# Surds and fractional indicies

• Apr 2nd 2010, 08:39 AM
AliceFisher
Surds and fractional indicies
I'm starting with basic simplification and while I thought I understood the rules I cannot seem to get the same answers as the book I am learning from.

The two questions that are vexing me are as follows:

Simplify:

1. $(3^3)^(1/2) * 9^(1/4)$

2. $CubeRoot(a^4b^5) * (b^(1/3))/a$

The answer for the first is 3^2 but I get 27^(7/4).

The second really makes me doubt by understanding as I swapped the first part around to read:

a^(4/3)b^(5/3) yet the book states this should be a^(3/4)b^(5/3).

Can anyone see where I am going wrong?
• Apr 2nd 2010, 09:38 AM
craig
For the first one.

$(3^3)^{\frac{1}{2}} * 9^(1/4)$
=
$3^{\frac{3}{2}} * (3^2)^{\frac{1}{4}}$
=
$3^{\frac{3}{2}} * 3^{\frac{1}{2}}$

which equals?

Read through this example looking at what I have done and attempt the second one again.

If you do not get it right post your working and we can go through where you went wrong.
• Apr 2nd 2010, 11:14 AM
AliceFisher
Thanks for the help Craig, answering the first has let me answer most of the other questions but I still seem to be drawing a blank on the second.

So the question is:

$\sqrt[3]{a^4b^5}*\sqrt[3]b/a$

which could be written as:

$(a^4)^\frac{1}{3}(b^5)^\frac{1}{3}*b^\frac{1}{3}*a ^{-1})$

So

$(a^\frac{4}{3})*(b^\frac{5}{3})*(b^\frac{1}{3})*(a ^{-1})$

To get the above answer would mean that:

Using the multiplication rule on the powers of a and b respectively gives:

$(a^\frac{1}{3})*(b^\frac{6}{3})$

or

$(a^\frac{1}{3})*(b^2)$

However the answer in the book is:

$(a^\frac{-1}{4})(b^2)$

To get this answer the first simplification would need to be:

$(a^\frac{3}{4})(b^\frac{5}{3})*b^\frac{1}{3}*a^{-1})$

but I have no idea how you would get this result for the first a...
• Apr 2nd 2010, 11:59 AM
harish21
Quote:

Originally Posted by AliceFisher
Thanks for the help Craig, answering the first has let me answer most of the other questions but I still seem to be drawing a blank on the second.

So the question is:

$\sqrt[3]{a^4b^5}*\sqrt[3]b/a$

which could be written as:

$(a^4)^\frac{1}{3}(b^5)^\frac{1}{3}*b^\frac{1}{3}*a ^{-1})$

So

$(a^\frac{4}{3})*(b^\frac{5}{3})*(b^\frac{1}{3})*(a ^{-1})$

To get the above answer would mean that:

Using the multiplication rule on the powers of a and b respectively gives:

$(a^\frac{1}{3})*(b^\frac{6}{3})$

or

$(a^\frac{1}{3})*(b^2)$

However the answer in the book is:

$(a^\frac{-1}{4})(b^2)$

To get this answer the first simplification would need to be:

$(a^\frac{3}{4})(b^\frac{5}{3})*b^\frac{1}{3}*a^{-1})$

but I have no idea how you would get this result for the first a...

As far as the question you have posted goes (if your question is $\sqrt[3]{a^4b^5}*\frac{\sqrt[3]b}{a}$),

the answer you have got , that is, $(a^\frac{1}{3})*(b^2)$
looks correct.

If the answer given by your book is $(a^\frac{-1}{4})(b^2)$, the question should look somewhat like this: $\sqrt[3]{a^4b^5}*\frac{\sqrt[3]b}{\sqrt[12]{a^{19}}}$
• Apr 2nd 2010, 12:02 PM
AliceFisher
Well that would certainly explain why I have had so many problems trying to reconcile my answer with the books!

Thanks hashish!
• Apr 2nd 2010, 12:09 PM
harish21
Quote:

Originally Posted by AliceFisher
Well that would certainly explain why I have had so many problems trying to reconcile my answer with the books!

Thanks hashish!

AliceFisher

Its Harish, not hashish!

hahah
• Apr 2nd 2010, 01:11 PM
AliceFisher
Oh dear :)