e
    • 3x>0    x>03x > 0 \implies x > 0

    Combining these conditions, the solution must be 2e<x<4e2e < x < 4e.

Final Answer:

The solution to the inequality is:

2e<x<4e2e < x < 4e

Would you like a more detailed explanation or further clarifications?

Here are 5 related questions:

  1. How do you solve other inequalities involving logarithms?
  2. What happens if the quadratic discriminant is negative?
  3. Can logarithmic inequalities have complex number solutions?
  4. How do logarithmic properties help in simplifying equations?
  5. What is the domain of natural logarithms?

Tip: Always check the domain of logarithmic functions to avoid undefined values in the final solution.

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Math Problem Analysis

Mathematical Concepts

Logarithmic Inequalities
Quadratic Equations
Properties of Logarithms

Formulas

Logarithmic Property: ln(a) - ln(b) = ln(a/b)
Quadratic Formula: x = (-b ± √(b^2 - 4ac)) / 2a

Theorems

Quadratic Formula
Logarithmic Properties

Suitable Grade Level

Grades 11-12