Solve for $x$ in the equation $(16x^6)^{\frac{1}{2}} = x$.

1. Problem Description

Solve for $x$ in the equation $(16x^6)^{\frac{1}{2}} = x$.

2. Solution Steps

We are given the equation $(16x^6)^{\frac{1}{2}} = x$.

First, we simplify the left side of the equation using the power of a product rule:

$(ab)^n = a^n b^n$

$(16x^6)^{\frac{1}{2}} = 16^{\frac{1}{2}} (x^6)^{\frac{1}{2}}$

Since $16 = 4^2$, we have $16^{\frac{1}{2}} = (4^2)^{\frac{1}{2}} = 4$.

Also, we use the power of a power rule $(a^m)^n = a^{mn}$, so $(x^6)^{\frac{1}{2}} = x^{6 \cdot \frac{1}{2}} = x^3$.

Therefore, we have:

$(16x^6)^{\frac{1}{2}} = 4x^3$

So the original equation becomes $4x^3 = x$.

Rearrange the equation:

$4x^3 - x = 0$

Factor out $x$:

$x(4x^2 - 1) = 0$

Further factor the term in parenthesis as a difference of squares:

$x(2x - 1)(2x + 1) = 0$

This gives us three possible solutions for $x$:

$x = 0$

$2x - 1 = 0 \implies 2x = 1 \implies x = \frac{1}{2}$

$2x + 1 = 0 \implies 2x = -1 \implies x = -\frac{1}{2}$

Now, we check these solutions in the original equation $(16x^6)^{\frac{1}{2}} = x$.

If $x = 0$, then $(16(0)^6)^{\frac{1}{2}} = (0)^{\frac{1}{2}} = 0$, so $x = 0$ is a solution.

If $x = \frac{1}{2}$, then $(16(\frac{1}{2})^6)^{\frac{1}{2}} = (16(\frac{1}{64}))^{\frac{1}{2}} = (\frac{16}{64})^{\frac{1}{2}} = (\frac{1}{4})^{\frac{1}{2}} = \frac{1}{2}$, so $x = \frac{1}{2}$ is a solution.

If $x = -\frac{1}{2}$, then $(16(-\frac{1}{2})^6)^{\frac{1}{2}} = (16(\frac{1}{64}))^{\frac{1}{2}} = (\frac{16}{64})^{\frac{1}{2}} = (\frac{1}{4})^{\frac{1}{2}} = \frac{1}{2}$.

Since $\frac{1}{2} \ne -\frac{1}{2}$, $x = -\frac{1}{2}$ is not a solution.

3. Final Answer

$x = 0, \frac{1}{2}$