Question

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

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

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

x
=
9
x equals 9

x
=
19
6

Answer 1

To solve the equation

\[
729^{4x - 1} = 9^{3x + 8},
\]

we can express both sides in terms of a common base. Notice that both 729 and 9 are powers of 3:

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

Now we can rewrite the equation using these bases:

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

Applying the law of exponents \((a^m)^n = a^{mn}\), we get:

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

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

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

Expanding both sides:

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

Now, we'll rearrange the equation to isolate \(x\). First, subtract \(6x\) from both sides:

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

Next, add 6 to both sides:

\[
18x = 22
\]

Finally, divide by 18:

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

Thus, the solution is:

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