1. ## simplifying expressions

could someone help me with practice for my math placement in a few days. and sorry i didnt realize until after i posted that this is the elem/middle school section.

Use the properties of exponents to simplify the expression

2. Originally Posted by rafaeli
could someone help me with practice for my math placement in a few days.

Use the properties of exponents to simplify the expression

$\left(\frac{3a^{\frac12} b^{-\frac13}c} {a^{-\frac12}b^{\frac13}
c^{\frac72}}\right)^{-3}$

As the bracket is to the power of -3, a minus means that that bracket is the reciprocal to 1.

$\frac{1}{\left(\frac{3a^{\frac12} b^{-\frac13}c} {a^{-\frac12}b^{\frac13} c^{\frac72}}\right)^3}$

This is equal to the bracket being flipped hence:

$\left(\frac{a^{-\frac12}b^{\frac13}
c^{\frac72}}{3a^{\frac12} b^{-\frac13}c}\right)^{3}$

The general rule is $(x^a)^b = x^{(a)(b)} = x^ {ab}$

$\frac {a^{-\frac32} b
c^{\frac{21}{2}}}{27 a^{\frac32} b^{-1} c^3}$

The general rule for simplifying fraction with variable is $\frac{x^a}{x^b} = x^{a-b}$ hence:

$\frac{a^{-3} b^2
c^{\frac{15}{2}}}{27}$

For indices which is in the form $x^{-a}$, this is equal to it being $\frac{1}{x^a}$, hence:

$\frac{b^2 c^{\frac{15}{2}}}{27a^3}$

3. ## Here it is

Hi!rafaeli
Recall that (a^n)/(b^m)=a^(n-m)
so for a we have (a^(1/2))/(a^(-1/2))=a^((1/2)-(-1/2)=a^((1/2)+(1/2))=a
similarly for b we have
(b^(-1/3))/(b^(1/3))=b^(-2/3)
similarly for c we have
c/c^(7/2)=c^(-5/2)

So the question can be written as T=[3ab^(-2/3)c^(-5/2)]^-3
Now as x^(-n)=1/x^n and
(x^a)^b=x^ab
so T=3^(-3)a^(-3)b^2c^(15/2)
which is =[b^2c^(15/2)]/[27a^3]

4. Originally Posted by rafaeli
could someone help me with practice for my math placement in a few days. and sorry i didnt realize until after i posted that this is the elem/middle school section.

Use the properties of exponents to simplify the expression

Edit: I know I'm late, but this is just another approach.

First, let's make sue you know a couple of exponent rules:

1. $\frac{x^m}{x^n}=x^{m-n}$

2. $(x^m)^n=x^{mn}$

3. $\left(\frac{x}{y}\right)^{-m}=\left(\frac{y}{x}\right)^m$

Now,

$\left(\frac{3a^\frac{1}{2}b^\frac{-1}{3}c}{a^\frac{-1}{2}b^\frac{1}{3}c^\frac{7}{2}}\right)^{-3}$

Use rule 1 to get:

$\left(\frac{3a^1}{b^\frac{2}{3}c^\frac{5}{2}}\righ t)^{-3}$

Use rule 3 to get:

$\left(\frac{b^\frac{2}{3}c^\frac{5}{2}}{3a}\right) ^3$

Use rule 2 to finish:

$\frac{b^2 c^\frac{15}{2}}{27a^3}$