Answer:
a. 3u + 2v
Step-by-step explanation:
To solve this problem, we need to apply some properties of logarithms. Properties are useful to simplify complicated expressions. Here we need to use a very useful property of logarithms called the logarithm of a product is the sum of the logarithms, that is:
[tex]log_{b}(MN)=log_{b}(M)+log_{b}(N)[/tex]
From the function, it is then true that:
[tex]ln(x^{3}y^{2})=ln(x^{3})+ln(y^{2})[/tex]
The other property we must use is Logarithm of a Power:
[tex]log_{b}M^{n}=nlog_{b}M[/tex]
Then:
[tex]ln(x^{3}y^{2})=ln(x^{3})+ln(y^{2}) \\ \\ ln(x^{3}y^{2})=3ln(x)+2ln(y)[/tex]
Since:
[tex]u=ln(x) \\ v=ln(y)[/tex]
Then:
[tex]ln(x^{3}y^{2})=3u+2v[/tex]
Finally, the correct option is:
a. 3u + 2v
