# [SOLVED] Indice Operations Simplification Problem

• Dec 16th 2009, 06:22 AM
immengin101
[SOLVED] Indice Operations Simplification Problem
okay, so here's the problem. I wish I had tutoring when I was younger or had more help in math, now I'm trying to cram algebra revision, trig and curves and functions in 9 days for a test so I can get into engineering related classes I need.

I get most problems, just here and there I stumble onto something I can't solve because I don't know the little points.

So this is the equation:

a^3 b^1/2 c^-1/2 (ab)^1/3
------------------------------ the "divided by" line

( squareroot(a^3 squareroot(b c)

ok, i'm trying to figure out if the (ab) portion becomes a^1/3 and b^1/3

and if the squareroot of a^3 is actually a^3/2 ,

I'm pretty confused.

p.s. I'm working from the Higher Engineering Mathematics Textbook (this problem specifically) and College Algebra Demystified and reviewing Algebra Demystified.

and the answer is also there in the textbook, but it doesn't show you how to get it, and I've been trying to work the problem backwards too.

Thanks.
• Dec 16th 2009, 07:46 AM
earboth
Quote:

Originally Posted by immengin101
...
So this is the equation:

a^3 b^1/2 c^-1/2 (ab)^1/3
------------------------------ the "divided by" line

( squareroot(a^3 squareroot(b c)

ok, i'm trying to figure out if the (ab) portion becomes a^1/3 and b^1/3

and if the squareroot of a^3 is actually a^3/2 ,

...

If I understand your question correctly you want to simplify

$\displaystyle \dfrac{a^3 b^{\frac12} c^{-\frac12} \cdot (a b)^{\frac13}}{\sqrt{a^3 \cdot \sqrt{b c}}}$

Transform into a product:

$\displaystyle a^3 \cdot b^{\frac12} \cdot c^{-\frac12} \cdot a^{\frac13} \cdot b^{\frac13} \cdot a^{-\frac32} \cdot b^{-\frac14} \cdot c^{-\frac14}$

Collect like terms:

$\displaystyle a^{3+\frac13 - \frac32} \cdot b^{\frac12+\frac13-\frac14} \cdot c^{-\frac12-\frac34}$

I'll leave the rest for you.
• Dec 16th 2009, 07:46 AM
Stroodle
Quote:

Originally Posted by immengin101
ok, i'm trying to figure out if the (ab) portion becomes a^1/3 and b^1/3

and if the squareroot of a^3 is actually a^3/2 ,

Hi there.

$\displaystyle (ab)^{\frac{1}{3}}=a^{\frac{1}{3}}\times b^{\frac{1}{3}}$

and

$\displaystyle \sqrt{a^3}=(a^3)^{\frac{1}{2}}=a^{\frac{3}{2}}$
• Dec 16th 2009, 08:11 AM
immengin101
the
"c" isn't squarerooted under the divided by line

it looks like this

http://img130.imageshack.us/img130/9594/solve.jpg
• Dec 16th 2009, 08:13 AM
immengin101
Quote:

Originally Posted by Stroodle
Hi there.

$\displaystyle (ab)^{\frac{1}{3}}=a^{\frac{1}{3}}\times b^{\frac{1}{3}}$

and

$\displaystyle \sqrt{a^3}=(a^3)^{\frac{1}{2}}=a^{\frac{3}{2}}$

thank you.
• Dec 16th 2009, 08:19 AM
Wilmer
Quote:

Originally Posted by immengin101
a^3 b^1/2 c^-1/2 (ab)^1/3
------------------------------ the "divided by" line
( squareroot(a^3) squareroot(b c)
ok, i'm trying to figure out if the (ab) portion becomes a^1/3 and b^1/3
and if the squareroot of a^3 is actually a^3/2 ,

To start, more brackets required, and squareroot is same as ^(1/2); so:

a^3 b^(1/2) c^(-1/2) (ab)^(1/3)
--------------------------------- the "divided by" line
(a^3)^(1/2) (bc)^(1/2)

On your 2 "trying to figure out":
(ab)^(1/3) = a^(1/3) b^(1/3)
sqrt(a^3) = (a^3)^(1/2) = a^(3/2)
Looks like you know what you're doing!

Next step is separate (ab) and (bc):
a^3 b^(1/2) c^(-1/2) a^(1/3) b^(1/3)
-------------------------------------- the "divided by" line
a^(3/2) b^(1/2) c^(1/2)

Now, using rules:
a^x a^y = a(x+y)
a^x / a^y = a^(x-y) or 1/a^(y-z)
you should be able to wrap that up in this neat little xmas package:
a^(11/6) b^(1/3) / c
• Dec 16th 2009, 08:34 AM
immengin101
Quote:

Originally Posted by Wilmer
To start, more brackets required, and squareroot is same as ^(1/2); so:

a^3 b^(1/2) c^(-1/2) (ab)^(1/3)
--------------------------------- the "divided by" line
(a^3)^(1/2) (bc)^(1/2)

On your 2 "trying to figure out":
(ab)^(1/3) = a^(1/3) b^(1/3)
sqrt(a^3) = (a^3)^(1/2) = a^(3/2)
Looks like you know what you're doing!

Next step is separate (ab) and (bc):
a^3 b^(1/2) c^(-1/2) a^(1/3) b^(1/3)
-------------------------------------- the "divided by" line
a^(3/2) b^(1/2) c^(1/2)

Now, using rules:
a^x a^y = a(x+y)
a^x / a^y = a^(x-y) or 1/a^(y-z)
you should be able to wrap that up in this neat little xmas package:
a^(11/6) b^(1/3) / c

i don't think that the c is square rooted. the answer is posted above. but I sort know how to get there now thanks to you guys' help.

the answer is somehow a^11/6 b^1/3 c^- 3/2, the negative -3/2 powered to base of c
• Dec 16th 2009, 12:31 PM
Wilmer
( squareroot(a^3 squareroot(b c)
mean?

You have 3 "("s but only 1 ")"
• Dec 18th 2009, 01:49 PM
immengin101
hello, it seems that using normal typing isn't seeming to help.

please refer to this picture for the problem.

I was using the software algebrator to simplify this problem, however, it is giving me a different result. http://img130.imageshack.us/img130/9594/solve.jpg

that is the solution in the brackets, however, when I insert it into algebrator, there are entirely different results.

and I think the problem is being read by the program differently and I don't think the "a" squarerooted isn't supposed to cover the "b" that is squarerooted also. ugh, I hate this, it all sounds so confusing, I'm just trying to understand the steps.
• Dec 18th 2009, 05:09 PM
immengin101
nevermind, solved it, thanks guys!