I know for a fact that it is not $\\log x + \\log y$, but Im unsure as to how to proceed.. I have checked the basic log properties but nowhere do they give an example of a statement like the one abov...

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:

Nov 7, 2016 · $\begingroup$ "another question , using the fact that y = log x to the base a is the inverse function of x = a^y does that mean that the identitiy mentioned at the begingin is the inverse of 10^x+y = 10^x times 10 ^ y ?" Syntactically, …

$\begingroup$ It is a case of the change of base formula: $\log_x(y)=\frac{\log_y(y)}{\log_y(x)}$. The change of base formula follows fairly easily from the familiar rules of exponents; you can ask about that if you like. $\endgroup$

Nov 12, 2022 · $\begingroup$ If $\log (x y)$ is defined opportunistically (for anything that makes sense) then it does not equal $\log x+\log y$ defined opportunistically (for anything that makes sense, ignoring imaginary numbers) $\endgroup$

Oct 28, 2022 · $\begingroup$ Well, if you want to do shortcuts. $-\log x -\log y = 0 - \log x - \log y =\log 1 - \log x -\log y = \log \frac 1x -\log y = \log \frac {\frac 1x}y ...

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Oct 29, 2012 · What is the graph of $x^{\log y}=y^{\log x}$? This question appears on GRE exam.

Oct 31, 2023 · (From here I will simply use $\log$ to mean a logarithm to any valid base.) Using the logarithm power rule from this list of logarithm rules $$ \log\left(x^{\frac{\log y}{\log x}}\right) = \log y$$ $$ \Rightarrow x^{\frac{\log y}{\log x}} = y$$ (The equation above can form a nice little identity in itself). This easily rearranges to

You can prove exp(x+y)=exp(x)exp(y) using the series then use the fact that log is the inverse of exp to get the result. $\endgroup$ – user58512 Commented Feb 23, 2013 at 23:08