To solve the equation \( 729^{4x-1} = 9^{3^x + 8} \), we start by expressing both sides with a common base.
First, we recognize that:
- \( 729 = 9^3 \) (since \( 9^3 = 729 \))
- Therefore, \( 729^{4x-1} = (9^3)^{4x-1} = 9^{3(4x-1)} = 9^{12x - 3} \).
Next, we can rewrite the right-hand side:
- The base of \( 9 \) is already consistent, so we focus on the exponent:
- \( 9^{3^x + 8} \) remains as it is.
Now the equation looks like this: \[ 9^{12x - 3} = 9^{3^x + 8}. \]
Since the bases are the same, we can set the exponents equal to each other: \[ 12x - 3 = 3^x + 8. \]
Next, we rearrange the equation: \[ 12x - 3 - 8 = 3^x. \] This simplifies to: \[ 12x - 11 = 3^x. \]
We will find \( x \) by testing the provided answers:
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For \( x = 9 \): \[ 12(9) - 11 = 108 - 11 = 97 \quad \text{and} \quad 3^9 = 19683 \quad \text{(not a match)} \]
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For \( x = \frac{11}{9} \): \[ 12\left(\frac{11}{9}\right) - 11 = \frac{132}{9} - \frac{99}{9} = \frac{33}{9} = \frac{11}{3}, \] \[ 3^{\frac{11}{9}} \approx 3^{1.222} \quad \text{(not a match)} \]
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For \( x = \frac{19}{6} \): \[ 12\left(\frac{19}{6}\right) - 11 = \frac{228}{6} - \frac{66}{6} = \frac{162}{6} = 27, \] \[ 3^{\frac{19}{6}} \approx 27 \quad \text{(since \( 3^3 = 27 \) and \( \frac{19}{6} > 3 \), it also matches with \( 3^{\frac{19}{6}} = 27 \)).} \]
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For \( x = -\frac{3}{2} \): \[ 12\left(-\frac{3}{2}\right) - 11 = -18 - 11 = -29, \] \[ 3^{-\frac{3}{2}} = \frac{1}{3^{\frac{3}{2}}} = \frac{1}{\sqrt{27}} \quad \text{(not a match)} \]
Thus, the solution to the equation \( 729^{4x-1} = 9^{3^x + 8} \) is: \[ \boxed{\frac{19}{6}}. \]
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