Solve for $x$ in the equation $(\frac{1}{3})^{\frac{x^2 - 2x}{16 - 2x^2}} = \sqrt[4x]{9}$.

1. Problem Description

Solve for $x$ in the equation $(\frac{1}{3})^{\frac{x^2 - 2x}{16 - 2x^2}} = \sqrt[4x]{9}$.

2. Solution Steps

The given equation is $(\frac{1}{3})^{\frac{x^2 - 2x}{16 - 2x^2}} = \sqrt[4x]{9}$.

We can rewrite $\frac{1}{3}$ as $3^{-1}$ and $9$ as $3^2$. So the equation becomes:

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

Using the power of a power rule, $(a^m)^n = a^{mn}$, we have:

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

$3^{\frac{-x(x - 2)}{2(8 - x^2)}} = 3^{\frac{1}{2x}}$

Since the bases are equal, we equate the exponents:

$\frac{-x(x - 2)}{2(8 - x^2)} = \frac{1}{2x}$

$\frac{-x(x - 2)}{2(8 - x^2)} = \frac{1}{2x}$

Cross-multiply:

$-2x^2(x - 2) = 2(8 - x^2)$

$-2x^3 + 4x^2 = 16 - 2x^2$

$-2x^3 + 6x^2 - 16 = 0$

Divide by -2:

$x^3 - 3x^2 + 8 = 0$

Let $f(x) = x^3 - 3x^2 + 8$. We look for integer roots of this cubic equation. By the rational root theorem, any rational root must be a factor of

8. Testing $x = -1$: $(-1)^3 - 3(-1)^2 + 8 = -1 - 3 + 8 = 4 \ne 0$

Testing $x = 1$: $(1)^3 - 3(1)^2 + 8 = 1 - 3 + 8 = 6 \ne 0$

Testing $x = -2$: $(-2)^3 - 3(-2)^2 + 8 = -8 - 12 + 8 = -12 \ne 0$

Testing $x = 2$: $(2)^3 - 3(2)^2 + 8 = 8 - 12 + 8 = 4 \ne 0$

Testing $x = -4$: $(-4)^3 - 3(-4)^2 + 8 = -64 - 48 + 8 = -104 \ne 0$

Testing $x = 4$: $(4)^3 - 3(4)^2 + 8 = 64 - 48 + 8 = 24 \ne 0$

However, trying $x = -1.64$:

$f(-1.64) = (-1.64)^3 - 3(-1.64)^2 + 8 \approx -4.39 - 8.07 + 8 \approx -4.46$

Let's try x = -1.

6. $f(-1.6) = (-1.6)^3 - 3(-1.6)^2 + 8 = -4.096 - 3(2.56) + 8 = -4.096 - 7.68 + 8 = -3.776$

Since $x$ cannot be 0, we have

$-x(x-2) / 2(8-x^2) = 1/(2x)$. So, $16 - 2x^2 \ne 0$ which means $x^2 \ne 8$, so $x \ne \pm \sqrt{8}$.

Since $4x$ is the index of a root, $x$ must be positive. Also, $x \ne 0$.

If we let $x = 2$, the original equation becomes:

$(\frac{1}{3})^{\frac{2^2 - 2(2)}{16 - 2(2^2)}} = (\frac{1}{3})^{\frac{4 - 4}{16 - 8}} = (\frac{1}{3})^{\frac{0}{8}} = (\frac{1}{3})^0 = 1$

$\sqrt[4x]{9} = \sqrt[4(2)]{9} = \sqrt[8]{9} = 9^{\frac{1}{8}} = (3^2)^{\frac{1}{8}} = 3^{\frac{2}{8}} = 3^{\frac{1}{4}} = \sqrt[4]{3}$

So $x = 2$ is not a solution.

Re-examine $x^3 - 3x^2 + 8 = 0$. Numerical methods suggest that x is approximately $-1.65$. However we can't have non-integer value.

The cubic equation $x^3 - 3x^2 + 8 = 0$ has one real root approximately -1.

6

5. Since $4x$ is the index of the root, $4x$ must be an integer and $4x > 1$. Therefore $x > 0$. Also, we need to ensure that $16 - 2x^2 \ne 0$ (denominator cannot be zero) and $x \ne 0$ (division by zero). Hence $x \ne \pm \sqrt{8}$.

3. Final Answer

No real solution for $x$.