# Another powers problem

#### 12345678

##### Junior Member
Firstly, I'd like to say sorry for posting another problem- feel slightly rude posting more than one problem in a day.
'Solve the follow equation: X^2/3= 2X -1/3'
Firstly, I changed the equation to : Cube root (X^2) = 2 * 1/X^3
I then cubed each side to get rid of the cube root : X^2 = 8 * 1/X^9
I then got rid of the fraction by multiplying by X^9: X^11 = 8(X^9)
I then divided by X^9: X^2 = 8
Meaning X = square root 8 or 2 root 2
I'm wondering if I have made an error with the powers: as the answer is x=2, which could be gotten from X^3=8.
Any suggestions?

#### Subhotosh Khan

##### Super Moderator
Staff member
Firstly, I'd like to say sorry for posting another problem- feel slightly rude posting more than one problem in a day.

No problem ........... as long as you show your work (as you have done) - you can post as many problems as you like...

'Solve the follow equation: X^(2/3)= 2X^(-1/3)'......................................................Is this what you meant?
Firstly, I changed the equation to : Cube root (X^2) = 2 * 1/X^3
I then cubed each side to get rid of the cube root : X^2 = 8 * 1/X^9
I then got rid of the fraction by multiplying by X^9: X^11 = 8(X^9)
I then divided by X^9: X^2 = 8
Meaning X = square root 8 or 2 root 2
I'm wondering if I have made an error with the powers: as the answer is x=2, which could be gotten from X^3=8.
Any suggestions?
My way:

X^(2/3)= 2X^(-1/3)

X^(2/3) * X^(1/3) = 2 *X^(-1/3) * X^(1/3) ...................... multiply both sides by X^(-1/3), assuming x$$\displaystyle \ne 0$$

x = 2

$$\displaystyle \displaystyle \sqrt{x^2} \ = \ \frac{2}{\sqrt{x}}$$ ..................... here was your mistake

$$\displaystyle \displaystyle x^2 \ = \ \frac{8}{x}$$

$$\displaystyle \displaystyle x^3 \ = \ 8$$

x = 2

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#### 12345678

##### Junior Member
My way:

X^(2/3)= 2X^(-1/3)

X^(2/3) * X^(1/3) = 2 *X^(-1/3) * X^(1/3) ...................... multiply both sides by X^(-1/3), assuming x$$\displaystyle \ne 0$$

x = 2

$$\displaystyle \displaystyle \sqrt{x^2} \ = \ \frac{2}{\sqrt{x}}$$ ..................... here was your mistake
$$\displaystyle \displaystyle x^2 \ = \ \frac{8}{x}$$
$$\displaystyle \displaystyle x^3 \ = \ 8$$
Oh, I did X^3 instead of cube root. Makes sense now, thank you for helping 