We are tasked with solving the equation for the lowest value of $x$:

$\frac{2\log_3(x) - 4}{\log_3\left(\frac{x}{9}\right)} = \log_3(x^7) - \left(\frac{1}{\log_x(3)}\right)^2 + 8$

Step 1: Simplify logarithmic terms

We'll begin by simplifying both sides of the equation.

Left-hand side:

$\frac{2\log_3(x) - 4}{\log_3\left(\frac{x}{9}\right)} = \frac{2\log_3(x) - 4}{\log_3(x) - \log_3(9)}$ Since $\log_3(9) = 2$, we can rewrite the denominator: $= \frac{2\log_3(x) - 4}{\log_3(x) - 2}$

Right-hand side:

$\log_3(x^7) = 7\log_3(x)$ Also, recall that $\log_x(3) = \frac{1}{\log_3(x)}$, so: $\left(\frac{1}{\log_x(3)}\right)^2 = \log_3(x)^2$ Thus, the right-hand side becomes: $7\log_3(x) - \log_3(x)^2 + 8$

Step 2: Rewrite the equation

The equation now is: $\frac{2\log_3(x) - 4}{\log_3(x) - 2} = 7\log_3(x) - \log_3(x)^2 + 8$

Step 3: Let $y = \log_3(x)$

To simplify the equation, substitute $y = \log_3(x)$. This gives us: $\frac{2y - 4}{y - 2} = 7y - y^2 + 8$

Step 4: Solve the equation

Multiply both sides by $y - 2$ to eliminate the denominator on the left-hand side: $2y - 4 = (7y - y^2 + 8)(y - 2)$ Expand the right-hand side: $2y - 4 = (7y - y^2 + 8)(y - 2) = 7y^2 - 14y - y^3 + 2y^2 + 8y - 16$ Simplify the equation: $2y - 4 = -y^3 + 9y^2 - 6y - 16$ Move all terms to one side: $0 = -y^3 + 9y^2 - 8y - 12$ Rearrange: $y^3 - 9y^2 + 8y + 12 = 0$

Step 5: Solve the cubic equation

We now solve $y^3 - 9y^2 + 8y + 12 = 0$. One way to find the roots of a cubic equation is by trial and error, or using the Rational Root Theorem.

Using trial and error, testing possible rational roots like $y = 1$, $y = 2$, and $y = 3$, we find that $y = 3$ is a root.

Step 6: Factor the cubic equation

Since $y = 3$ is a root, we can factor the cubic equation as: $y^3 - 9y^2 + 8y + 12 = (y - 3)(y^2 - 6y - 4) = 0$ Now, solve the quadratic $y^2 - 6y - 4 = 0$ using the quadratic formula: $y = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-4)}}{2(1)} = \frac{6 \pm \sqrt{36 + 16}}{2} = \frac{6 \pm \sqrt{52}}{2} = \frac{6 \pm 2\sqrt{13}}{2} = 3 \pm \sqrt{13}$

Thus, the roots are $y = 3$, $y = 3 + \sqrt{13}$, and $y = 3 - \sqrt{13}$.

Step 7: Convert back to $x$

Recall $y = \log_3(x)$, so: $x = 3^y$ For $y = 3$, we have $x = 3^3 = 27$.

For the other roots $y = 3 + \sqrt{13}$ and $y = 3 - \sqrt{13}$, $x = 3^{3 \pm \sqrt{13}}$ would give irrational values, but since we're asked for the lowest value of $x$, the lowest occurs when $y = 3 - \sqrt{13}$, resulting in a very small $x$.

Conclusion:

The lowest value of $x$ is $3^{3 - \sqrt{13}}$, which is approximately a very small value.

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Would you like a deeper explanation of any step?

Related questions:

1. How do you solve cubic equations using the Rational Root Theorem?
2. Can you explain the properties of logarithmic equations in more detail?
3. How do you apply the quadratic formula to solve quadratic equations?
4. Why is $\log_x(3) = \frac{1}{\log_3(x)}$?
5. What are some practical applications of solving logarithmic equations?

Tip: When solving logarithmic equations, it's helpful to use logarithmic identities like change of base and log-product/sum rules to simplify expressions before solving.