1. ## negative fractional exponents

hey guys i needed help on understanding how to do these questions?

a) (49)^(-3/2)

b) (27/8)^(-1/3)

c) -(25)^(1/2)

2. $\displaystyle 49^{-3/2} = \frac{1}{49^{3/2}}$
$\displaystyle 49^{3/2}=\sqrt[3]{49^2}=\sqrt[3]{49^2}=13.39052$
Finally:
$\displaystyle 49^{-3/2} =\frac{1}{13.39052}=0.07467971$

You can solve the others the same way. I hope it has been helpful.

3. a.) (49)^(-3/2) is the same as writing 1 / [(49)^(3/2)]
The 3/2 part is the same as the cube root of 49^2.

b.) (27/8)^(1/3) = (8/27)^(1/3)

c.) The exponent part takes precendence over the negative, which counts as a multiplication by -1. Therefore - [25^(1/2)], which just means to take the square root of 25 and multiply by -1. Answer is -5.

4. thanks guy that helped me out alot !!!

5. Originally Posted by Californiaboy
a.) (49)^(-3/2) is the same as writing 1 / [(49)^(3/2)]
The 3/2 part is the same as the cube root of 49^2.
not quite. it's the square root of 49 cubed ...

$\displaystyle \frac{1}{49^{\frac{3}{2}}} = \frac{1}{7^3} = \frac{1}{343}$