Answer
Explanation
Given:
[tex]\log_au-\log_av+9\log_aw[/tex]
To write the above given into a single logarithm, we first apply the log rule:
[tex]b\cdot\log_a(x)=\log_a(x^b)[/tex]
So,
[tex]9\log_aw=\log_aw^9[/tex]
Hence,
[tex]\operatorname{\log}_au-\operatorname{\log}_av+9\operatorname{\log}_aw=\log_au-\log_av+\log_aw^9[/tex]
Next, we apply the log rule:
[tex]\log_a(x)-\log_a(y)=\log_a(\frac{x}{y})[/tex]
This means that:
[tex]\operatorname{\log}_au-\operatorname{\log}_av=\log_a(\frac{u}{v})[/tex]
Hence,
[tex]\log_au-\log_av+\log_aw^9=\log_a(\frac{u}{v})+\log_aw^9[/tex]
Then, we apply the log rule:
[tex]\log_a(x)+\log_a(y)=\log_a(xy)[/tex]
So,
[tex]\log_a(\frac{u}{v})+\log_aw^9=\log_a(\frac{uw^9}{v})[/tex]
Therefore, the answer is:
[tex]\begin{equation*} \log_a(\frac{uw^9}{v}) \end{equation*}[/tex]
