PTT 3 Day 12: Toplc 6: Assessment vAl (LMS graded) Solve the equation \( \left(\frac{1}{3}\right)^{x+1}=\left(\frac{1}{9}\right)^{x+3} \) A. \( x=5 \) B. \( x=-\frac{7}{3} \) C. \( x=-5 \) D. The equation has no solution.

To solve the equation \( \left(\frac{1}{3}\right)^{x+1}=\left(\frac{1}{9}\right)^{x+3} \), first recall that \( \frac{1}{9} \) can be rewritten as \( \left(\frac{1}{3}\right)^{2} \). Thus, we can rewrite the equation as: \[ \left(\frac{1}{3}\right)^{x+1} = \left(\left(\frac{1}{3}\right)^{2}\right)^{x+3} \] This simplifies to: \[ \left(\frac{1}{3}\right)^{x+1} = \left(\frac{1}{3}\right)^{2(x+3)} \] Now that the bases are the same, we can set the exponents equal to each other: \[ x + 1 = 2(x + 3) \] Distributing the 2 gives us: \[ x + 1 = 2x + 6 \] Rearranging gives us: \[ 1 - 6 = 2x - x \] \[ -5 = x \] Therefore, the solution to the equation is \( x = -5 \). So, the correct answer is **C. \( x = -5 \)**.