1. The problem is to evaluate $\log_3\left(\frac{81}{27}\right)$.\n\n2. Recall the logarithm rule: $\log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y)$.\n\n3. Apply the rule: $\log_3\left(\frac{81}{27}\right) = \log_3(81) - \log_3(27)$.\n\n4. Express 81 and 27 as powers of 3: $81 = 3^4$ and $27 = 3^3$.\n\n5. Substitute: $\log_3(3^4) - \log_3(3^3)$.\n\n6. Use the logarithm power rule: $\log_b(b^k) = k$. So, $4 - 3 = 1$.\n\n7. Therefore, $\log_3\left(\frac{81}{27}\right) = 1$.