Solve the equation: $\;$ $1 - \log 5 = \dfrac{1}{3} \left(\log \dfrac{1}{2} + \log x + \dfrac{1}{3} \log 5\right)$
Given equation: $\;\;$ $1 - \log 5 = \dfrac{1}{3} \left(\log \dfrac{1}{2} + \log x + \dfrac{1}{3} \log 5\right)$
i.e. $\;$ $1 - \log 5 = \dfrac{1}{3} \log \left(\dfrac{5^{\frac{1}{3}} \times x}{2}\right)$
i.e. $\;$ $1 - \log 5 = \log \left(\dfrac{5^{\frac{1}{3}} \times x}{2}\right)^{\frac{1}{3}}$
i.e. $\;$ $1 = \log \left(\dfrac{5^{\frac{1}{9}} \times x^{\frac{1}{3}}}{2^{\frac{1}{3}}}\right) + \log 5$
i.e. $\;$ $1 = \log \left(\dfrac{5^{\frac{1}{9}} \times x^{\frac{1}{3}} \times 5}{2^{\frac{1}{3}}}\right)$
i.e. $\;$ $\dfrac{5^{\frac{10}{9}} \times x^{\frac{1}{3}}}{2^{\frac{1}{3}}} = 10^1 = 10$
i.e. $\;$ $\dfrac{5^{\frac{10}{9}} \times x^{\frac{1}{3}}}{2^{\frac{1}{3}}} = 5 \times 2$
i.e. $\;$ $x^{\frac{1}{3}} = \dfrac{5 \times 2 \times 2^{\frac{1}{3}}}{5^{\frac{10}{9}}}$
i.e. $\;$ $x^{\frac{1}{3}} = 5^{1 - \frac{10}{9}} \times 2^{1 + \frac{1}{3}}$
i.e. $\;$ $x^{\frac{1}{3}} = 5^{\frac{-1}{9}} \times 2^{\frac{4}{3}}$
i.e. $\;$ $x^{\frac{1}{3}} = \left(5^{\frac{-1}{3}} \times 2^4\right)^{\frac{1}{3}}$
i.e. $\;$ $x = 2^4 \times 5^{\frac{-1}{3}}$
i.e. $\;$ $x = \dfrac{16}{\sqrt[3]{5}}$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\dfrac{16}{\sqrt[3]{5}} \right\}$