[tex]\bf \textit{Logarithm of rationals}\\\\
log_a\left( \frac{x}{y}\right)\implies log_a(x)-log_a(y)
\\\\\\
\textit{Logarithm Cancellation Rules}\\\\
log_a a^x= x\qquad \qquad \boxed{a^{log_ax}=x}\\\\
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log(3x)-log(13)=2\implies log_{10}(3x)-log_{10}(13)=2
\\\\\\
log_{10}\left( \frac{3x}{13} \right)=2\implies 10^{log_{10}\left( \frac{3x}{13} \right)}=10^2\implies \cfrac{3x}{13}=2
\\\\\\
3x=26\implies x=\cfrac{26}{3}\implies x=8\frac{2}{3}[/tex]
