Math Help - Surds and fractional indicies

1. 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?

2. 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.

3. 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...

4. 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: $\sqrt[3]{a^4b^5}*\frac{\sqrt[3]b}{\sqrt[12]{a^{19}}}$

5. Well that would certainly explain why I have had so many problems trying to reconcile my answer with the books!

Thanks hashish!

6. 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

7. Oh dear