Question

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

Answer 1

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

First, we notice that \( 729 \) and \( 9 \) can both be expressed as powers of \( 3 \):
- \( 729 = 3^6 \) (since \( 3^6 = 729 \))
- \( 9 = 3^2 \) (since \( 3^2 = 9 \))

Now we can rewrite the original equation using these expressions:
\[
(3^6)^{4x-1} = (3^2)^{3x+8}
\]

Using the laws of exponents, we simplify both sides:
\[
3^{6(4x-1)} = 3^{2(3x+8)}
\]

This simplifies to:
\[
3^{24x - 6} = 3^{6x + 16}
\]

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

Now, we will solve for \( x \). Start by moving \( 6x \) to the left side and \( -6 \) to the right side:
\[
24x - 6x = 16 + 6
\]
\[
18x = 22
\]

Next, divide both sides by 18:
\[
x = \frac{22}{18} = \frac{11}{9}
\]

Thus, the solution for \( x \) is:
\[
\boxed{\frac{11}{9}}
\]