The following question is a Math Olympiad Problem:

Find $$x>0$$ that solves $$x\sqrt{x\sqrt x}=2$$

The answer is $$\sqrt[7]{16}$$ but I got $$2\sqrt[3]{2}$$ by: \begin{align}x\sqrt{x\sqrt x}=2&\to \sqrt{x\sqrt{x}}=\frac2x\\&\to x\sqrt{x}=\frac{4}{x^{2}}\\&\to \sqrt{x}=\frac{4}{x}\to x=\frac{16}{x^{2}}\\&\to x^{3}=16\\&\to x=\sqrt[3]{16}\\&\to x=2\sqrt[3]2\end{align}

What have I done wrong?

• Welcome to Math SE. One issue is going from $x\sqrt{x}=\frac{4}{x^2}$ to your $\sqrt{x}=\frac{4}{x}$. You apparently divided both sides by $x$ to get rid of the $x$ on the LHS, so the result should then be $\sqrt{x}=\frac{4}{x^{\color{red}{3}}}$ instead. Apr 23 at 18:04
• Do you mean $\sqrt[7]{16}$? Apr 23 at 18:05
• You were fine up until $$x\sqrt{x}=\frac{4}{x^2},$$ then you made a mistake. Apr 23 at 18:05
• Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.
– Community Bot
Apr 23 at 18:09

$$\color{red}{x\sqrt{x}=\frac{4}{x^{2}}}\to \color{purple}{\sqrt{x}=\frac{4}{x}}$$

I think you can understand your mistake.

• Thanks, found my error, i guess my mind isnt working today. Apr 23 at 18:15

The answer by MathStackexchangeisNotSoBad identifies the error in the method that you have used.

Here is a simpler alternative approach to the one you have taken which uses the rules of exponentiation to simplify the problem:

$$x \sqrt{x \sqrt{x}} = x \sqrt{x \cdot x^{1/2}} = x \sqrt{x^{3/2}} = x(x^{3/2})^{1/2} = x \cdot x^{3/4} = x^{7/4}$$

So we substitute the above into your problem as follows:

$$x \sqrt{x \sqrt{x}} = 2 \iff x^{7/4} = 2 \iff x^7 = 16$$

Therefore $$x= \sqrt[7]{16}$$ as required.