Use the laws of exponents to solve for x in the equation 7294x−1=93x+8

Question

.(1 point)
Responses

x=196
x equals Start Fraction 19 over 6 End Fraction

x=−32
x equals negative Start Fraction 3 over 2 End Fraction

x=119
x equals Start Fraction 11 over 6 End Fraction

x=9
x equals 9

Answer 1

To solve the equation \( 729^{4x-1} = 9^{3x+8} \) using the laws of exponents, we can first express both sides in terms of powers of the same base.

We know:

• \( 729 = 9^3 \)
• Therefore, \( 729^{4x-1} = (9^3)^{4x-1} = 9^{3(4x-1)} = 9^{12x - 3} \)

Now we rewrite the equation: \[ 9^{12x - 3} = 9^{3x + 8} \]

Since the bases are the same, we can set the exponents equal to each other: \[ 12x - 3 = 3x + 8 \]

Next, we solve for \( x \):

1. Subtract \( 3x \) from both sides: \[ 12x - 3x - 3 = 8 \] \[ 9x - 3 = 8 \]

2. Add 3 to both sides: \[ 9x = 11 \]

3. Divide by 9: \[ x = \frac{11}{9} \]

None of the provided responses (196, \(\frac{19}{6}\), -32, \(-\frac{3}{2}\), 119, \(\frac{11}{6}\), or 9) match the solution \( x = \frac{11}{9} \). However, based on your options and calculations, please check if the equation has been transcribed correctly. If so, the solution is \( x = \frac{11}{9} \). Please let me know if there is anything else I can assist you with!