Simplifying a logarithmic expression.

Question

I have:

$\log xy + \log 2x^2 - 0.5\log 4y^2$

The unlike terms make it hard to see what can be done?

Thanks.

Answer 1

$\textbf{hint}$ $$ \log (xy) = \log x + \log y $$

Answer 2

Considering $x,y>0$ we can write the given expression as $$\log xy+\log 2x^2−0.5 \log 4y^2$$ $$=\log xy+\log 2x^2− \log (4y^2)^{\frac{1}{2}}$$ $$=\log xy+\log 2x^2−\log 2y$$ $$=\log \left(\frac{xy \times 2x^2}{2y}\right)$$ $$=\log x^3$$ $$=3\log x$$

Answer 3

To simplify this we need to use the following rules for $x,y,z \in \mathbb{R}^+ \setminus \{ 0 \}$ $$ \ln(xy) = \ln x + \ln y, \ \ln(x^z) = z \ln x$$ For instance $$0.5 \ln (4y^2) = \ln((4y^2)^{0.5}) = \ln (2y) = \ln 2 + \ln y$$ Can you take it from here?

Answer 4

Here's a hint. You can use the following rules of logarithms:

$$\log(ab) = \log(a)+\log(b),$$ $$\log(a/b) = \log(a)-\log(b),$$ $$\log(a^b) = b\log(a).$$

For instance, $\log(xy) + \log(2x^2) = \log((xy)(2x^2)) = \log(2x^3y)$.