algebra help?

Question

My calculus derivative is:
5/3x^2/3=1

How do you solve for x?
It's supposed to be .46 or (3/5)^3/2

5/3 times x to the 2/3 power.
does that help?

Answer 1

5/3x^2/3=1
multiply both sides by 3/5 to get
x^2/3 = 3/5
cube both sides to get
x^2 = (3/5)^3
square root both sides to get
x = +/-(3/5)^(3/2)

If you want this in simplest radical form, then
x = +/- (3/5)sqrt(3/5)
rationalize the denominator by multiplying by sqrt(5)/sqrt(5)
x = +/- [3 sqrt(15)]/25

I hope this helps!

Answer 2

I solved it as follows:
5/3x^2/3=1
5/3x^2=3
Scissor way: 5=3(3x^2)
5=9x^2
5/9=9x^2/9
5/9=x^2
√5/9=x = about 0.7 (It's an irrational number. So, I think √5/9 is more like the way I think would be correct.)

but no response from the above ones is √5/9, maybe I did something wrong.

Answer 3

5/3x^2/3=1
multiply both sides by 3/5 to get
x^2/3 = 3/5
cube both sides to get
x^2 = 27/125 .... (or (3/5)^3)
square root both sides to get
x = +/- sqrt(27/125) ..... (or +/-(3/5)^(3/2))
x = +/- (3/5)sqrt(3/5)
rationalize the denominator by multiplying by sqrt(5)/sqrt(5)
x = +/- [3 sqrt(15)]/25
this is what you do

Answer 4

it would be
x^2/3=1/(5/3)

x=5/3^3/2 0r .46
___________________________________________
becasue 1/(5/3) is equal to 5/3^-1
therefore
x^2/3=5/3^-1

then it would be
x=2/3sqrt{5/3^-1}

which equals to: x=5/3^3/2

if u don't understand this, ask ur teacher

Answer 5

5/(3x^(2/3))=1
3x^2/3 =5
x^2/3= 5/3
x = (5/3)^(3/2)

Answer 6

i dont understand the symbol ^ what does that mean i may be able to help

Answer 7

x= [-+] 1.34164 .......real