Question

revisiting exponents & their functions quick check
use the laws of exponents to solve for x in the equation $729^{4x - 1} = 9^{3x + 8}$. (1 point)
$\circ$ $x = \frac{11}{9}$
$\circ$ $x = \frac{19}{6}$
$\circ$ $x = 9$
$\circ$ $x = -\frac{3}{2}$

Explanation:

Step1: Express 729 as a power of 9

Since \( 729 = 9^3 \), we can rewrite the left - hand side of the equation \( 729^{4x - 1}\) as \( (9^3)^{4x-1}\).
Using the power - of - a - power rule \((a^m)^n=a^{mn}\), we have \((9^3)^{4x - 1}=9^{3\times(4x - 1)}=9^{12x-3}\).
So the original equation \( 729^{4x - 1}=9^{3x + 8}\) becomes \( 9^{12x-3}=9^{3x + 8}\).

Step2: Set the exponents equal

If \( a^m=a^n\) (where \( a>0,a
eq1\)), then \( m = n\). Since \( a = 9>0,a
eq1\), we can set the exponents equal to each other:
\(12x-3=3x + 8\)

Step3: Solve for x

Subtract \( 3x\) from both sides of the equation:
\(12x-3x-3=3x-3x + 8\)
\(9x-3 = 8\)
Add 3 to both sides:
\(9x-3 + 3=8 + 3\)
\(9x=11\)
Divide both sides by 9:
\(x=\frac{11}{9}\)

Answer:

\(x=\frac{11}{9}\) (corresponding to the option \(x = \frac{11}{9}\))