For some positive real number $x$, if $\log_{\sqrt{3}}(x) + \frac{\log_x (25)}{\log_x (0.008)} = \frac{16}{3}$, then the value of $\log_3(3x^2)$ is

First, let us identify what we're working with:

$25 = 5^2$

$0.008 = \frac{8}{1000} = \frac{2^3}{10^3} = \frac{2^3}{(2 \times 5)^3} = \frac{1}{5^3} = 5^{-3}$

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Now we can rewrite the second term:

$\frac{\log_x (25)}{\log_x (0.008)} = \frac{\log_x (5^2)}{\log_x (5^{-3})}$

Using the logarithm property $\log_a(b^n) = n\log_a(b)$:

$= \frac{2\log_x 5}{-3\log_x 5} = \frac{2}{-3} = -\frac{2}{3}$

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For $\log_{\sqrt{3}}(x)$, we'll use the change of base formula:

$\log_{\sqrt{3}}(x) = \frac{\log x}{\log \sqrt{3}} = \frac{\log x}{\frac{1}{2}\log 3} = \frac{2\log x}{\log 3} = 2\log_3 x$

Substituting into our equation:

$2\log_3 x + \left(-\frac{2}{3}\right) = \frac{16}{3}$

$2\log_3 x - \frac{2}{3} = \frac{16}{3}$

$2\log_3 x = \frac{16}{3} + \frac{2}{3} = \frac{18}{3} = 6$

$\log_3 x = 3$

Therefore: $x = 3^3 = 27$

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Now we need to find $\log_3(3x^2)$:

$\log_3(3x^2)$

$= \log_3(3 \times (27)^2)$

$= \log_3(3 \times (3^3)^2)$

$= \log_3(3 \times 3^6)$

$= \log_3(3^7) = 7$