Answer
The solution is, $3{{\log }_{b}}x-2{{\log }_{b}}5-\frac{1}{3}{{\log }_{b}}y={{\log }_{b}}\frac{{{x}^{3}}{{y}^{\frac{-1}{3}}}}{25}$.
Work Step by Step
First use the power rule for logarithms to make all the coefficients 1.
That is, $a{{\log }_{b}}x={{\log }_{b}}{{x}^{a}}$
Thus,
${{\log }_{b}}{{x}^{3}}-{{\log }_{b}}{{5}^{2}}-{{\log }_{b}}{{y}^{\frac{1}{3}}}$
Simplifying it further, we get,
${{\log }_{b}}{{x}^{3}}-{{\log }_{b}}25-{{\log }_{b}}{{y}^{\frac{1}{3}}}$
${{\log }_{b}}{{x}^{3}}-\left( {{\log }_{b}}25+{{\log }_{b}}{{y}^{\frac{1}{3}}} \right)$
Use the product rule for logarithms,
${{\log }_{b}}M+{{\log }_{b}}N={{\log }_{b}}\left( M\cdot N \right)$
Thus,
${{\log }_{b}}{{x}^{3}}-\left( {{\log }_{b}}25{{y}^{\frac{1}{3}}} \right)$
Now, we will use the quotient rule for logarithms, ${{\log }_{b}}\left( \frac{M}{N} \right)={{\log }_{b}}M-{{\log }_{b}}N$
Thus,
${{\log }_{b}}\frac{{{x}^{3}}}{25{{y}^{\frac{1}{3}}}}={{\log }_{b}}\frac{{{x}^{3}}{{y}^{\frac{-1}{3}}}}{25}$
Hence,
$3{{\log }_{b}}x-2{{\log }_{b}}5-\frac{1}{3}{{\log }_{b}}y={{\log }_{b}}\frac{{{x}^{3}}{{y}^{\frac{-1}{3}}}}{25}$
Hence, $3{{\log }_{b}}x-2{{\log }_{b}}5-\frac{1}{3}{{\log }_{b}}y={{\log }_{b}}\frac{{{x}^{3}}{{y}^{\frac{-1}{3}}}}{25}$.
