To solve the equation
\[ \frac{3^{x+1} - 3^x}{3^{x-1}} = 2 \left(\frac{1}{9}\right)^{x-1}, \]
let's start by simplifying the left side of the equation.
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Simplifying the left side: The left side can be rewritten as:
\[ \frac{3^{x+1} - 3^x}{3^{x-1}} = \frac{3 \cdot 3^x - 3^x}{3^{x-1}} = \frac{(3-1)3^x}{3^{x-1}} = \frac{2 \cdot 3^x}{3^{x-1}}. \]
This simplifies to:
\[ 2 \cdot 3. \]
Therefore, the left side is simply \(6\).
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Simplifying the right side: The term \(\frac{1}{9}\) can be written as \(3^{-2}\), so we have:
\[ \left(\frac{1}{9}\right)^{x-1} = (3^{-2})^{x-1} = 3^{-2(x-1)} = 3^{-2x + 2}. \]
Thus, the right side becomes:
\[ 2 \cdot 3^{-2x + 2}. \]
We can simplify this to:
\[ 2 \cdot \frac{9}{3^{2x}} = \frac{18}{3^{2x}}. \]
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Setting the two sides equal: Now we rewrite the original equation:
\[ 6 = \frac{18}{3^{2x}}. \]
To eliminate the fraction, we can multiply both sides by \(3^{2x}\):
\[ 6 \cdot 3^{2x} = 18. \]
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Dividing both sides by 6: We can further simplify this equation:
\[ 3^{2x} = 3. \]
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Solving for \(x\): The equation \(3^{2x} = 3\) can be written as:
\[ 3^{2x} = 3^1. \]
Since the bases are equal, we equate the exponents:
\[ 2x = 1. \]
Dividing both sides by 2 gives:
\[ x = \frac{1}{2}. \]
Thus, the solution for \(x\) is
\[ \boxed{\frac{1}{2}}. \]