# Algebra Math B level

• Jun 10th 2009, 10:53 AM
Mouseman
Algebra Math B level
Hi, i would like to recieve some help with this math problem. I have been working on it for a while but seem to be getting nowhere due to the powered negative one after the powered 3n in the first part of the fraction.
Hopefully someone can help me and explain the steps taken. :)
Thank you!

((x^(3n-1))^n*x^n)/((x^n)^(3n))
• Jun 10th 2009, 11:25 AM
TheEmptySet
Quote:

Originally Posted by Mouseman
Hi, i would like to recieve some help with this math problem. I have been working on it for a while but seem to be getting nowhere due to the powered negative one after the powered 3n in the first part of the fraction.
Hopefully someone can help me and explain the steps taken. :)
Thank you!

((x^(3n-1))^n*x^n)/((x^n)^(3n))

With out LaTex I am not sure if which of these you mean

$\frac{(x^{(3n-1)^n})\cdot x^n}{(x^n)^{3n}}$

OR

$\frac{(x^{(3n-1)})^n\cdot x^n}{(x^n)^{3n}}$

and are you trying to simplify?
• Jun 10th 2009, 11:26 AM
Amer
Quote:

Originally Posted by Mouseman
Hi, i would like to recieve some help with this math problem. I have been working on it for a while but seem to be getting nowhere due to the powered negative one after the powered 3n in the first part of the fraction.
Hopefully someone can help me and explain the steps taken. :)
Thank you!

((x^(3n-1))^n*x^n)/((x^n)^(3n))

$(x^{3n-1})^{n} \times \frac{x^n}{(x^n)^{3n}}$

$(\frac{x^{3n}}{x})^{n} \times \frac{x^n}{(x^n)^{3n}}$

$\frac{{\color{red}(x^{3n})^{n}}}{{\color{blue}x^n} }\times \frac{{\color{blue}x^n}}{{\color{red}(x^n)^{3n}}}= 1$

the same color same value
• Jun 11th 2009, 11:38 AM
Mouseman
Quote:

Originally Posted by TheEmptySet
With out LaTex I am not sure if which of these you mean

$\frac{(x^{(3n-1)^n})\cdot x^n}{(x^n)^{3n}}$

OR

$\frac{(x^{(3n-1)})^n\cdot x^n}{(x^n)^{3n}}$

and are you trying to simplify?

It would be the second one and yes I'm trying to simplify.
• Jun 11th 2009, 11:48 AM
Kasper
$
\frac{(x^{(3n-1)})^n\cdot x^n}{(x^n)^{3n}}
$

= $
\frac{x^{(3n-1)(n)}\cdot x^n}{x^{(n)(3n)}}
$

= $
\frac{x^{(3n^2-n)+(n)}}{x^{3n^2}}
$

= $
\frac{x^{3n^2}}{x^{3n^2}}
$

= 1

Using the power law of exponents, $(x^n)^m=x^{nm}$ and the multiplication law of exponents, $x^{n} \cdot x^{m}=x^{n+m}$, I believe this is how it is simpified, someone correct me if I am wrong; Does this help?
• Jun 11th 2009, 12:08 PM
Amer
Quote:

Originally Posted by Kasper
$
\frac{(x^{(3n-1)})^n\cdot x^n}{(x^n)^{3n}}
$

= $
\frac{x^{(3n-1)(n)}\cdot x^n}{x^{(n)(3n)}}
$

= $
\frac{{\color{red}x^{(3n^2-n)(n)}}}{x^{3n^2}}
$

= $
\frac{x^{3n^3-n^2}}{x^{3n^2}}
$

Using the power law of exponents, $(x^n)^m=x^{nm}$, I believe this is how it is simpified, someone correct me if I am wrong; Does this help?

the red is wrong the correct should be like this

= $
\frac{x^{(3n-1)(n)}\cdot x^n}{x^{(n)(3n)}}
$

$\frac{{\color{blue}x^{(3n^2-n)+n} }}{x^{(n)(3n)}}$ the blue should be instead of the red

because in multiple the powers add not multiply right
• Jun 11th 2009, 12:11 PM
Kasper
Yeah I'd already corrected that after you started replying, haha. Sorry for the confusion, edited it 7 minutes before you posted, heh. Love forum timers.