3. 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 \(\left(\frac{1}{3}\right)^{x+1}=\left(\frac{1}{9}\right)^{x+3}\), we can rewrite \(\frac{1}{9}\) as \(\left(\frac{1}{3}\right)^2\). Thus, the equation can be rewritten as: \[ \left(\frac{1}{3}\right)^{x+1} = \left(\left(\frac{1}{3}\right)^2\right)^{x+3} \] This further simplifies to: \[ \left(\frac{1}{3}\right)^{x+1} = \left(\frac{1}{3}\right)^{2(x+3)} \] Since both sides have the same base, we can set the exponents equal to each other: \[ x + 1 = 2(x + 3) \] Now, solving for \(x\): \[ x + 1 = 2x + 6 \quad \Rightarrow \quad 1 - 6 = 2x - x \] \[ -5 = x \quad \Rightarrow \quad x = -5 \] Thus, the solution to the equation is \(x = -5\). Final answer: (C) \(x = -5\)