Question

Evaluate fourth root of 9 multiplied by square root of 9 over the fourth root of 9 to the power of 5.

Answer 1

@Jadeishere

Answer 2

@TheSmartOne

Answer 3

i dont know where to even start with this equation

Answer 4

\[\frac{ \sqrt[4]{9} \times \sqrt{9}}{ \sqrt[4]{9}^2 }\]

Does it look like this?

Answer 5

Show your work. That does not help in any way :)

Answer 6

it's not even one of the answers

Answer 7

Okay then :) What we can do is rewrite the terms :)
\[\sqrt[4]{9}\]
can be written as \[9^{1/4}\]
\[\sqrt{9} = 3\]
\[\sqrt[4]{9}^5\] can be rewritten as
\[9^{5/4}\]

Answer 8

Any questions so far, as to why it does that?

Answer 9

what happened to the fact that the two parts where being divided?

Answer 10

Oh, they still are, I'm just breaking down the terms and rewriting them

Answer 11

ok

Answer 12

But, do you understand why they're changing into the new form? without the radical

Answer 13

it looks like you're treating the square root sigh as if it was a = sign and moving the powers to the other side with division

Answer 14

Yes, pretty much :) So, when you're looking for the root of a number, whether it's a square root, cubed root, or any other root, it's to a fractional power.

Now we have this;

\[\frac{ 9^{1/4} \times3 }{ 9^{5/4} }\]

Answer 15

where did three come from?

Answer 16

the square root of 9 is 3. 3 multiplied by 3 is equal to 9 :)

Answer 17

ok

Answer 18

then what

Answer 19

okay, sorry, I'm helping multiple people right now >.<

Answer 20

Then, since the base of the exponents are the same
we subtract (as done when dividing exponents) the exponents, always the bigger number by the larger
so \[9^{5/4} - 9^{1/4} = 9^{4/4} = 9\]
then we end up with 3/9, which equals 1/3.

Obviously, something is incorrect, so what we can do is make the 3 back into the square root of 9.
\[\sqrt{9} = 9^{1/2}\]
now we have \[\frac{ 9^{1/2} }{ 9 }\]

Answer 21

transforming 9 into \[9^{1} \]
then subtract \[9^{1/1} - 9^{1/2}\]

Answer 22

@agent0smith I think I did something wrong, please help!!

Answer 23

\[\frac{\sqrt[4]{9} \sqrt{9}}{9^{5/4}}=\frac{9^{3/4}}{9^{5/4}}=9^{\frac{3}{4}-\frac{5}{4}}=9^{-2/4}=\frac{1}{\sqrt{9}}=\frac{1}{3} \]

Answer 24

@Jadeishere \[\large \frac{ 9^{1/2} }{ 9^1 } = 9^{\frac{ 1 }{ 2} - 1}\]

Answer 25

ohhhh, Okay.