# Math Help - Evaluate 9^(-3/2)

1. ## Evaluate 9^(-3/2)

Evaluate:

9^(-3/2)

Got a mental blank, please guys help me quickly!
How is it 1/27? Please show complete working out.

2. $
9^{1/2}\;=\;3
$

4. 9^(-3/2)

9^(-3/2) = 1/(square root of 9^3 )
= 1/27

5. For any number $a\ne 0$, the number $a^{-1}$ is defined as such a number that $a\cdot{a^{-1}} = a^{-1}\cdot{a} = 1.$ If we for example take $3$, then the number $3^{-1}$ such that $3\cdot{3^{-1}} = 3^{-1}\cdot{3} = 1$ is clearly $\dfrac{1}{3}$, as $3\cdot\left(\dfrac{1}{3}\right) = \left(\dfrac{1}{3}\right)\cdot{3} = 1.$ It can be easily seen that $a^{-1}$ is always $\dfrac{1}{a}$ (with the exception of $a = 0$, of course). So, for your case, we have $9^{-\frac{3}{2}} = \dfrac{1}{9^\frac{3}{2}}$ Remember that $X^{\frac{Y}{Z}} = (X^{\frac{1}{Z}})^Y$, so you have $\dfrac{1}{9^\frac{3}{2}} = \dfrac{1}{(9^{\frac{1}{2}})^3}.$ Since $9^{\frac{1}{2}} = 3$, we get $\dfrac{1}{3^3},$ which is of course $27$.

6. Originally Posted by Joker37
Evaluate:

9^(-3/2)

Got a mental blank, please guys help me quickly!
How is it 1/27? Please show complete working out.
You need to thoroughly review your basic index laws (and that's you job, not ours).

$9^{-3/2} = \frac{1}{9^{3/2}} = \frac{1}{(9^{1/2})^3}$.