I see the image contains three math problems. Let's go through each one step by step.

Problem 17:

Given:

• $\log 2 = x$
• $\log 3 = y$

Find the value of $\log 0.45$.

Solution: We start by expressing 0.45 as a product of prime factors: $0.45 = \frac{45}{100} = \frac{9 \times 5}{100} = \frac{3^2 \times 5}{2^2 \times 5^2}$ Thus, $\log 0.45 = \log \left( \frac{9 \times 5}{100} \right) = \log (9) + \log(5) - \log(100)$.

This simplifies to:

log(0.45)->