Math Problem Statement

lnx1ln(x+1)3<0\frac{\ln x-1}{\ln\left(x+1\right)-3}<0

Solution

We are tasked with solving the inequality:

lnx1ln(x+1)3<0\frac{\ln x - 1}{\ln(x+1) - 3} < 0

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: lnx1=0\ln x - 1 = 0

Solving for xx:

lnx=1    x=e\ln x = 1 \implies x = e

So, the numerator is zero at x=ex = e.

1.2 Denominator: ln(x+1)3=0\ln(x+1) - 3 = 0

Solving for xx:

ln(x+1)=3    x+1=e3    x=e31\ln(x+1) = 3 \implies x + 1 = e^3 \implies x = e^3 - 1

So, the denominator is zero at x=e31x = e^3 - 1.

Step 2: Analyze the sign of the fraction

The fraction changes sign at the critical points x=ex = e and x=e31x = e^3 - 1. We can analyze the sign of the fraction in the intervals determined by these points: (,e)(-\infty, e), (e,e31)(e, e^3 - 1), and (e31,)(e^3 - 1, \infty).

2.1 Interval (,e)(-\infty, e)

  • xx must be greater than 0 for the logarithms to be defined. Hence, we only consider x>0x > 0.
  • In this interval, both lnx1\ln x - 1 and ln(x+1)3\ln(x+1) - 3 are negative because lnx<1\ln x < 1 and ln(x+1)<3\ln(x+1) < 3. Thus, the fraction is positive.

2.2 Interval (e,e31)(e, e^3 - 1)

  • In this interval, lnx1>0\ln x - 1 > 0 because x>ex > e, but ln(x+1)3<0\ln(x+1) - 3 < 0 because x+1<e3x + 1 < e^3. Therefore, the fraction is negative.

2.3 Interval (e31,)(e^3 - 1, \infty)

  • For x>e31x > e^3 - 1, both lnx1>0\ln x - 1 > 0 and ln(x+1)3>0\ln(x+1) - 3 > 0, so the fraction is positive.

Step 3: Solve the inequality

We want the fraction to be less than zero, which occurs in the interval (e,e31)(e, e^3 - 1).

Final Solution:

The solution to the inequality is:

x(e,e31)x \in (e, e^3 - 1)

Additional Notes:

  • The function is undefined at x=0x = 0, so xx must be positive.
  • The fraction is zero at x=ex = e and undefined at x=e31x = e^3 - 1.

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:

  1. How does the sign of the fraction change when analyzing different intervals of logarithmic functions?
  2. What happens if we consider a similar inequality involving square roots or exponents?
  3. Can you derive the values of xx when the fraction equals zero, and how would this affect the inequality?
  4. How can we generalize the approach for solving inequalities involving more complex logarithmic expressions?
  5. 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