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
 A: 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
$(3^{\frac{5}{2}})(3^{-2}) = 3^{\frac{5}{2} + -2} = 3^{\frac{5}{2} + \frac{-4}{2}} = 3^{\frac{1}{2}}$.
A: In general, 
$$(a^b)^c=a^{(bc)}.$$
 Let us apply that to your first example. We have
$$(7^3)^{\frac{5}{6}}=7^{\left(3\cdot \frac{5}{6}\right)}.$$
But $3\cdot \frac{5}{6}=\frac{5}{2}$.
A: The exponent rules are boookkeeping. You need to keep track of the number of powers being used. For example,
$$x^a\cdot x^b = x^{a+b}$$ 
because there are a total of $a+b$ factors of $x$ here.  The other important tool is
$$\left(x^a\right)^b = x^{ab}.$$
Again, this is bookkeeping.
Hence, in your example
$$\left(7^{3/5}\right)^{5/6} = 7^{{3/5}\cdot{5/6}} = 7^{1/2} = \sqrt{7}.  $$
