Известно, что lg 2 = а и lg 13 = b. Найдите log₅ 3,38.

Дано:

• \[ lg 2 = a \]
• \[ lg 13 = b \]

Найти:

• \[ log_5 3.38 \]

Решение:

1. \[ 3.38 = \frac{338}{100} = \frac{169}{50} \]
2. \[ log_5 3.38 = log_5 \frac{169}{50} = log_5 169 - log_5 50 \]
3. \[ log_5 169 = log_5 13^2 = 2 log_5 13 \]
4. \[ log_5 50 = log_5 (5^2 \times 2) = log_5 5^2 + log_5 2 = 2 + log_5 2 \]
5. \[ log_5 13 = \frac{lg 13}{lg 5} = \frac{b}{lg 5} \]
6. \[ log_5 2 = \frac{lg 2}{lg 5} = \frac{a}{lg 5} \]
7. \[ lg 5 = lg \frac{10}{2} = lg 10 - lg 2 = 1 - a \]
8. \[ log_5 13 = \frac{b}{1 - a} \]
9. \[ log_5 2 = \frac{a}{1 - a} \]
10. \[ log_5 3.38 = 2 \left( \frac{b}{1 - a} \right) - \left( 2 + \frac{a}{1 - a} \right) = \frac{2b}{1 - a} - 2 - \frac{a}{1 - a} = \frac{2b - a}{1 - a} - 2 \]
11. \[ \frac{2b - a - 2(1 - a)}{1 - a} = \frac{2b - a - 2 + 2a}{1 - a} = \frac{2b + a - 2}{1 - a} \]

Ответ: \( \frac{a + 2b - 2}{1 - a} \)