Question

If x^4 = y^16, then y =

Answer 1

you dont have any choices

Answer 2

?

Answer 3

@asnaseer

Answer 4

i mean like a b c or d

Answer 5

i have the answer i just need to know how and why of the process in obtaining the answer

Answer 6

it looks like you first need to learn about the various laws of exponents.

I suggest you look here first: http://www.mathsisfun.com/exponent.html

and then come back to this question.

If you are still stuck after this then ping me.

Answer 7

yea still stuck....

Answer 8

which parts do you not understand?

Answer 9

when you asked what is the 4th root of x^4

Answer 10

if you are given \(x^a\) then the a'th root of this is given by:\[\left(x^a\right)^{\frac{1}{a}}=x^{a\times\frac{1}{a}}=x^1=x\]

Answer 11

yea ... i've seen that before but im not gettin the logic of that.... why do you do that?

Answer 12

so the square root of \(x^2\) is given by:\[\left(x^2\right)^\frac{1}{2}=x^{2\times\frac{1}{2}}=x^1=x\]

Answer 13

square root can be written in two forms:\[\sqrt{x}\text{ or }x^{\frac{1}{2}}\]

Answer 14

yea i know that part

Answer 15

what logic do you not get?

Answer 16

so okay if theres 4 on the left side and 4^16 on the right how is it multiplied by 1/16 on both sides and why?

Answer 17

let me try to explain in another way...

Answer 18

sorry i meant 4(1/16) and 16(1/16)

Answer 19

I assume you that:\[x^2=x\times x\]and\[x^3=x\times x\times x\]etc...

Answer 20

yea i know that of course lol

Answer 21

good :)

now square root is called the "inverse" of squaring and cube root is called the "inverse" of cubing.

so, if you have \(x^2\), you can get back to \(x\) by taking its square root, i.e.:\[\sqrt{x^2}=(x^2)^\frac{1}{2}=x\]

Answer 22

similarly, if you have \(x^3\) you can get back yo \(x\) by taking its cube root, i.e.:\[\sqrt[3]{x^3}=(x^3)^\frac{1}{3}=x\]

Answer 23

oh .... hollup damn i think u just reminded me of something so a \[\sqrt{x} \] is equal to \[x ^{1/2}\]

Answer 24

and, in general, if you have the a'th power of \(x\), then you can get back to \(x\) by taking its a'th root, i.e.:\[\sqrt[a]{x^a}=(x^a)^\frac{1}{a}=x\]

Answer 25

yes - what you just said is correct

Answer 26

so even if there isn't a x^2 and just an x inside the square root it'll still be a x^1/2?

Answer 27

yes:\[\sqrt{x}=x^{\frac{1}{2}}\]

Answer 28

oh okay lol

Answer 29

now, in your question you are given \(y^{16}\) and are asked to find \(y\)

Answer 30

so you effectively need to take the 16'th root of both sides of the equation you are given

Answer 31

so it'll be 1/16 on both sides right?

Answer 32

perfect! :)

Answer 33

is there another way to do this problem ? you don't make it (4)^2?

Answer 34

I am not sure what you are asking by saying (4)^2 ?

Answer 35

making the 16 on the right side a (4)^2

Answer 36

you have:\[x^4=y^{16}\]now we take the 16'th root of both sides to get:\[(x^4)^{\frac{1}{16}}=(y^{16})^{\frac{1}{16}}\]

Answer 37

but we're trying to yet y by itself so that's why we're multiplying it by 1/16 tho right?

Answer 38

get not yet*

Answer 39

yes - except we are not "multiplying" by 1/16 - we are taking the 16'th root

Answer 40

the indexes on \(x\) and \(y\) are being multiplied

Answer 41

or exponents

Answer 42

yea okay i think i get it now

Answer 43

thanks !!

Answer 44

yw :)