1. **State the problem:** We need to evaluate $\log_9 0.48$ given $\log_2 2 = 0.3010$, $\log_3 3 = 0.4771$, and $\log_5 5 = 0.6990$. Also, we have a right triangle with sides 3 cm (adjacent), $x$ cm (opposite), and hypotenuse 5 cm, and we want to find $x$. 2. **Evaluate $\log_9 0.48$:** Recall the change of base formula: $$\log_a b = \frac{\log_c b}{\log_c a}$$ We can use base 3 since $9 = 3^2$: $$\log_9 0.48 = \frac{\log_3 0.48}{\log_3 9}$$ Since $9 = 3^2$, $$\log_3 9 = 2$$ Now, express $0.48$ in terms of known logs or approximate using base 10 logs: Given $\log_{10} 2 = 0.3010$, $\log_{10} 3 = 0.4771$, and $\log_{10} 5 = 0.6990$ (from the problem's given values, assuming these are base 10 logs), Express $0.48$ as $\frac{48}{100} = \frac{16 \times 3}{100} = \frac{2^4 \times 3}{10^2}$. So, $$\log_{10} 0.48 = \log_{10} (2^4 \times 3) - \log_{10} 10^2 = 4 \log_{10} 2 + \log_{10} 3 - 2$$ Calculate: $$4 \times 0.3010 + 0.4771 - 2 = 1.204 + 0.4771 - 2 = 1.6811 - 2 = -0.3189$$ Now, $$\log_3 0.48 = \frac{\log_{10} 0.48}{\log_{10} 3} = \frac{-0.3189}{0.4771} \approx -0.6685$$ Therefore, $$\log_9 0.48 = \frac{-0.6685}{2} = -0.3343$$ 3. **Find $x$ in the right triangle:** Given a right triangle with hypotenuse 5 cm, adjacent side 3 cm, and opposite side $x$ cm. Use the Pythagorean theorem: $$x^2 + 3^2 = 5^2$$ $$x^2 + 9 = 25$$ $$x^2 = 25 - 9 = 16$$ $$x = \sqrt{16} = 4$$ **Final answers:** - $\log_9 0.48 \approx -0.3343$ - Opposite side $x = 4$ cm