Question

Use the laws of exponents to solve for x in the equation 7294x−1=93x+8 .(1 point) Responses x=196 x equals Start Fraction 19 over 6 End Fraction x=119 x equals Start Fraction 11 over 6 End Fraction x=−32 x equals negative Start Fraction 3 over 2 End Fraction x=9

Answer 1

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

We know that:
- \( 729 = 3^6 \) (since \( 9^3 = (3^2)^3 = 3^6 \))
- \( 9 = 3^2 \)

We can rewrite the left side:
\[
729^{4x-1} = (3^6)^{4x-1} = 3^{6(4x-1)} = 3^{24x - 6}
\]

Now, we'll rewrite the right side:
\[
9^{3x+8} = (3^2)^{3x+8} = 3^{2(3x+8)} = 3^{6x + 16}
\]

Now we have the equation:
\[
3^{24x - 6} = 3^{6x + 16}
\]

Since the bases are the same, we can set the exponents equal to each other:
\[
24x - 6 = 6x + 16
\]

Now, we will solve for \(x\):
1. Subtract \(6x\) from both sides:
\[
24x - 6x - 6 = 16
\]
\[
18x - 6 = 16
\]

2. Add \(6\) to both sides:
\[
18x = 16 + 6
\]
\[
18x = 22
\]

3. Divide by \(18\):
\[
x = \frac{22}{18} = \frac{11}{9}
\]

None of the provided choices exactly match \( \frac{11}{9} \). However, let's confirm whether typographical errors exist regarding \(\frac{19}{6}\) and \(\frac{11}{6}\).

Since we derived \(x = \frac{11}{9}\), and it does not match the available options, please double-check the choices or consider if a revision or reevaluation is necessary.

Final value:
\[
x = \frac{11}{9}
\]