To solve the equation \( 729^{4x-1} = 9^{3x+8} \), we'll first express both sides with the same base.
First, we note that:
- \( 729 = 9^3 \), since \( 9^3 = 729 \).
Now we can rewrite \( 729^{4x-1} \) using base 9: \[ 729^{4x-1} = (9^3)^{4x-1} = 9^{3(4x-1)} = 9^{12x - 3} \]
So the equation now is: \[ 9^{12x - 3} = 9^{3x + 8} \]
Since the bases are the same (base 9), we can equate the exponents: \[ 12x - 3 = 3x + 8 \]
Now, solve for \( x \):
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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 by 9: \[ x = \frac{11}{9} \]
None of the provided responses matches \( x = \frac{11}{9} \), so the correct answer doesn't seem to be present in the responses given.