Question

MEDAL AND FAN
PLEASE HELP ME!!

Evaluate.
5^square root of 8 times square root of 8 divided by
3^square root of 8 to the power of 5

Answer 1

something like 5*root8 * root8 / 3*root8^5 ?

Answer 2

no the 5 is like the cube root of square root of 8 and the 3 is the cube root of the square root of to the power of 5, i think its called cube root. Its like an exponent on the opposite side of the number

Answer 3

Here

Answer 4

It's late so I need to finish asap

Answer 5

P0sitr0n?

Answer 6

still need help

Answer 7

yes!

Answer 8

are you able to convert each radical to exponent form?

Answer 9

Well I know that the 8 on top becomes 8 to the power of 1/5 but after that I dont know what to do

Answer 10

I think you mean \[\Large \sqrt[5]{8} = 8^{1/5}\] right?

Answer 11

yes

Answer 12

how about the square root of 8?

Answer 13

I don't how to simplify it from that point

Answer 14

which is pretty sad i know

Answer 15

no, it's not

Answer 16

\[\Large \sqrt{8} = 8^{1/2}\]

Answer 17

then how about the cube root?

Answer 18

where did the 2 come from

Answer 19

it turns out that square roots are simply radicals with index of 2

\[\Large \sqrt{8} = \sqrt[2]{8}\]

Answer 20

oh ok

Answer 21

but you don't need to write that little "2" because it's understood that square roots are radicals with index 2

it's like x/1 = x in a way

Answer 22

how about the cube root?

Answer 23

it becomes 8^3/5?

Answer 24

oh wait, no, I mixed up the two

Answer 25

so right now we have 8^1/5 and 8^1/2 over 8^3/5

Answer 26

what?

Answer 27

it should be

\[\Large \sqrt[3]{8^5} = 8^{5/3}\]

Answer 28

so we go from this

\[\Large \frac{\sqrt[5]{8}\sqrt{8}}{\sqrt[3]{8^5}}\]

to this

\[\Large \frac{8^{1/5}*8^{1/2}}{8^{5/3}}\]

Answer 29

alright so the top simplifies to 64^7/10

Answer 30

or is it 64^1/10?

Answer 31

no you don't say 8*8 = 64 in the numerator

the base stays 8 the whole time

Answer 32

oh

Answer 33

so its 8^7/10

Answer 34

yeah up top

Answer 35

bye

Answer 36

ok so what do we do from here

Answer 37

we have this now