indices, how do i solve the equations; x^4/3=16, and x^2/3=4?

Answer 1

OK!

1) x^4/3 = 16?

raise to the 3rd power:
x^4 = 16^3 = 2^4^3 = 2^12

take the 4th root
x= 2^(12/4 )= 2^3 = +8 and -8


2) x^2/3=4?

raise to 3rd power:
x^2= 4^3 = 2^2^3 = 2^6

take the second root
x= + or - 2^3 = +8 and -8

NOTE: the correct answer must include *both* +8 ands -8

See: http://answers.yahoo.com/question/index?qid=20061230093751AAwpfUo

...

Answer 2

X=8

Answer 3

1)
x^(4/3) = 16

x^2 = 16^(3/2) = 64

x = ±8

2)
x^2/3 = 4

x^2 = 4^3 = 64

x = ±8

-----------------
If you don't believe -8 is also a solution, you can use a calculator to check (-8)^(4/3). You'll see that (-8)^(4/3) = 16

Answer 4

you can slove this equation with the help of logarithams.
take log to the base 2:
x^4/3=16,takin log,
4/3logx=4;
logx=3;
x=2^3

Answer 5

X^(4/3) = 16
X=16^(3/4)
X= 2^(3)
X=8

X^(2/3) = 4
X = 4^(3/2)
X = 2^3
X = 8

Answer 6

x^4/3=16
the cuberoot of x^4 = 16
x^4 = 16^3
x^4= 4096
x=8

x^2/3=4
the cuberoot of x^2 = 4
x^2 = 4^3
x^2= 64
x=8

Answer 7

logarithams are your best best
take log to the base 2:
x^4/3=16,takin log,
4/3logx=4;
logx=3;
x=2^3

Answer 8

F*cking real. i in my opinion love math, algebra, calculus, geometry, etc. and that i might take a seat and do this each and each sometime if i had to. in certainty I even locate new uncomplicated ways to be sure new formula, and equations. i think it is why I have been given all "A's" at college