Question

Solve log 3x -log 13=2. Round to the nearest thousandth if necessary.

Answer 1

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