To solve the equation \( 729^{4x-1} = 9^{3x+8} \), we can start by expressing both sides of the equation in terms of the same base.
First, we notice that \( 729 \) and \( 9 \) can both be expressed as powers of \( 3 \):
- \( 729 = 3^6 \) (since \( 3^6 = 729 \))
- \( 9 = 3^2 \) (since \( 3^2 = 9 \))
Now we can rewrite the original equation using these expressions: \[ (3^6)^{4x-1} = (3^2)^{3x+8} \]
Using the laws of exponents, we simplify both sides: \[ 3^{6(4x-1)} = 3^{2(3x+8)} \]
This simplifies to: \[ 3^{24x - 6} = 3^{6x + 16} \]
Since the bases are the same, we can set the exponents equal to each other: \[ 24x - 6 = 6x + 16 \]
Now, we will solve for \( x \). Start by moving \( 6x \) to the left side and \( -6 \) to the right side: \[ 24x - 6x = 16 + 6 \] \[ 18x = 22 \]
Next, divide both sides by 18: \[ x = \frac{22}{18} = \frac{11}{9} \]
Thus, the solution for \( x \) is: \[ \boxed{\frac{11}{9}} \]
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