Evaluate b. ( 4- 25)/(1+2 1)5

Using the logarithm property $\log a - \log b = \log \frac{a}{b}$:

$$\log 4 - \log 25 = \log \frac{4}{25}$$

We can also write $\frac{4}{25}$ as $\left(\frac{2}{5}\right)^2$.
So, the numerator is $\log \left(\frac{2}{5}\right)^2$.

First, use the property $\log \frac{1}{x} = -\log x$:

$$\log \frac{1}{5} = -\log 5$$

Substitute this into the denominator:

$$1 + 2(-\log 5) = 1 - 2 \log 5$$

Now, express '1' as a logarithm with base 10: $1 = \log 10$.

$$\log 10 - 2 \log 5$$

Using the property $b \log a = \log a^b$:

$$\log 10 - \log (5^2) = \log 10 - \log 25$$

Using the property $\log a - \log b = \log \frac{a}{b}$:

$$\log 10 - \log 25 = \log \frac{10}{25}$$

Simplify the fraction:

$$\log \frac{10}{25} = \log \frac{2}{5}$$

So, the denominator simplifies to $\log \frac{2}{5}$.

The expression becomes:

$$\frac{\log \left(\frac{2}{5}\right)^2}{\log \frac{2}{5}}$$

Using the property $\log a^b = b \log a$ for the numerator:

$$\frac{2 \log \frac{2}{5}}{\log \frac{2}{5}}$$

Since $\log \frac{2}{5}$ is a common term in both the numerator and the denominator, and $\frac{2}{5} \neq 1$, $\log \frac{2}{5} \neq 0$. We can cancel it out.

$$\frac{2 \cancel{\log \frac{2}{5}}}{\cancel{\log \frac{2}{5}}} = 2$$

The final answer is $\text{2}$.