Solve: log₁/₃(e²ln x) < 3log₁/₃(2) Logarithmic Inequality

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 $\sqrt{8} < x$.