Solve the equation sqrt(x sqrt(x)) = 2x.

Step-by-step solution

1. 1

   Convert the nested radical to a single power. Using $\sqrt{x}=x^{1/2}$ and the product rule for exponents:

   $x\sqrt x=x^{1}\cdot x^{1/2}=x^{3/2}\quad\Longrightarrow\quad \sqrt{x^{3/2}}=\left(x^{3/2}\right)^{1/2}=x^{3/4}$

   so the equation is $x^{3/4}=2x$.

2. 2

   Fix the domain. Both $\sqrt{x}$ and the outer root require $x\ge 0$, and the right-hand side $2x$ must then be $\ge 0$ too — consistent. So we search only in $[0,\infty)$.

3. 3

   Handle $x=0$ separately before dividing. At $x=0$ the left side is $\sqrt{0}=0$ and the right side is $0$, so $x=0$ is a solution. Dividing by a power of $x$ in the next step would destroy it, which is exactly why it is checked first.

4. 4

   Divide by $x^{3/4}$ for $x>0$ and solve.

   $1=2x^{1-3/4}=2x^{1/4}\quad\Longrightarrow\quad x^{1/4}=\frac12\quad\Longrightarrow\quad x=\left(\frac12\right)^4=\frac{1}{16}$

5. 5

   Verify both solutions in the original equation. At $x=\tfrac1{16}$: $\sqrt{x}=\tfrac14$, so $x\sqrt x=\tfrac1{16}\cdot\tfrac14=\tfrac1{64}$ and $\sqrt{\tfrac1{64}}=\tfrac18$; meanwhile $2x=\tfrac18$ $\checkmark$. At $x=0$: $0=0$ $\checkmark$. Since $x^{3/4}$ grows more slowly than $2x$ for large $x$, there can be no further crossings.