Suppose you had:


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.


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 $$z=\frac{\log_x(y)}{b}$$

  • 3
    $\begingroup$ 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

You can calculate $$\frac{\ln(y)}{\ln(x^b)}=\frac{\ln(y)}{b\cdot \ln(x)}=\frac{\log_x(y)}{b}$$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.