[tex]\Large\displaystyle\text{$\begin{gathered} \log\left(\frac{27}{64}\right)=3(b-2a)\end{gathered}$}[/tex]
Bom, para resolver sua questão, deveremos utilizar algumas propriedades dos logaritmos. As que eu irei utilizar são:
[tex]\Large\displaystyle\text{$\begin{gathered}\log_a \left(\frac{b}{c} \right)=\log_a(b)-\log_a(c)\ \ \red{(I)}.\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered}\log_a \left(b^c\right)=c\cdot \log_a(b)\ \ \ \red{(II)}.\end{gathered}$}[/tex]
Sabendo disso, vamos então resolver sua questão. Aplicando então as propriedades I e II, temos que:
[tex]\Large\displaystyle\text{$\begin{gathered}\log \left(\frac{27}{64}\right)=\log(27)-\log(64) \end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered}\log \left(\frac{27}{64}\right)=\log(3^3)-\log(2^6) \end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered}\log \left(\frac{27}{64}\right)=3\log(3)-6\log(2) \end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered}\log \left(\frac{27}{64}\right)=3b-6a\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} \therefore \green{\underline{\boxed{\log\left(\frac{27}{64}\right)=3(b-2a)}}}\ \ (\checkmark).\end{gathered}$}[/tex]
