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:

Question

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:

Accepted Answer

To solve the given equation:

1. **Simplify $$ \log_x(0.008) $$:**  
   Since $$ 0.008 = \frac{8}{1000} = 2^3 \cdot 10^{-3} = 5^{-3} $$, we have:  
   $$
   \log_x(0.008) = -3\log_x(5).
   $$

2. **Substitute into the equation:**  
   The equation becomes:  
   $$
   \log_{\sqrt{3}}(x) + \frac{\log_x(25)}{-3\log_x(5)} = \frac{16}{3}.
   $$  
   Using $$ \log_x(25) = 2\log_x(5) $$, this simplifies to:  
   $$
   \log_{\sqrt{3}}(x) - \frac{2}{3} = \frac{16}{3}.
   $$

3. **Solve for $$ \log_{\sqrt{3}}(x) $$:**  
   Adding $$ \frac{2}{3} $$ to both sides:  
   $$
   \log_{\sqrt{3}}(x) = 6.
   $$

4. **Convert the base:**  
   Using the property $$ \log_{\sqrt{3}}(x) = 2\log_3(x) $$, we get:  
   $$
   2\log_3(x) = 6, \quad 	ext{so } \log_3(x) = 3.
   $$

5. **Find $$ x^2: $$**  
   Since $$ x = 3^3, x^2 = (3^3)^2 = 3^6. $$

6. **Calculate $$ 3x^2: $$**  
   Multiply by 3:  
   $$
   3x^2 = 3^1 \cdot 3^6 = 3^7.
   $$

7. **Find $$ \log_3(3x^2):**  
   Using the property of logarithms:  
   $$
   \log_3(3x^2) = \log_3(3^7) = 7.
   $$

Thus, the value of $$ \log_3(3x^2) $$ is **7**.