Solve this equation for x, using exponents

[hank]

Trying to solve the following equation for x:

x^(-2/3) + 2 = 0

Here's what I do:

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

This is where I get stuck, because (-1/2)^(1/3) is a negative, and the next step would require me to take the square root of a negative, which I can't do.

What am I missing?

 

[Denis]

Keep it simple; rule: if a^x = b, then a = b^(1/x)

 

[galactus]

\(\displaystyle \L\\\frac{1}{x^{\frac{2}{3}}}=-2\)


Not true for any x.

 

[pka]

Agreed! No real solution.