Fraction exponents in division if I have $\frac{a^{6/5}}{b^{1/5}}$, I know you subtract exponents when dividing so $6/5 - 1/5$ is $5/5$, so since that's just one, is this equal to $a/b$?
 A: $$\dfrac{a^{6/5}}{b^{1/5}} = a^{6/5}b^{-1/5} = (a^6b^{-1})^{1/5}= \left(\dfrac{a^6}{b}\right)^{1/5}$$
This expression is different from $\dfrac{a}{b}$... Take $a = b = 2$ to convince yourself:
$$\left(\dfrac{a^6}{b}\right)^{1/5}=(a^{5})^{1/5}=a = 2 \neq 1 = \frac{a}{b}$$
A: You substract exponents when dividing only if there is the same base. If it was $\frac{a^{6/5}}{a^{1/5}}$, then your approach would be correct.
The right rule: $\frac{a^{b}}{a^{c}} = a^{b - c}$.
Wrong rule which you used and which makes no sense: $\frac{a^b}{c^d} = (\frac{a}{c})^{b - d}$. It is similarily wrong like $ab - cd = (a - c)(b - d)$ whereas the right rule is $ab - ac = a(b - c)$.
A: Long story short, that expression is $\ne \frac{a}{b}$. Why? Because $a \ne b $. In other words, those bases are different. You can only subtract exponents when bases are same.
Take for example two numbers 16 and 8 if you divide them, the answer is 2. Numerically, $$\frac{16}{8}=\frac{2^4}{2^3}=2^{4-3}=2$$
If you even want to imply the above rule in your question, you need log to convert them in the same base and I am unsure if you know log. So let it be. :)
