Let's solve the inequality:

$3 \cdot \log_5(x - 4) > \frac{6}{\log_5(x - 4) + 1}.$

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Step 1: Let $y = \log_5(x - 4)$

This substitution simplifies the inequality. Recall that $\log_5(x - 4)$ is valid only when $x - 4 > 0$, so:

$x > 4.$

The inequality becomes:

$3y > \frac{6}{y + 1}.$

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Step 2: Eliminate the fraction

Multiply both sides by $y + 1$, assuming $y \neq -1$ (to avoid division by zero):

$3y(y + 1) > 6.$

Simplify:

$3y^2 + 3y - 6 > 0.$

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Step 3: Solve the quadratic inequality

Divide through by 3:

$y^2 + y - 2 > 0.$

Factorize the quadratic:

$(y + 2)(y - 1) > 0.$

The critical points are $y = -2$ and $y = 1$. Using a sign test, determine where the inequality is true:

• For $y < -2$, $(y + 2)(y - 1) > 0$: false (both terms are negative).
• For $-2 < y < 1$, $(y + 2)(y - 1) > 0$: false (one term is positive, the other is negative).
• For $y > 1$, $(y + 2)(y - 1) > 0$: true (both terms are positive).

Thus, the solution for $y$ is:

$y > 1.$

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Step 4: Back-substitute $y = \log_5(x - 4)$

We now solve:

$\log_5(x - 4) > 1.$

Recall that $\log_5(a) > b$ implies $a > 5^b$. Therefore:

$x - 4 > 5^1 \quad \Rightarrow \quad x - 4 > 5 \quad \Rightarrow \quad x > 9.$

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Step 5: Combine with initial condition

Initially, we had $x > 4$. Combining this with $x > 9$, the solution is:

$x > 9.$

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Final Answer:

$\boxed{x > 9.}$

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Let me know if you'd like a detailed explanation of any step or if you have more questions!

Related Questions:

1. What are the properties of logarithmic inequalities?
2. How do you perform a sign test for quadratic inequalities?
3. How do you handle equations with multiple logarithms?
4. Can logarithmic equations have extraneous solutions?
5. How do transformations of logarithmic graphs affect inequality solutions?

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Tip: Always check the domain of a logarithmic inequality before solving to ensure all expressions are valid!