Solution:
Given:
[tex]\log (w-6)-\log (w+2)+4\log n[/tex]
Applying the law of logarithm below;
[tex]a\log x=\log x^a[/tex]
Then, the expression becomes;
[tex]\log (w-6)-\log (w+2)+4\log n=\log (w-6)-\log (w+2)+\log n^4[/tex]
Also, applying the product and quotient law of logarithms,
[tex]\begin{gathered} \log A+\log B=\log AB \\ \log A-\log B=\log (\frac{A}{B}) \end{gathered}[/tex]
Applying these laws to the expression, to make it become a single logarithm, the expression becomes;
[tex]\begin{gathered} \log (w-6)-\log (w+2)+\log n^4=\log (\frac{w-6}{w+2})\times n^4 \\ \log (\frac{w-6}{w+2})\times n^4=\log (\frac{w-6}{w+2})n^4 \\ \log (\frac{w-6}{w+2})n^4=\log (\frac{n^4(w-6)}{w+2}) \\ \\ =\log (\frac{n^4(w-6)}{w+2}) \end{gathered}[/tex]
Therefore, the expression simplified as a single logarithm is;
[tex]\log \frac{n^4(w-6)}{w+2}[/tex]
