do the same for $$ \log_{b}\left(.04\right) $$

To express $\log_{b}(0.04)$ in terms of $p$, $q$, and $r$, let's start by expressing $0.04$ in a form that relates to the given logarithms.

Notice that:

\[
0.04 = \frac{4}{100} = \frac{4}{10^2}
\]

We can also express 4 as:

\[
4 = 2^2
\]

So:

\[
0.04 = \frac{2^2}{10^2} = \left(\frac{2}{10}\right)^2
\]

Now, let's take the logarithm:

\[
\log_{b}(0.04) = \log_{b}\left(\left(\frac{2}{10}\right)^2\right)
\]

Using the logarithmic property $\log_{b}(a^c) = c \cdot \log_{b}(a)$, we get:

\[
\log_{b}(0.04) = 2 \cdot \log_{b}\left(\frac{2}{10}\right)
\]

Next, apply the property of logarithms for division $\log_{b}\left(\frac{a}{c}\right) = \log_{b}(a) - \log_{b}(c)$:

\[
\log_{b}\left(\frac{2}{10}\right) = \log_{b}(2) - \log_{b}(10)
\]

We know that $\log_{b}(10) = p + r$, as previously derived. Therefore:

\[
\log_{b}(0.04) = 2 \cdot \left(\log_{b}(2) - \log_{b}(10)\right)
\]

Substitute the known values:

\[
\log_{b}(0.04) = 2 \cdot \left(p - (p + r)\right)
\]

Simplify the expression:

\[
\log_{b}(0.04) = 2 \cdot (-r) = -2r
\]

So, the expression for $\log_{b}(0.04)$ in terms of $p$ and $r$ is:

\[
\log_{b}(0.04) = -2r
\]

Would you like more details or have any further questions?

Here are some related questions:
1. How do you express $\log_{b}(0.2)$ in terms of $p$ and $r$?
2. What is the logarithm of a fraction in terms of its numerator and denominator?
3. Can you express $\log_{b}(0.25)$ in terms of $p$?
4. How would you express $\log_{b}(0.5)$ in terms of $p$?
5. What is the general method for expressing logarithms of decimal numbers in terms of known logarithms?

**Tip:** Remember that expressing numbers as fractions or products of simpler numbers can make logarithmic expressions easier to handle using basic logarithmic identities.

Learn how to express $\log_{b}(0.04)$ in terms of $p$ and $r$ using logarithmic properties and the change of base formula. This problem involves detailed manipulation of logarithmic identities and is suitable for advanced high school students and beyond.

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