Can $\log(x)\log(y)$ be reduced? I'm currently taking Pre-Calc and am learning about logs. I know that $\log(xy) = \log(x) + \log(y)$, but can $\log(x)\log(y)$ be reduced further?
 A: The answer to your question is, technically, yes: $$\log(x)\log(y) = \log(x^{\log(y)}) = \log(y^{\log(x)})$$
The above follows from the logarithm property that $\log(x^y)=y\log(x)$.
I think, though, it's worth an explanation of why these rules exist, and for that, I'm going to dive into a bit of a long-winded derivation, but bear with me:
The logarithm is defined to be the inverse of exponentiation; that is, $\log_a(b)$ is defined explicitly to be some value $x$ for which $a^x=b$. So when you have statements like $$\log_a(x) + \log_a(y) = \log_a(xy)$$ what you're really saying is (considering $\log$ as $\log_a$): 
$$
\begin{array}{rcl}
xy &=& a^{\log(x)}a^{\log(y)} = a^{\log(x)+\log(y)} \\
&=& a^{\log(xy)}\end{array}
$$
and since we know that $a^b=a^c$ happens if and onlf if $b=c$, then we get that $\log(x)+\log(y)=\log(xy)$. So we derive the multiplication rule like so:
$$x^y = (a^{\log(x)})^y = a^{y\log(x)}$$
and in some way, this is intuitive: the sum of logs becomes the log of a product (combining addition into multiplication) and the product of logs becomes the log of a power (combining multiplication into exponentiation). Unfortunately it's not quite as pretty, but that's the way the cookie crumbles.
A: Here is one way to represent the product of the logs.
Note that $(x+y)^2=x^2+2xy+y^2$ and $(x-y)^2=x^2-2xy+y^2$.  
Subtracting the respective sides of these equalities reveals 
$$2xy=\frac12\left((x+y)^2-(x-y)^2\right)$$
Then, letting $x=\log a$ and $y=\log b$, we obtain
$$2\log(a)\,\log(b)=\frac12 \log^2(ab)-\frac12 \log^2(a/b)$$
While this form might be preferential over the original for some fit-for-purpose reason, we have not gained much, and might have complicated things further.  Note that we now have to square two logarithms - and this is equivalent to two products.
