# How do you simplify (3a^-1)^-1 (9a^2 b^3)^-2?

Oct 18, 2015

$\frac{1}{243} \cdot \frac{1}{a} ^ 3 \cdot \frac{1}{b} ^ 6$

#### Explanation:

Start by recognizing that a negative exponent can be written as

${n}^{- a} = \frac{1}{n} ^ a$

Now, notice that you can rewrite the first term of the expression as

${\left(3 {a}^{- 1}\right)}^{- 1} = {3}^{- 1} \cdot {\left({a}^{- 1}\right)}^{- 1} = \frac{1}{3} ^ 1 \cdot {\left({a}^{- 1}\right)}^{- 1}$

You can actually bypass the negative exponent for $a$ by using the power of a power property of exponents

$\textcolor{b l u e}{{\left({n}^{a}\right)}^{b} = {n}^{a \cdot b}}$

In this case, you have

${\left(3 {a}^{- 1}\right)}^{- 1} = \frac{1}{3} \cdot {a}^{\left(- 1\right) \cdot \left(- 1\right)} = \frac{1}{3} \cdot {a}^{1} = \frac{1}{3} \cdot a$

The second term of the expression will be

${\left(9 {a}^{2} {b}^{3}\right)}^{- 2} = \frac{1}{9 {a}^{2} {b}^{3}} ^ 2 = \frac{1}{9} ^ 2 \cdot \frac{1}{{a}^{2}} ^ 2 \cdot \frac{1}{{b}^{3}} ^ 2$

$= \frac{1}{81} \cdot \frac{1}{a} ^ 4 \cdot \frac{1}{b} ^ 6$

This means that you have

${\left(3 {a}^{- 1}\right)}^{- 1} \cdot {\left(9 {a}^{2} {b}^{3}\right)}^{- 2} = \frac{1}{3} \cdot a \cdot \frac{1}{81} \cdot \frac{1}{a} ^ 4 \cdot \frac{1}{b} ^ 6$

This can be simplifed to

$\frac{1}{3} \cdot \frac{1}{81} \cdot \frac{1}{a} ^ 3 \cdot \frac{1}{b} ^ 6 = \textcolor{g r e e n}{\frac{1}{243} \cdot \frac{1}{a} ^ 3 \cdot \frac{1}{b} ^ 6}$