# Some confusion over some indices questions

• Jun 16th 2009, 06:05 AM
Duckers
Some confusion over some indices questions
I am currently doing some algebra practice for English pre A Level Maths. I am doing examples of a worksheet and I am finding a few problems that have cropped up so far that I would greatly appreciate if someone could explain how to get the right solution as my algebra is pretty patchy in places and needs to be mastered! (Happy)

Here are they are:

Simplify the following expressions into their simplest form

1,

$\displaystyle {\frac {ab^2}{cd}} \times {\frac {ac^2}{bd}}$

I simplfied to:

$\displaystyle ={\frac {a^2bc}{d^2}}$

The answer in the back of the book:

$\displaystyle = a^2bcd^2$ ?

2,

$\displaystyle {[(ab)^3]^{-\frac {1}{2}}}$

I simplified to:

$\displaystyle =[a^3b^3]^{-\frac {3}{2}}$

Answer in the back of the book:

$\displaystyle =a^{-\frac {1}{2}}b^{-\frac {3}{2}}$ ?

3,

$\displaystyle (abc)^{\frac{1}{2}} \div (a^3b^3c^3)^{\frac {2}{3}}$

I simplified to:

$\displaystyle ={\frac {a^{\frac{1}{2}}b^{\frac{1}{2}}c^{\frac{1}{2}}} {a^2b^2c^2}}$

The answer im the back of the book:

$\displaystyle = a^{-\frac{3}{2}} b^{-\frac{3}{2}} c^{-\frac{3}{2}}$ ?

David
• Jun 16th 2009, 06:18 AM
Quote:

Originally Posted by Duckers
I am currently doing some algebra practice for English pre A Level Maths. I am doing examples of a worksheet and I am finding a few problems that have cropped up so far that I would greatly appreciate if someone could explain how to get the right solution as my algebra is pretty patchy in places and needs to be mastered! (Happy)

Here are they are:

Simplify the following expressions into their simplest form

1,

$\displaystyle {\frac {ab^2}{cd}} \times {\frac {ac^2}{bd}}$

I simplfied to:

$\displaystyle ={\frac {a^2bc}{d^2}}$

The answer in the back of the book:

$\displaystyle = a^2bcd^2$ ?

2,

$\displaystyle {[(ab)^3]^{-\frac {1}{2}}}$

I simplified to:

$\displaystyle =[a^3b^3]^{-\frac {3}{2}}$

Answer in the back of the book:

$\displaystyle =a^{-\frac {1}{2}}b^{-\frac {3}{2}}$ ?

3,

$\displaystyle (abc)^{\frac{1}{2}} \div (a^3b^3c^3)^{\frac {2}{3}}$

I simplified to:

$\displaystyle ={\frac {a^{\frac{1}{2}}b^{\frac{1}{2}}c^{\frac{1}{2}}} {a^2b^2c^2}}$

The answer im the back of the book:

$\displaystyle = a^{-\frac{3}{2}} b^{-\frac{3}{2}} c^{-\frac{3}{2}}$ ?

David

(1) Your ans is correct .

(2)$\displaystyle a^{-\frac{3}{2}}b^{-\frac{3}{2}}$

(3)$\displaystyle a^{\frac{1}{2}-2}b^{\frac{1}{2}-2}c^{\frac{1}{2}-2}$
• Jun 16th 2009, 06:25 AM
Duckers
Quote:

(1) Your ans is correct .

(2)$\displaystyle a^{-\frac{3}{2}}b^{-\frac{3}{2}}$

(3)$\displaystyle a^{\frac{1}{2}-2}b^{\frac{1}{2}-2}c^{\frac{1}{2}-2}$

Thank you for confirming the answers, as I am unsure of how you got the last two answers would you mind shown how you solved them?

Thanks again

David
• Jun 16th 2009, 06:37 AM
Quote:

Originally Posted by Duckers
Thank you for confirming the answers, as I am unsure of how you got the last two answers would you mind shown how you solved them?

Thanks again

David

Ok sure .

(2) $\displaystyle [(ab)^3]^{-\frac{1}{2}}=(ab)^{-\frac{3}{2}}$

$\displaystyle =a^{-\frac{3}{2}} b^{-\frac{3}{2}}$

(3) division of indices -- ie $\displaystyle \frac{a^m}{a^n}=a^{m-n}$

so follow my previous post , and simplify it .