Question
use the laws of exponents to solve for x 729^4x-1=9^3x+8
Answers
Answered by
GPT-4o mini
To solve the equation \( 729^{4x - 1} = 9^{3x + 8} \), we can start by expressing both sides with a common base. Both \( 729 \) and \( 9 \) can be expressed as powers of \( 3 \):
1. \( 729 = 3^6 \) (since \( 3^6 = 729 \))
2. \( 9 = 3^2 \) (since \( 3^2 = 9 \))
Now we can rewrite the equation using these expressions:
\[
(3^6)^{4x - 1} = (3^2)^{3x + 8}
\]
Next, we apply the power of a power property, which states \( (a^m)^n = a^{m \cdot n} \):
\[
3^{6(4x - 1)} = 3^{2(3x + 8)}
\]
This simplifies to:
\[
3^{24x - 6} = 3^{6x + 16}
\]
Now, since the bases are the same, we can set the exponents equal to each other:
\[
24x - 6 = 6x + 16
\]
Next, we solve for \( x \). First, we will move \( 6x \) to the left side and \( 6 \) to the right side:
\[
24x - 6x = 16 + 6
\]
This simplifies to:
\[
18x = 22
\]
Now, divide both sides by \( 18 \):
\[
x = \frac{22}{18}
\]
We can simplify this fraction:
\[
x = \frac{11}{9}
\]
Thus, the solution is:
\[
\boxed{\frac{11}{9}}
\]
1. \( 729 = 3^6 \) (since \( 3^6 = 729 \))
2. \( 9 = 3^2 \) (since \( 3^2 = 9 \))
Now we can rewrite the equation using these expressions:
\[
(3^6)^{4x - 1} = (3^2)^{3x + 8}
\]
Next, we apply the power of a power property, which states \( (a^m)^n = a^{m \cdot n} \):
\[
3^{6(4x - 1)} = 3^{2(3x + 8)}
\]
This simplifies to:
\[
3^{24x - 6} = 3^{6x + 16}
\]
Now, since the bases are the same, we can set the exponents equal to each other:
\[
24x - 6 = 6x + 16
\]
Next, we solve for \( x \). First, we will move \( 6x \) to the left side and \( 6 \) to the right side:
\[
24x - 6x = 16 + 6
\]
This simplifies to:
\[
18x = 22
\]
Now, divide both sides by \( 18 \):
\[
x = \frac{22}{18}
\]
We can simplify this fraction:
\[
x = \frac{11}{9}
\]
Thus, the solution is:
\[
\boxed{\frac{11}{9}}
\]