Solving x^(2/3)=4: Logarithms or What?

[Mozart]

I don't remember how to do this x^(2/3)=4 solve for x Do you use logarithm or something?

 

[matrixx333]

I'm going to try my best to explain. Whenever you have a fractional exponent, that is another way of talking about a radical...


The "3" in the denominator of the fraction is what "root" your radical will be:

Ex. square root of 2 = 2^1/2
cubed root of 2 = 2^1/3

now, the numerator is what power "X" is going to be:

Ex. square root of 2¹ = 2^1/2
cubed root of 2² = 2^2/3

That should hopefully get you started in the right direction...

 

[Mozart]

After reading what you said I did the following.

wrote x^2 with the third root around it and made it equal 4.

I then took the square root of 4 to get rid of the square sign on x

I then cubed the 2 and got 8

After pugging the 8 into x^(2/3) I got 4

Please tell me I got the answer using a proper method and this isn't a cruel coincidence. Thanks by the way for the help!

 

[d_leet]

After reading what you said I did the following.
wrote x^2 with the third root around it and made it equal 4.
I then took the square root of 4 to get rid of the square sign on x
I then cubed the 2 and got 8
After pugging the 8 into x^(2/3) I got 4
Please tell me I got the answer using a proper method and this isn't a cruel coincidence. Thanks by the way for the help!

sounds like you did it right.

 

[Mozart]

Thanks.