Multiplication of Logarithms: Inequality

To solve this problem, we'll apply logarithmic properties and transformations:

Step 1: Adjust each term with logarithm properties to a common base. Start with the property that for any positive number $a$, $\log_b a = \frac{1}{\log_a b}$.

Step 2: We know:
$\log_{\frac{1}{7}} 9 = -\frac{\log_7 9}{\log_7 \frac{1}{7}} = \frac{-1}{\log_9 7}$ and
$\log_{\frac{1}{7}} 4 = -\frac{\log_7 4}{\log_7 \frac{1}{7}} = \frac{-1}{\log_4 7}$.

Step 3: Viewing $\log_3 5x$ in the canonical form, $\log_3 5x$.

Step 4: The inequality becomes $\log_3 5x \times \frac{-1}{\log_9 7} \ge \frac{-1}{\log_4 7}$.

Step 5: Multiply through by $-1$ (reversing inequality):
$\log_3 5x \times \frac{1}{\log_9 7} \le \frac{1}{\log_4 7}$.

Step 6: Cross multiply to clear fractions because all log values are positive:

$\log_3 5x \cdot \log_4 7 \le \log_9 7.$

Step 7: Reorganize: $\log_3 5x \le \frac{\log_9 7}{\log_4 7}$.

Step 8: Use fact $\log_3 5x = \log_3 5 + \log_3 x$.
$\log_3 x \le \frac{\log_9 7}{\log_4 7} - \log_3 5$

Step 9: Explicit values for simplification:
- $\log_3 5 = \frac{\log 5}{\log 3}$ (base conversion)
- $\log_9 7 = \frac{\log 7}{2\log 3}$ because $9 = 3^2$
- $\log_4 7 = \frac{\log 7}{2\log 2}$ because $4 = 2^2$.

Step 10: Reevaluate the inequality considering numeric values extracted:
Solve $3^{(\text{net inequality from above conditions})}$, leading inevitably:
$\log_3 x \le -5$.

Step 11: Evaluating to exponential expression $x = 3^{-5}: \leq \frac{1}{3^5} = \frac{1}{243}$.

From logarithmic inequality recalibration, the condition holds:
$0 < x \le \frac{1}{245}$

The solution is $0 < x \le \frac{1}{245}$.

To solve this problem, we'll follow these key steps:

• Separate the components inside the logarithm using the property: $\log_b(a \cdot c) = \log_b(a) + \log_b(c)$.
• Apply the power property: $\log_b(a^c) = c\log_b(a)$.
• Simplify the inequality and solve it.

Consider the inequality given:

$\log_{\frac{1}{3}}(e^2\ln x) < 3\log_{\frac{1}{3}}(2)$

Using the product property of logarithms, we can rewrite this as:

$\log_{\frac{1}{3}}(e^2) + \log_{\frac{1}{3}}(\ln x) < 3\log_{\frac{1}{3}}(2)$

Next, apply the power property to simplify $\log_{\frac{1}{3}}(e^2)$:

$2\log_{\frac{1}{3}}(e) + \log_{\frac{1}{3}}(\ln x) < 3\log_{\frac{1}{3}}(2)$

Let $a = \log_{\frac{1}{3}}(e)$ and $b = \log_{\frac{1}{3}}(2)$. The inequality becomes:

$2a + \log_{\frac{1}{3}}(\ln x) < 3b$

Rearrange to isolate $\log_{\frac{1}{3}}(\ln x)$:

$\log_{\frac{1}{3}}(\ln x) < 3b - 2a$

Since $\frac{1}{3}$ is less than 1, meaning the inequality reverses when converting back to exponential form:

$\ln x > \left(\frac{1}{3}\right)^{(3b - 2a)}$

Converting the expression on the right-hand side to exponential form:

$\ln x > (\frac{1}{3})^{\log_{\frac{1}{3}}(8)}$

This simplifies to:

$\ln x > \frac{1}{8}$

Take the exponential of both sides to solve for $x$:

$x > e^{\frac{1}{8}}$

Simplifying gives:

$x > \sqrt{8}$

Therefore, the solution to the problem is $\sqrt{8} < x$.

To solve this problem, we need to compare the expressions $\log_4 x \times \log_5 64$ and $\log_5 (x^3 + x^2 + x + 1)$.

First, calculate $\log_5 64$. We know that $64 = 4^3 = 2^6$. Therefore:
$\log_5 64 = \frac{\log_5 2^6}{\log_5 4} = \frac{6 \log_5 2}{2 \log_5 2} = 3$

Next, simplify the left-hand side expression $\log_4 x$. Using the change of base formula:
$\log_4 x = \frac{\log_5 x}{\log_5 4}$

Therefore, the left-hand side becomes:
$\frac{\log_5 x}{\log_5 4} \times 3 = \frac{3 \log_5 x}{2 \log_5 2}$

For the inequality:
$\frac{3 \log_5 x}{2 \log_5 2} \ge \log_5 (x^3 + x^2 + x + 1)$

We can now equate the right-hand side:
$\log_5 x^{3/2\log_5 2} \ge \log_5 (x^3 + x^2 + x + 1)$

This implies:
$x^{3/2\log_5 2} \ge x^3 + x^2 + x + 1$

Testing and analyzing this expression results in no valid $x$ satisfying the inequality within real values since exponential growth and polynomial terms do not align. Thus, the inequality cannot be satisfied, and no solution satisfies the given conditions.

Therefore, the solution to the problem is: No solution.