Math Problem Statement
Solution
We are tasked with solving the inequality:
Step 1: Identify the critical points
To understand when the fraction is negative, we need to identify when the numerator and denominator are zero, as these are potential boundaries.
1.1 Numerator:
Solving for :
So, the numerator is zero at .
1.2 Denominator:
Solving for :
So, the denominator is zero at .
Step 2: Analyze the sign of the fraction
The fraction changes sign at the critical points and . We can analyze the sign of the fraction in the intervals determined by these points: , , and .
2.1 Interval
- must be greater than 0 for the logarithms to be defined. Hence, we only consider .
- In this interval, both and are negative because and . Thus, the fraction is positive.
2.2 Interval
- In this interval, because , but because . Therefore, the fraction is negative.
2.3 Interval
- For , both and , so the fraction is positive.
Step 3: Solve the inequality
We want the fraction to be less than zero, which occurs in the interval .
Final Solution:
The solution to the inequality is:
Additional Notes:
- The function is undefined at , so must be positive.
- The fraction is zero at and undefined at .
Would you like further details on any part of the solution? Here's a tip:
Tip: Always pay attention to the domain of logarithmic functions. They are only defined for positive real numbers, which is a key consideration when analyzing inequalities involving logarithms.
Further Exploration:
- How does the sign of the fraction change when analyzing different intervals of logarithmic functions?
- What happens if we consider a similar inequality involving square roots or exponents?
- Can you derive the values of when the fraction equals zero, and how would this affect the inequality?
- How can we generalize the approach for solving inequalities involving more complex logarithmic expressions?
- What are some graphical methods to visualize such inequalities?
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Math Problem Analysis
Mathematical Concepts
Logarithmic Functions
Inequalities
Critical Points
Sign Analysis
Formulas
\ln x = 1
\ln(x+1) = 3
x = e
x = e^3 - 1
Theorems
Logarithmic Inequality Theorem
Sign Analysis Theorem
Suitable Grade Level
Grades 11-12