Bearing in mind that when the base is omitted, base 10 is assumed, or namely log(x) = log₁₀(x).
[tex]\begin{array}{llll} \textit{Logarithm of rationals} \\\\ \log_a\left( \frac{x}{y}\right)\implies \log_a(x)-\log_a(y) \end{array}~\hfill \begin{array}{llll} \textit{Logarithm of exponentials} \\\\ \log_a\left( x^b \right)\implies b\cdot \log_a(x) \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}[/tex]
[tex]\log_{10}(2)=0.3010\qquad \qquad \log_{10}(3)=0.4771 \\\\[-0.35em] ~\dotfill\\\\ \log_{10}\left( \cfrac{4}{9} \right)\implies \log\left( \cfrac{4}{9} \right)\implies \log\left[ \cfrac{2^2}{3^2} \right]\implies \log\left[ \left( \cfrac{2}{3} \right)^2 \right]\implies 2\log\left[ \cfrac{2}{3}\right] \\\\\\ 2[\log(2)-\log(3)]\implies 2[0.3010~~ - ~~0.4771]\implies -0.3522[/tex]
