I don't remember how to do this x^(2/3)=4 solve for x Do you use logarithm or something?
I'm gonna 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 = 21/2
cubed root of 2 = 21/3
now, the numerator is what power "X" is going to be:
Ex. square root of 21 = 21/2
cubed root of 22 = 22/3
That should hopefully get you started in the right direction.....
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.
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