# How do you simplify a log with an exponent in the base?

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

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.

$$\log_{x^b}(y)=z$$ 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 $$z=\frac{\log_x(y)}{b}$$
• 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. – Hugh Nov 11 '16 at 22:17
• Actually, you want $x>1$ for this to make sense. (Log base 1 is not nice.) – Mario Carneiro Nov 12 '16 at 6:25
• There's no need for $z > 0$. Reciprocals are permissible. – Eric Towers Nov 12 '16 at 17:40
You can calculate $$\frac{\ln(y)}{\ln(x^b)}=\frac{\ln(y)}{b\cdot \ln(x)}=\frac{\log_x(y)}{b}$$