I need to simplify this:
$\displaystyle {(b^{-4/3}c^{1/2})}{(b^{-2/3}c^{-3/2})}$
My answer:
$\displaystyle {b^{2}c}$
Is it correct?
When terms are multiplied together, their exponents add.
Consider $\displaystyle a^3a^3$ each term is really a*a*a, so you can rewrite it as
$\displaystyle a^3a^3~~~~=~~~~a*a*a*a*a*a~~~~=~~~~a^6$
so what happens is that the exponents add together. so you can simply say
$\displaystyle a^3a^3~~~~=~~~~a^{3+3}~~~~=~~~~a^6$
Now onto your problem, since all these terms are multiplied, you can reorder the parentheses like this:
$\displaystyle (b^{-4/3}b^{-2/3})(c^{-3/2}c^{1/2})$
Then adding together:
=$\displaystyle ~b^{-4/3+(-2/3)}~~c^{-3/2+1/2}$
Simplify
=$\displaystyle ~b^{-4/3-2/3}~~c^{-3/2+1/2}$
And since -4/3-2/3 = -6/3 = -2
And also since -3/2 + 1/2 = -2/2 = -1
we get:
=$\displaystyle ~b^{-2}~c^{-1}$
Hello, mt_lapin!
Simplify: .$\displaystyle {(b^{-4/3}c^{1/2})}{(b^{-2/3}c^{-3/2})}$
My answer: .$\displaystyle {b^{2}c}$
Is it correct? . . . . no
$\displaystyle \left(b^{-\frac{4}{3}}c^{\frac{1}{2}}\right)\left(b^{-\frac{2}{3}}c^{-\frac{3}{2}}\right) \;=\;\left(b^{-\frac{4}{3}}b^{-\frac{2}{3}}\right)\left(c^{\frac{1}{2}}c^{-\frac{3}{2}}\right) \;=\;\left(b^{-2}\right)\left(c^{-1}\right) \;=\;\frac{1}{b^2c}$
They are the same
$\displaystyle x^{-1} = \frac 1x$
the negative sign in the exponent means 1 over that number, so
$\displaystyle b^{-2}c^{-1} ~~~~=~~~~\frac 1{b^2}*\frac 1c~~~~=~~~~ \frac{1}{b^{2}c}$
You can write it either way. Choose whichever your instructor prefers. If you don't know, then keep it in the same form as the original problem. Yours started out with negative exponents, so this might be what they are looking for, though I would find it very unusual if you were demerited for writing it the other way.