Can $\log(x)\log(y)$ be reduced?

Question

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?

Answer 1

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.

Answer 2

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.