To prove $xyz + 1 = 2yz$, let's analyze the given conditions and expressions.

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Given:

1. $\log_{2a}(z) = x$
2. $\log_{2a}(z) = y$
3. $\log_{4a}(z) = z$

First, notice that $x = y$, since both represent $\log_{2a}(z)$. Let's denote $\log_{2a}(z)$ as $x$, so $x = y$.

Thus: $z = \log_{4a}(z).$

Using the change of base formula for logarithms: $\log_{4a}(z) = \frac{\log_{2a}(z)}{\log_{2a}(4a)}.$

Now calculate $\log_{2a}(4a)$: $\log_{2a}(4a) = \log_{2a}(4) + \log_{2a}(a) = 2 + 1 = 3.$

Therefore: $z = \frac{\log_{2a}(z)}{3}.$

Substituting $\log_{2a}(z) = x$: $z = \frac{x}{3}.$

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Substituting into $xyz + 1 = 2yz$:

Since $x = y$ and $z = \frac{x}{3}$, substitute these into the equation:

$xyz + 1 = 2yz.$

1. Left-hand side: $xyz + 1 = x \cdot x \cdot \frac{x}{3} + 1 = \frac{x^3}{3} + 1.$

2. Right-hand side: $2yz = 2 \cdot x \cdot \frac{x}{3} = \frac{2x^2}{3}.$

Equating both sides: $\frac{x^3}{3} + 1 = \frac{2x^2}{3}.$

Multiply through by 3 to eliminate the fraction: $x^3 + 3 = 2x^2.$

Rearrange into standard form: $x^3 - 2x^2 + 3 = 0.$

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Verifying the Equation:

The derived cubic equation ($x^3 - 2x^2 + 3 = 0$) doesn't seem to factor nicely over real numbers, implying that $x$ might have complex solutions. This suggests the original relationship $xyz + 1 = 2yz$ holds under specific conditions, particularly for $x$, $y$, and $z$ defined logarithmically.

If you'd like, I can numerically or symbolically solve $x^3 - 2x^2 + 3 = 0$ to determine exact values. Let me know how to proceed.

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Follow-Up Questions:

1. Would you like a numerical solution for the cubic equation?
2. Shall we explore alternative approaches to validate $xyz + 1 = 2yz$?
3. Would you like a deeper explanation of the logarithmic identities used?
4. Do you want a visual graph of $x^3 - 2x^2 + 3 = 0$?
5. Shall we explore specific cases where $xyz + 1 = 2yz$ holds true?

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Tip:

Always simplify logarithmic expressions by using the change-of-base formula to manage different bases effectively.