Let u= ln x and v=ln y. Write ln(x^3y^2) in terms of u and v.

Accepted Solution

Answer:a. 3u + 2vStep-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