Certainly! Here's the provided solution written in inline math using \(\):
\(6\log_x 2 - 3\log_x 3 = 3\log_5 0.2 = \log_x^{2^6} - 3\log_x^{3^3} = \log_5^{(0.2)^3} = \log_x \left(\frac{64}{27}\right) = \log_5 \left(\frac{1}{5}\right)^3\)

\(\log_x \left(\frac{64}{27}\right) = \log_5 \left(\frac{1}{125}\right) \text{, Let } \log_x \left(\frac{64}{27}\right) = y\)

\(\therefore x^y = \frac{64}{27} \text{, and } \log_5 \left(\frac{1}{125}\right) = y \therefore 5^y = \frac{1}{125} = 125^{-1} = 5^{-3}\)

\(\therefore y = -3 \text{, Substitute } y = -3 \text{ in } x^y = \frac{64}{27} \text{ implies } x^{-3} = \frac{64}{27} \text{, } \frac{1}{x^3} = \frac{64}{27} \text{, } 64x^3 = 27 \text{, } x^3 = \frac{27}{64}\)

\(x^3 = \sqrt[3]{\frac{27}{64}} \text{, } x = \frac{3}{4}\)

So, the solution is \(x = \frac{3}{4}\).