l\(\displaystyle{o}{g}{\left[{4}\right]}{\left({16}\right)}\cdot{\log{{\left[{16}\right]}}}{4}={1}\) where [] denotes base, is an example of a fundamental rule in logarithms: \(\displaystyle{\log{{\left[{a}\right]}}}{\left({b}\right)}\cdot{\log{{\left[{b}\right]}}}{\left({a}\right)}={1}\), so, since \(\displaystyle{\log{{\left[{4}\right]}}}{\left({16}\right)}={2}{\log{{\left[{4}\right]}}}{4}={2},{\log{{\left[{16}\right]}}}{4}=\frac{{1}}{{2}}\), or, alternatively, \(log[16]4=log[16]\frac{16^{1}}{2}=\frac{1}{2}log[16]16=\frac{1}{2}.\)

Also, \(\displaystyle{\log{{\left[{a}\right]}}}{\left({x}\right)}={\log{{\left[{b}\right]}}}\frac{{{x}}}{{\log{{\left[{b}\right]}}}}\)

(a). \(\displaystyle{\log{{\left[{16}\right]}}}{\left({3}{x}-{1}\right)}={\log{{\left[{4}\right]}}}{\left({3}{x}\right)}+{\log{{\left[{4}\right]}}}{\left({0.5}\right)}⇒{2}{\log{{\left[{4}\right]}}}{\left({3}{x}-{1}\right)}={\log{{\left[{4}\right]}}}{\left({3}{x}\cdot{0.5}\right)}\). We can equate the logs: \(\displaystyle{\left({3}{x}-{1}\right)}^{{2}}={3}\frac{{x}}{{2}}\rightarrow{9}{x}^{{2}}-{6}{x}+{1}={3}\frac{{x}}{{2}}⇒{18}{x}^{{2}}-{12}{x}-{3}{x}+{2}={0}\)

\(\displaystyle\rightarrow{18}{x}^{{2}}-{15}{x}+{2}={0}\rightarrow{\left({3}{x}-{2}\right)}{\left({6}{x}-{1}\right)}={0}\)

From this \(\displaystyle{x}=\frac{{2}}{{3}}{\quad\text{or}\quad}\frac{{1}}{{6}}.\) Substitute these values in the original equation: \(\displaystyle{0}=\frac{{1}}{{2}}-\frac{{1}}{{2}}\). The value \(\displaystyle\frac{{1}}{{6}}\) cannot be used because it would require the log of a negative number so the only solution is \(Lx=\frac{2}{3}\)