To solve the equation \( 729^{4x-1} = 9^{3x+8} \), we can express both sides of the equation using a common base.
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Recognize that:
- \( 729 = 9^3 \), so we can rewrite \( 729^{4x-1} \) as \( (9^3)^{4x-1} = 9^{3(4x-1)} = 9^{12x-3} \).
- Keeping \( 9^{3x+8} \) as is, we get the equation: \[ 9^{12x-3} = 9^{3x+8} \]
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Since the bases are the same, we can set the exponents equal to each other: \[ 12x - 3 = 3x + 8 \]
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Now, solve for \( x \):
- Subtract \( 3x \) from both sides: \[ 12x - 3x - 3 = 8 \] \[ 9x - 3 = 8 \]
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Add 3 to both sides: \[ 9x = 11 \]
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Divide both sides by 9: \[ x = \frac{11}{9} \]
Thus, the solution is \( x = \frac{11}{9} \), which corresponds to the option "x equals Start Fraction 11 over 6 End Fraction", which appears to be a formatting error as the correct amount is \( \frac{11}{9} \).
Therefore, the answer is: \[ x = \frac{11}{9} \]