Math Problem Statement

Solve the logarithmic inequality: log_7(1/7) * log_8(1/8) * log_-x(1/7) / (log_8(1/8) * log_7(1/7) * log_-x(1/8)) > 8.

Solution

To solve the given inequality, let's carefully analyze it step by step. The inequality is:

log7(17)log8(18)log(x)(17)log8(18)log7(17)log(x)(18)>8\frac{\log_{7} \left( \frac{1}{7} \right) \cdot \log_{8} \left( \frac{1}{8} \right) \cdot \log_{(-x)} \left( \frac{1}{7} \right)}{\log_{8} \left( \frac{1}{8} \right) \cdot \log_{7} \left( \frac{1}{7} \right) \cdot \log_{(-x)} \left( \frac{1}{8} \right)} > 8

Key Steps:

  1. Simplify logarithmic expressions:

    • log7(17)=1\log_{7} \left( \frac{1}{7} \right) = -1, since 17=71\frac{1}{7} = 7^{-1}.
    • Similarly, log8(18)=1\log_{8} \left( \frac{1}{8} \right) = -1.
    • So, the numerator becomes: (1)(1)log(x)(17)=log(x)(17).(-1) \cdot (-1) \cdot \log_{(-x)} \left( \frac{1}{7} \right) = \log_{(-x)} \left( \frac{1}{7} \right).
    • The denominator becomes: (1)(1)log(x)(18)=log(x)(18).(-1) \cdot (-1) \cdot \log_{(-x)} \left( \frac{1}{8} \right) = \log_{(-x)} \left( \frac{1}{8} \right).

    Hence, the inequality simplifies to: log(x)(17)log(x)(18)>8.\frac{\log_{(-x)} \left( \frac{1}{7} \right)}{\log_{(-x)} \left( \frac{1}{8} \right)} > 8.

  2. Apply the change of base formula: Using the change of base formula logab=lnblna\log_{a} b = \frac{\ln b}{\ln a}, rewrite the inequality: ln(17)ln(x)ln(18)ln(x)>8.\frac{\frac{\ln \left( \frac{1}{7} \right)}{\ln (-x)}}{\frac{\ln \left( \frac{1}{8} \right)}{\ln (-x)}} > 8.

    Since ln(x)\ln (-x) is common in the numerator and denominator, it cancels out, leaving: ln(17)ln(18)>8.\frac{\ln \left( \frac{1}{7} \right)}{\ln \left( \frac{1}{8} \right)} > 8.

  3. Simplify the natural logarithms: ln(17)=ln(1)ln(7)=ln(7)\ln \left( \frac{1}{7} \right) = \ln(1) - \ln(7) = -\ln(7), and similarly, ln(18)=ln(8)\ln \left( \frac{1}{8} \right) = -\ln(8). Substituting these: ln(7)ln(8)>8.\frac{-\ln(7)}{-\ln(8)} > 8.

    The negatives cancel out: ln(7)ln(8)>8.\frac{\ln(7)}{\ln(8)} > 8.

  4. Interpret the inequality: This implies: ln(7)>8ln(8).\ln(7) > 8 \cdot \ln(8).

  5. Solve for xx:

    • The natural logarithms ln(7)\ln(7) and ln(8)\ln(8) are constants. Substituting their approximate values: ln(7)1.945\ln(7) \approx 1.945 and ln(8)2.079\ln(8) \approx 2.079.
    • The inequality becomes: 1.945>82.079.1.945 > 8 \cdot 2.079.
    • Calculate the right-hand side: 82.079=16.632.8 \cdot 2.079 = 16.632.
    • Clearly, 1.945<16.6321.945 < 16.632, so the inequality is never true.

Conclusion:

The inequality has no solution because it is impossible for ln(7)>8ln(8)\ln(7) > 8 \cdot \ln(8).

Would you like me to clarify any step or discuss logarithmic properties further?


Related Questions:

  1. What is the change of base formula and how is it applied in logarithmic problems?
  2. Can a logarithmic base ever be negative, and what does it imply for the domain of the function?
  3. How do you simplify logarithmic expressions involving fractions, like loga(1b)\log_{a} \left( \frac{1}{b} \right)?
  4. What are some real-world applications of logarithmic inequalities?
  5. How can you numerically solve logarithmic inequalities using software tools?

Tip:

Always check the domain of the logarithmic functions in inequalities, as they impose constraints on the values of xx.

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Math Problem Analysis

Mathematical Concepts

Logarithmic Inequalities
Properties of Logarithms
Change of Base Formula

Formulas

log_a(b) = ln(b) / ln(a)
log_a(1/b) = -log_a(b)

Theorems

Properties of Logarithms
Logarithmic Inequality Theorem

Suitable Grade Level

Undergraduate or advanced high school (Grades 11-12)