Problem

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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Answer

So, the final answer is \(\boxed{A = 1}\) and \(\boxed{B = \frac{14}{3}}\).

Steps

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 \(\boxed{A = 1}\) and \(\boxed{B = \frac{14}{3}}\).

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