# Math Help - Changin Root Forms

1. ## Changin Root Forms

Hi

How would you change the following root into the second expression. I am completely lost.

$\sqrt[5]{33}$ into $2(1+\frac{1}{32})^\frac{1}{5}$

I am kind of cheating by saying $\sqrt[5]{33} = 2.0.12346617$
and working back from that, but I am sure there is a better way.

Any help much appreciated!

Thanks!

2. Originally Posted by romjke
Hi

How would you change the following root into the second expression. I am completely lost.

$\sqrt[5]{33}$ into $2(1+\frac{1}{32})^\frac{1}{5}$

I am kind of cheating by saying $\sqrt[5]{33} = 2.924$
and working back from that, but I am sure there is a better way.

Any help much appreciated!

Thanks!
hi

$\sqrt[5]{32+1}$

$\sqrt[5]{32(1+\frac{1}{32})}$

$\sqrt[5]{32}\cdot (1+\frac{1}{32})^{\frac{1}{5}}$

and $\sqrt[5]{32}=2$

3. hmmm

I actually want to change the original form into the second I had listed.

This doesnt do that.

4. Originally Posted by romjke
hmmm

I actually want to change the original form into the second I had listed.

This doesnt do that.
what i did is from $\sqrt[5]{33}$ to $2(1+\frac{1}{32})^\frac{1}{5}$ ,

is this what you want ?

5. hmmm

I actually want to change the original form into the second I had listed.

This doesnt do that.
I don't want to start yet another conflict but "Any help much appreciated", right ?

However, the fact that $\sqrt[b]{a} = a^{\frac{1}{b}}$ will probably help you.

what i did is from $\sqrt[5]{33}$ to $2(1+\frac{1}{32})^\frac{1}{5}$ ,

is this what you want ?

Yes, sorry. My mistake. I see now what you did. Thank you.

But could you explain how you would choose the first step to split it into 32+1.

For example, what would be your first step in converting $\sqrt[3]{25}$ into a similar format?

7. $\sqrt[3]{25}$

$25^{\frac{1}{3}}$

$(27 - 2)^{\frac{1}{3}}$

$\left (27 \left ( 1 - \frac{2}{27} \right ) \right )^{\frac{1}{3}}$

$3 \left ( 1 - \frac{2}{27} \right )^{\frac{1}{3}}$

Is this what you want as a format ? If yes, then I chose 27 because it is the nearest number to 25 that is a perfect cube (that has an integer as its cubic root, because we are considering the cubic root here). In the previous question, mathaddict chose 32 because it is the nearest number to 33 that has an integer as its fifth root. Do you get the idea ?

8. Yes this makes sense now

thanks