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Suppose you had:

$$\log_{x^b}(y)$$

How can you simplify this? Do you use the change of base formula?

Note: I tried to come up with something similar to a homework problem without actually being a homework problem. I think this is the most simple form.

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We have your original function

$$\log_{x^b}(y)=z$$ Following basic rules for logarithms, assuming $x,y,z>0$ $$(x^b)^z=y$$ $$x^{bz}=y$$ $$\log_x(y)=bz$$ Thus $z$ can be expressed as $$z=\frac{\log_x(y)}{b}$$

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    $\begingroup$ I was just writing up this exact answer! @Jeff note that the laws of logs used here only apply when $x,y,b$ are all positive. $\endgroup$ – Hugh Nov 11 '16 at 22:17
  • $\begingroup$ I edited the answer and added your remark. $\endgroup$ – Žiga Sajovic Nov 11 '16 at 22:19
  • $\begingroup$ I find it useful often to rewrite log expressions involving various bases as exponentials - it saves me some remembering of things like base change formulae, because it is easy to work out what they should be. $\endgroup$ – Mark Bennet Nov 11 '16 at 22:27
  • $\begingroup$ Actually, you want $x>1$ for this to make sense. (Log base 1 is not nice.) $\endgroup$ – Mario Carneiro Nov 12 '16 at 6:25
  • $\begingroup$ There's no need for $z > 0$. Reciprocals are permissible. $\endgroup$ – Eric Towers Nov 12 '16 at 17:40
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You can calculate $$\frac{\ln(y)}{\ln(x^b)}=\frac{\ln(y)}{b\cdot \ln(x)}=\frac{\log_x(y)}{b}$$

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  • $\begingroup$ this is also a very effective way $\endgroup$ – Steve Aug 2 '18 at 5:25

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