use the laws of exponents to solve for x 729^4x-1=9^3x+8

1 answer

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:

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