Question

Use the laws of exponents to solve for x in the equation 7294x−1=93x+8
729
4
𝑥
−
1
=
9
3
𝑥
+
8
.(1 point)
Responses

x=119
𝑥
=
11
9
x equals Start Fraction 11 over 6 End Fraction

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

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

x=9

Answer 1

To solve the equation \( 729^{4x-1} = 9^{3x+8} \), we can express both sides of the equation using a common base.

1. Recognize that:
- \( 729 = 9^3 \), so we can rewrite \( 729^{4x-1} \) as \( (9^3)^{4x-1} = 9^{3(4x-1)} = 9^{12x-3} \).
- Keeping \( 9^{3x+8} \) as is, we get the equation:
\[
9^{12x-3} = 9^{3x+8}
\]

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

3. Now, solve for \( x \):
- Subtract \( 3x \) from both sides:
\[
12x - 3x - 3 = 8
\]
\[
9x - 3 = 8
\]

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

5. Divide both sides by 9:
\[
x = \frac{11}{9}
\]

Thus, the solution is \( x = \frac{11}{9} \), which corresponds to the option "x equals Start Fraction 11 over 6 End Fraction", which appears to be a formatting error as the correct amount is \( \frac{11}{9} \).

Therefore, the answer is:
\[
x = \frac{11}{9}
\]