We need to solve for $x$ in the equation $(\frac{1}{3})^{\frac{x^2-2x}{16-2x^2}} = 4^x \sqrt{9}$.

1. Problem Description

We need to solve for $x$ in the equation $(\frac{1}{3})^{\frac{x^2-2x}{16-2x^2}} = 4^x \sqrt{9}$.

2. Solution Steps

First, rewrite the equation. Since $\frac{1}{3} = 3^{-1}$ and $\sqrt{9} = 3$, we have:

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

$3^{\frac{-x^2+2x}{16-2x^2}} = 4^x \cdot 3^1$

$3^{\frac{-x^2+2x}{16-2x^2}} = (2^2)^x \cdot 3^1$

$3^{\frac{-x^2+2x}{16-2x^2}} = 2^{2x} \cdot 3^1$

Taking the logarithm base 3 on both sides is not helpful because of the term $2^{2x}$.

Let's try rewriting the original equation as follows:

$(\frac{1}{3})^{\frac{x^2-2x}{16-2x^2}} = 4^x \cdot \sqrt{9}$

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

Rewrite the equation by taking logarithms with base 10 on both sides:

$\log((\frac{1}{3})^{\frac{x^2-2x}{16-2x^2}}) = \log(4^x \cdot 3)$

$\frac{x^2-2x}{16-2x^2} \log(\frac{1}{3}) = \log(4^x) + \log(3)$

$\frac{x^2-2x}{16-2x^2} \log(3^{-1}) = x \log(4) + \log(3)$

$\frac{x^2-2x}{16-2x^2} (- \log(3)) = x \log(4) + \log(3)$

$-\frac{x^2-2x}{16-2x^2} \log(3) = x \log(4) + \log(3)$

Divide both sides by $\log(3)$:

$-\frac{x^2-2x}{16-2x^2} = x \frac{\log(4)}{\log(3)} + 1$

$-\frac{x^2-2x}{16-2x^2} = x \log_3(4) + 1$

Another approach: let's try to get rid of $4^x$:

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

If $x=0$, then $(\frac{1}{3})^{\frac{0}{16}} = 4^0 \cdot 3$, so $1 = 1 \cdot 3$, which means $1 = 3$, which is false.

If $x = -1$, then $(\frac{1}{3})^{\frac{1+2}{16-2}} = (\frac{1}{3})^{\frac{3}{14}} = 4^{-1} \cdot 3 = \frac{3}{4}$

$(\frac{1}{3})^{\frac{3}{14}} \approx 0.7534$, so $\frac{3}{4} = 0.75$. Therefore, $x=-1$ could be an approximate solution.

Let $x=-2$: $(\frac{1}{3})^{\frac{4+4}{16-8}} = (\frac{1}{3})^{\frac{8}{8}} = \frac{1}{3} = 4^{-2} \cdot 3 = \frac{1}{16} \cdot 3 = \frac{3}{16}$. This is also false.

Now let's simplify our original equation:

$3^{\frac{2x-x^2}{16-2x^2}} = 4^x \cdot 3^1$

$3^{\frac{2x-x^2}{16-2x^2} - 1} = 4^x$

$3^{\frac{2x-x^2-16+2x^2}{16-2x^2}} = 4^x$

$3^{\frac{x^2+2x-16}{16-2x^2}} = 4^x$

If $x = -4$, $3^{\frac{16-8-16}{16-32}} = 3^{\frac{-8}{-16}} = 3^{\frac{1}{2}} = \sqrt{3}$. Then $4^{-4} = \frac{1}{256}$. So $x=-4$ is not a solution.

The equation can be rewritten as:

$(\frac{1}{3})^{\frac{x^2-2x}{16-2x^2}} = 3 \cdot (2^2)^x$

$(\frac{1}{3})^{\frac{x^2-2x}{16-2x^2}} = 3 \cdot 2^{2x}$

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

$3^{\frac{2x-x^2}{16-2x^2}} = 3^1 \cdot 4^x$

Take the natural logarithm of both sides:

$\ln(3^{\frac{2x-x^2}{16-2x^2}}) = \ln(3^1 \cdot 4^x)$

$\frac{2x-x^2}{16-2x^2} \ln(3) = \ln(3) + x \ln(4)$

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

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

$\frac{2x-x^2 - (16-2x^2)}{16-2x^2} = x \log_3(4)$

$\frac{2x-x^2 - 16+2x^2}{16-2x^2} = x \log_3(4)$

$\frac{x^2+2x-16}{16-2x^2} = x \log_3(4)$

$x^2+2x-16 = x \log_3(4) (16-2x^2)$

$x^2+2x-16 = 16x \log_3(4) - 2x^3 \log_3(4)$

$2 \log_3(4)x^3 + x^2 + (2 - 16\log_3(4))x - 16 = 0$

We can observe that if $x = 2$, the left side of the original equation is $(\frac{1}{3})^{\frac{4-4}{16-8}} = (\frac{1}{3})^0 = 1$. The right side is $4^2\sqrt{9} = 16 \cdot 3 = 48$.

Consider the case where the exponent $\frac{x^2-2x}{16-2x^2} = 0$. This requires $x^2-2x = 0$, meaning $x(x-2) = 0$, so $x=0$ or $x=2$. We have already shown $x=0$ is not a solution. If $x=2$, we have $(\frac{1}{3})^0 = 4^2 \sqrt{9}$, or $1 = 16 \cdot 3 = 48$. Thus, $x=2$ is also not a solution.

Let's test $x = -2$. $(\frac{1}{3})^{\frac{4+4}{16-8}} = (\frac{1}{3})^1 = \frac{1}{3}$. The right side is $4^{-2} \sqrt{9} = \frac{1}{16} \cdot 3 = \frac{3}{16}$.

If $x = \frac{1}{2}$, then $\frac{x^2-2x}{16-2x^2} = \frac{\frac{1}{4}-1}{16-\frac{1}{2}} = \frac{-\frac{3}{4}}{\frac{31}{2}} = -\frac{3}{4} \cdot \frac{2}{31} = -\frac{3}{62}$. Then the left side is $(\frac{1}{3})^{-\frac{3}{62}} = 3^{\frac{3}{62}} \approx 1.0552$. The right side is $4^{1/2} \cdot 3 = 2 \cdot 3 = 6$.

Consider x = -1: $(\frac{1}{3})^{\frac{1+2}{16-2}} = (\frac{1}{3})^{\frac{3}{14}} \approx (0.3333)^{0.214} = 0.7534$. $4^{-1} \sqrt{9} = \frac{1}{4} \cdot 3 = \frac{3}{4} = 0.75$.

3. Final Answer

x = -1