how do you solve x^(2/3)+13=17?

Start by subtracting 13 off of both sides so you have the X on one side and the normal on the other side of the equal sign.

so then you have x^(2/3)=4
Then you multiply x by the recipricle so that X is by itself. Whenever you do that, you have to do that to both sides.
so x^(2/3)=4 becomes x^(2/3)(3/2)=4(3/2)
x=8
8 to the 2/3 power = 4, which is from the 17-13 step we did earlier.

given that

x^(2/3) +13 = 17

put x^(1/3) = a

=> a^2 = 4

=> a = + or - 2

for a =2

x^(1/3) = 2

on cubing on both sides

=> x = 2^3

=> x=8

for a = -2

=. x^(1/3) = -2

on cubing on both sides

=> x = (-2)^3

=> x= -8

solution set is x= { -8 ,+8}

x^(2/3)+13=17
x^(2/3)=17-13
x^(2/3)=5
x=5^3/2
x=+/- 8

if you cube both sides
you should get
x^2=64
x=+or-8

x^(2/3) + 13 = 17
x^(2/3) = 17 - 13
x^(2/3) = 4
x = 4^(3/2)
x = ±√4^3
x = ±√64
x = ±8

x^(2/3)=17-13=4
now take the logarithm of both sides
log(x^(2/3)) = log(4)
and use the logarithm law log(a^b) = b log(a)
=> (2/3) log(x) = log(4)
=> log(x) = (3/2) log(4) = log(4^(3/2))
= log(sqrt(4)³) = log(2³) = log(8)
=> x = 8
Remark : x=-8 is also a solution but normally
we define powers only for positive numbers !

x^(2/3) = 4
x² = 4³
x² = 64
x = ± 8