How to simplify logs and powers? Is there any way to simplify $(\log a)^{\log b} = c$?
And even this $(\log x)^y = z$?
And also this $(\log m)(\log n) = p$ (which is essentially $\log m^{\log n} = p$)
I was trying to simplify some other equations when I encountered this sort of thing.
I'm only 17, so I'm not extremely experienced at maths. You will probably have to explain whatever answer you have to me.
Also, by simplifying I do mean arrange in terms on each variable.
 A: A word of warning is to remember first that $a^{b^c}=a^{(b^c)}\neq (a^b)^{c}$, a common mistake when being faced with questions like this.
In truth, the ways you have written them already are about as pretty as you can make them look, however you are able to isolate one or more of the variables as you wish.  I will also make the assumption that here we are using the natural logarithm unless base is specified as a subscript, i.e. $\log x = y \Leftrightarrow x = e^y$.  In what way you want to simplify the expressions depends on what variable you wish to isolate in the end.
$$(\log a)^{\log b} = c\\ \log((\log a)^{\log b}) = \log c \\ \log b \cdot (\log \log a) = \log c\\
\log \log a = \frac{\log c}{\log b} = \log_b c\\
\log a = e^{\log_b c}\\
a = e^{(e^{\log_b c})}$$

For the second, $(\log x)^y=z$
$$(\log x)^y = z\\ y\cdot\log\log x = \log z\\ x = e^{e^{\frac{\log z}{y}}}\\
x = e^{e^{\log z^{\frac{1}{y}}}} = e^{(z^{\frac{1}{y}})}$$

As for $(\log m)(\log n) = p$
$$(\log m)(\log n) = p\\ \log m = \frac{p}{\log n}\\ m = e^{\frac{p}{\log n}}$$

Again, I doubt that these will be any cleaner or easier to use than what you already had.  There are no "nice" simplifications such as $\log(xy) = \log x + \log y$ or $a \log b = \log (b^a)$.
