Question

Use the laws of exponents to solve for x in the equation 729
4x−1
=9
3x+8
.

A. x=
6
19
​

B. x=
9
11
​

C. x=−
2
3
​

D. x=9

Answer 1

To solve the equation \( 729^{4x-1} = 9^{3x+8} \) using the laws of exponents, we start by expressing both numbers on the left and right in terms of powers of the same base.

First, we note the prime factorization of \( 729 \) and \( 9 \):

\[
729 = 3^6 \quad \text{(since \( 3^6 = 729 \))}
\]
\[
9 = 3^2 \quad \text{(since \( 3^2 = 9 \))}
\]

Now we can rewrite the equation in terms of base \( 3 \):

\[
(3^6)^{4x-1} = (3^2)^{3x+8}
\]

Next, we apply the exponent multiplication rule \( (a^m)^n = a^{m \cdot n} \):

\[
3^{6(4x-1)} = 3^{2(3x+8)}
\]

Now we simplify the exponents:

\[
3^{24x - 6} = 3^{6x + 16}
\]

Since the bases are the same, we can set the exponents equal to each other:

\[
24x - 6 = 6x + 16
\]

Now, we solve for \( x \):

1. Subtract \( 6x \) from both sides:

\[
24x - 6x - 6 = 16
\]
\[
18x - 6 = 16
\]

2. Add \( 6 \) to both sides:

\[
18x = 22
\]

3. Divide by \( 18 \):

\[
x = \frac{22}{18} = \frac{11}{9}
\]

Thus, the solution is:

\[
\boxed{\frac{11}{9}}
\]

This corresponds to option B in your provided choices.