Question

I need to know how to collapse this log problem

Answer 1

Given:

`\log (w-6)-\log (w+2)+4\log n`

Applying the law of logarithm below;

`a\log x=\log x^a`

Then, the expression becomes;

`\log (w-6)-\log (w+2)+4\log n=\log (w-6)-\log (w+2)+\log n^4`

Also, applying the product and quotient law of logarithms,

`\begin{gathered} \log A+\log B=\log AB \\ \log A-\log B=\log ((A)/(B)) \end{gathered}`

Applying these laws to the expression, to make it become a single logarithm, the expression becomes;

`\begin{gathered} \log (w-6)-\log (w+2)+\log n^4=\log ((w-6)/(w+2))* n^4 \\ \log ((w-6)/(w+2))* n^4=\log ((w-6)/(w+2))n^4 \\ \log ((w-6)/(w+2))n^4=\log ((n^4(w-6))/(w+2)) \\ \\ =\log ((n^4(w-6))/(w+2)) \end{gathered}`

Therefore, the expression simplified as a single logarithm is;

`\log (n^4(w-6))/(w+2)`