I need to simplify:
(49c^6)^(-3/2)
I know that the answer is 1/343c^9. But I have noo idea how they got it?
Thank you!
$\displaystyle (49c^6)^{-\frac {3}{2}}$
Recall the rule of negative exponents:
$\displaystyle x^{-a} = \frac {1}{x^a}$
So we have:
$\displaystyle
(49c^6)^{-\frac {3}{2}} = \frac {1}{(49c^6)^{\frac {3}{2}}}$
Also recall that:
$\displaystyle x^{\frac {a}{b}} = (x^{\frac {1}{b}})^a$
So we have that:
$\displaystyle \frac {1}{(49c^6)^{\frac {3}{2}}} = \frac {1}{\left((49c^6)^{\frac {1}{2}}\right)^3}$
Recall that:
$\displaystyle x^{\frac {1}{2}} = \sqrt {x}$
This gives us:
$\displaystyle \frac {1}{\left((49c^6)^{\frac {1}{2}}\right)^3} = \frac {1}{(7c^3)^3} = \frac {1}{343c^9}$
Ok, I understand what happened with my 1st question... now i'm stuck on another
How does (4^-2)+(3^-1)/(3^-2)+(2^-3) become 1(23/34)?
This is waht I'm doing:
= (1/4^2+3^-1)/(1/3^2+2^3)
= (1/16+3)/(1/9+8)
= (1/19)/(1/17)
= (1/19)*(17/1)
= 323.
I have no idea where I am going wrong. thank you for any tips you might have!
Two things.
1) New questions should go in new threads.
2) Use parenthesis! This is a mess. Is your problem
$\displaystyle \frac{4^{-2} + 3^{-1}}{3^{-2} + 2^{-3}}$
If so then this is
$\displaystyle = \frac{\frac{1}{16} + \frac{1}{3}}{\frac{1}{9} + \frac{1}{8}}$
You appear to be losing sight of which fractions are which because you aren't using the parenthesis properly.
-Dan