Thread: Two exponent and root problems

1. Two exponent and root problems

I just need help with one problem on my homework...

Both the directions are to simplify
First one:
Fifth root of x cubed all over Seventh root of x to the 4th

I didnt know how to do that one at all...

Second one
(Fourth root of x cubed times Fourth root of x to the fifth) with a -2 exponent outside of the paranthesis

For that one, my answer was 1 over square root of x to the 8th.

Thanks so much to anyone who responds.... any help would be appreciated, let me know if anything needs to be further/more clearly explained.

2. 1. $\frac{\sqrt[5]{x}}{\sqrt[7]{x^{4}}}$

Roots can be expressed as exponents:
$a^{\frac{m}{n}} = \sqrt[n]{x^{m}} = \left(\sqrt[n]{x}\right)^{m}$

Then simplify as you would with whole number exponents.

2. Show us your work because I'm not sure how you arrived at that answer.

3. Originally Posted by o_O
1. $\frac{\sqrt[5]{x}}{\sqrt[7]{x^{4}}}$

Roots can be expressed as exponents:
$a^{\frac{m}{n}} = \sqrt[n]{x^{m}} = \left(\sqrt[n]{x}\right)^{m}$

Then simplify as you would with whole number exponents.

2. Show us your work because I'm not sure how you arrived at that answer.
Thank you!

For the second one...
First I multiplied what was in the paranthesis.... so that would be fourth root of x to the 8th. I didnt really know what to do after this. I know that when you have a negative exponent, if it is on the top of a fraction it will go on the bottom and vice versa. But I wasnt really sure how to simplify the fourth root of x to the 8th... maybe when you multiply 8 and -2 you get the fourth root of x to the -16th... and to reduce that further it could possibly be x to the -4th? So maybe it would be 1 over x to the fourth power? I'm not sure, I'm really lost on this one.

Sorry if all the text is confusing by the way, I'm not sure how to format my posts to use fractions and roots.

4. Hello, ZBomber

We must change the roots and powers into fractional exponents . . .

$1)\;\;\frac{\sqrt[5]{x^3}}{\sqrt[7]{x^4}}$
We have: . $\frac{x^{\frac{3}{5}}}{x^{\frac{4}{7}}}$

Then (as expected) subtract exponents: . $x^{\frac{3}{5}-\frac{4}{7}} \;=\;x^{\frac{21}{35} - \frac{20}{35}} \;=\;x^{\frac{1}{35}}$

$2)\;\;\left(\sqrt[4]{x^3}\cdot\sqrt[4]{x^5}\right)^{-2}$
We have: . $\left(x^{\frac{3}{4}}\cdot x^{\frac{5}{4}}\right)^{-2} \;=\;\left(x^{\frac{3}{4}+\frac{5}{4}}\right)^{-2} \;=\;\left(x^{\frac{8}{4}}\right)^{-2} \;=\;\left(x^2\right)^{-2} \;= \;x^{-4} \;=\;\frac{1}{x^4}$

5. Originally Posted by Soroban
Hello, ZBomber

We must change the roots and powers into fractional exponents . . .

We have: . $\frac{x^{\frac{3}{5}}}{x^{\frac{4}{7}}}$

Then (as expected) subtract exponents: . $x^{\frac{3}{5}-\frac{4}{7}} \;=\;x^{\frac{21}{35} - \frac{20}{35}} \;=\;x^{\frac{1}{35}}$

We have: . $\left(x^{\frac{3}{4}}\cdot x^{\frac{5}{4}}\right)^{-2} \;=\;\left(x^{\frac{3}{4}+\frac{5}{4}}\right)^{-2} \;=\;\left(x^{\frac{8}{4}}\right)^{-2} \;=\;\left(x^2\right)^{-2} \;= \;x^{-4} \;=\;\frac{1}{x^4}$

So I got the second one right the second time! Ok, thanks a lot! Those answers look spot on to me!