Answer
$log_{7}\frac{18}{4}$
Work Step by Step
We know that $log_{b}xy=log_{b}x+log_{b}y$ (where $x$, $y$, and $b$ are positive real numbers and $b\ne1$).
Therefore, $log_{7}6+log_{7}3-log_{7}4=log_{7}(6\times3)-log_{7}4=log_{7}18-log_{7}4$.
We know that $log_{b}\frac{x}{y}=log_{b}x-log_{b}y$ (where $x$, $y$, and $b$ are positive real numbers and $b\ne1$).
Therefore, $log_{7}18-log_{7}4=log_{7}\frac{18}{4}$.