Manipulating Exponents

Question

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

Answer 1

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

$(7^{\frac{3}{5}})^{\frac{5}{6}} = 7^{\frac{3}{5} \cdot \frac{5}{6}} = 7^{\frac{3}{6}} = 7^{\frac{1}{2}}$.

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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

$(3^{\frac{5}{2}})(3^{-2}) = 3^{\frac{5}{2} + -2} = 3^{\frac{5}{2} + \frac{-4}{2}} = 3^{\frac{1}{2}}$.

Answer 2

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 $3\cdot \frac{5}{6}=\frac{5}{2}$.

Answer 3

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 $$\left(7^{3/5}\right)^{5/6} = 7^{{3/5}\cdot{5/6}} = 7^{1/2} = \sqrt{7}. $$