<s>[INST] Manipulating Exponents I'm doing my homework and there are a couple of things that I am having trouble grasping. All my homework asks is that I simplify the exponents. For example: 6^5 * 6^3 = 6^8
There are 2 problems I am unsure on what to do. They have to do with multiplying fractions: 
(7^3)^5/6 = 7^5/2 is the answer but I don't understand how they got to that. I know there is a certain example I was shown where you had to get common demonators and such but if I apply that it doesn't turn out well. Same for this problem:
(7^3/5)^5/6 = 7^2
3^5/2 * 3^-2 = 3^1/2
 [/INST] For any $x$, $a$, and $b$. 
$(x^a)^b = x^{ab}$. 
Hence for the first one, letting $x = 7$, $a = 3$, $b = \frac{5}{6}$, you have
$(7^3)^{\frac{5}{6}} = 7^{3 \cdot \frac{5}{6}} = 7^{\frac{5}{2}}$
For the second letting $x = 7$, $a = \frac{3}{5}$ and $b = \frac{5}{6}$, you have
$(7^{\frac{3}{5}})^{\frac{5}{6}} = 7^{\frac{3}{5} \cdot \frac{5}{6}} = 7^{\frac{3}{6}} = 7^{\frac{1}{2}}$. 

For the last one, for any $x$, $a$ and $b$
$(x^a)(x^b) = x^{a + b}$. 
Letting $x = 3$, $a = \frac{5}{2}$ and $b = -2$, you have
$(3^{\frac{5}{2}})(3^{-2}) = 3^{\frac{5}{2} + -2} = 3^{\frac{5}{2} + \frac{-4}{2}} = 3^{\frac{1}{2}}$.
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