I know $\log(a^b)=b\log(a)$.
However, Wolfram Alpha tells me that $\frac{\log(a)}{\log(b)}$ does not equal $\log(a^\frac{1}{\log(b)})$.
Is Wolfram Alpha correct? If it is, why is it correct?
I'm using base 10 logs instead of natural logs, although I doubt it makes a difference.
2 Answers
Actually you are correct.
However, you must take note that there are some constraints (which are pretty self-explanatory).
$$\log(a^{\left(\frac{1}{\log(b)}\right)}) = \frac{\log a}{\log b}$$
As long as,
$$a > 0, \text{because you cannot take the logarithm of a negative number or 0}$$ $$b \ne 1\ \text{and}\ b >0, \text{because then} \log{b} = 0, \text{which means that the fraction} \frac{1}{\log b} \text{would be undefined}$$
Hope this helps you understand why.
Comment if you have any questions.
I think your issue might be that $\log(x)$ is a multivalued function if you allow complex outputs. This is due to the fact that $e^{2\pi i n}=1$ for $n\in\mathbb{Z}$. The complex logarithm can give out different values as follows.
$$\log(z)=\ln|\,z\,|+i\arg(z+2\pi n)$$
So when you take one value for the left hand side of your original equation, you could take a completely different one for the right hand side (though I'm not entirely certain that this is the reason).
