So we have the equation: [tex] 5x=40x^{-\frac{1}{2} } [/tex]
Let's divide by [tex] x^{-\frac{1}{2}} [/tex] to get x on one side:
[tex] \frac{5x}{x^{-\frac{1}{2}}} =\frac{40x^{-\frac{1}{2}}}{x^{-\frac{1}{2}}} [/tex]
When we divide variables and they have different exponents, we subtract the numerator's exponent from the denominator's exponent, so in this question we have: [tex] 1-(-\frac{1}{2})=1+\frac{1}{2}=\frac{3}{2} [/tex]
So then we are left with:
[tex] 5x^{\frac{3}{2}} =40 [/tex]
Then we need to divide by 5 to solve for the x term:
[tex] x^{\frac{3}{2}} =8 [/tex]
This can be rewritten as:
[tex] \sqrt{x^{3}}=8 [/tex]
Then we square both sides to get the x term out of the square root:
[tex] (\sqrt{x^{3}})^{2} =8^{2} [/tex]
[tex] x^{3} =64 [/tex]
Then we take the cube-root of both sides to solve for x:
[tex] \sqrt[3]{x^{3}} =\sqrt[3]{64} [/tex]
[tex] x=4 [/tex]
So now we know that x is equal to 4.
