Solve: log₂(3x) × log₅(8) = log₅(a) + log₅(2a) for Variable X

Let's solve the problem step-by-step:

We start with the equation:

$\log_2 3x \times \log_5 8 = \log_5 a + \log_5 2a$

We simplify the right side using the product rule for logarithms:

$\log_5 a + \log_5 2a = \log_5 (a \cdot 2a) = \log_5 (2a^2)$

Next, we simplify $\log_5 8$ on the left side:

$\log_5 8 = \log_5 (2^3) = 3 \log_5 2$

Thus, we substitute into the original equation:

$\log_2 3x \times 3 \log_5 2 = \log_5 (2a^2)$

Now, divide both sides by $3 \log_5 2$:

$\log_2 3x = \frac{\log_5 (2a^2)}{3 \log_5 2}$

Using the change of base formula, express $\log_5 (2a^2)$ and $\log_5 2$ with base 2:

$\log_5 (2a^2) = \frac{\log_2 (2a^2)}{\log_2 5}$ $\log_5 2 = \frac{\log_2 2}{\log_2 5} = \frac{1}{\log_2 5}$

Substitute these into the equation:

$\log_2 3x = \frac{\log_2 (2a^2)}{3}$

This implies:

$\log_2 3x = \frac{1}{3} \log_2 (2a^2)$

Raising 2 to both sides of the equation to remove the logarithms:

$3x = (2a^2)^{\frac{1}{3}}$

Therefore, solving for $x$:

$x = \frac{1}{3} (2a^2)^{\frac{1}{3}} = \frac{1}{3} \cdot \sqrt[3]{2a^2}$

Thus, we conclude:

$x = \sqrt[3]{\frac{2a^2}{27}}$

Therefore, the value of $x$ in terms of $a$ is $\sqrt[3]{\frac{2a^2}{27}}$.