Question

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

Answer 1

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:

$log_{b}(MN)=log_{b}(M)+log_{b}(N)$

From the function, it is then true that:

$ln(x^{3}y^{2})=ln(x^{3})+ln(y^{2})$

The other property we must use is Logarithm of a Power:

$log_{b}M^{n}=nlog_{b}M$

Then:

$ln(x^{3}y^{2})=ln(x^{3})+ln(y^{2}) \\ \\ ln(x^{3}y^{2})=3ln(x)+2ln(y)$

Since:

$u=ln(x) \\ v=ln(y)$

Then:

$ln(x^{3}y^{2})=3u+2v$

Finally, the correct option is:

a. 3u + 2v

Answer 2

Answer:

A edge

Step-by-step explanation: