We can write ${\log }_3\left(x\sqrt[3]{y^{14}}\right)$ into the form $A{\log }_{3}x+B{\log }_{3}y$
where $A=$ and $B=$ Write $A$ and $B$ as integers or reduced fractions. Question Help:
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Step 1 ：Separate the terms inside the logarithm using the property that the logarithm of a product is the sum of the logarithms of its factors. So, ${\log }_3\left(x\sqrt[3]{y^{14}}\right)$ can be written as ${\log }_{3}x+{\log }_3\sqrt[3]{y^{14}}$.

Step 2 ：Rewrite the cube root in the second term as a power. So, ${\log }_3\sqrt[3]{y^{14}}$ can be written as ${\log }_3{y}^{14/3}$.

Step 3 ：Use the property that the logarithm of a number raised to a power is the product of the power and the logarithm of the number. So, ${\log }_3{y}^{14/3}$ can be written as $\frac{14}{3}{\log }_{3}y$.

Step 4 ：So, the original expression ${\log }_3\left(x\sqrt[3]{y^{14}}\right)$ can be written as ${\log }_{3}x+\frac{14}{3}{\log }_{3}y$.

Step 5 ：Comparing this with the form $A{\log }_{3}x+B{\log }_{3}y$, we can see that $A=1$ and $B=\frac{14}{3}$.

Step 6 ：So, the final answer is $\boxed{{\displaystyle A=1}}$ and $\boxed{{\displaystyle B=\frac{14}{3}}}$.