To solve the equation \( 729^{4x-1} = 9^{3x+8} \), we can start by expressing both sides with a common base.
First, we recognize that \( 729 \) and \( 9 \) can be rewritten as powers of \( 3 \): \[ 729 = 3^6 \quad \text{(since \( 3^6 = 729 \))}, \] \[ 9 = 3^2 \quad \text{(since \( 3^2 = 9 \))}. \]
Now we can rewrite the equation with base \( 3 \): \[ (3^6)^{4x-1} = (3^2)^{3x+8}. \]
Applying the laws of exponents: \[ 3^{6(4x-1)} = 3^{2(3x+8)}. \]
Since the bases are the same, we can set the exponents equal to each other: \[ 6(4x - 1) = 2(3x + 8). \]
Now, distribute both sides: \[ 24x - 6 = 6x + 16. \]
Next, we isolate \( x \) by moving all \( x \) terms to one side and constants to the other side: \[ 24x - 6x = 16 + 6, \] \[ 18x = 22. \]
Now, divide both sides by 18: \[ x = \frac{22}{18} = \frac{11}{9} = \frac{11}{9} = \frac{11}{9}. \]
None of the provided options seem to match \( \frac{11}{9} \). Let's check the response options you've given:
- \( x = -32 \)
- \( x = -\frac{3}{2} \)
- \( x = 119 \)
- \( x = \frac{11}{6} \)
- \( x = 196 \)
- \( x = \frac{19}{6} \)
- \( x = 9 \)
It looks like there may have been a miscommunication or error in the options compared to the solution we derived. It appears the answer is not listed in the responses you've provided.
If you're looking for \( x \) to be among those options, please double-check the problem statement or the options you've written down. The solution derived is \( x = \frac{11}{9} \).
1 answer