Properties of Logarithms worksheet with 12 problems for expanding and condensing logarithmic expressions.

Problem: Expand each logarithm.

We will use the properties of logarithms to expand each expression. The key properties are:
1. Product Rule: $\log_b (MN) = \log_b M + \log_b N$
2. Quotient Rule: $\log_b \left(\frac{M}{N}\right) = \log_b M - \log_b N$
3. Power Rule: $\log_b (M^p) = p \log_b M$

#### 1) $\log_4 (7 \cdot 9 \cdot 3)$
Using the Product Rule:
\[
\log_4 (7 \cdot 9 \cdot 3) = \log_4 7 + \log_4 9 + \log_4 3
\]
So, the expanded form is:
\[
\boxed{\log_4 7 + \log_4 9 + \log_4 3}
\]

#### 2) $\log_{16} (p \cdot s \cdot d)^{\frac{1}{4}}$
First, apply the Power Rule:
\[
\log_{16} (p \cdot s \cdot d)^{\frac{1}{4}} = \frac{1}{4} \log_{16} (p \cdot s \cdot d)
\]
Then, apply the Product Rule:
\[
\log_{16} (p \cdot s \cdot d) = \log_{16} p + \log_{16} s + \log_{16} d
\]
So, the expanded form is:
\[
\frac{1}{4} (\log_{16} p + \log_{16} s + \log_{16} d) = \frac{1}{4} \log_{16} p + \frac{1}{4} \log_{16} s + \frac{1}{4} \log_{16} d
\]
\[
\boxed{\frac{1}{4} \log_{16} p + \frac{1}{4} \log_{16} s + \frac{1}{4} \log_{16} d}
\]

#### 3) $\log_6 \left(\frac{m}{s}\right)$
Using the Quotient Rule:
\[
\log_6 \left(\frac{m}{s}\right) = \log_6 m - \log_6 s
\]
So, the expanded form is:
\[
\boxed{\log_6 m - \log_6 s}
\]

#### 4) $\log_5 \left(\frac{s^6}{m}\right)$
Using the Quotient Rule:
\[
\log_5 \left(\frac{s^6}{m}\right) = \log_5 (s^6) - \log_5 m
\]
Then, apply the Power Rule to $\log_5 (s^6)$:
\[
\log_5 (s^6) = 6 \log_5 s
\]
So, the expanded form is:
\[
6 \log_5 s - \log_5 m
\]
\[
\boxed{6 \log_5 s - \log_5 m}
\]

#### 5) $\log_{12} (d \cdot m)^2$
First, apply the Power Rule:
\[
\log_{12} (d \cdot m)^2 = 2 \log_{12} (d \cdot m)
\]
Then, apply the Product Rule:
\[
\log_{12} (d \cdot m) = \log_{12} d + \log_{12} m
\]
So, the expanded form is:
\[
2 (\log_{12} d + \log_{12} m) = 2 \log_{12} d + 2 \log_{12} m
\]
\[
\boxed{2 \log_{12} d + 2 \log_{12} m}
\]

#### 6) $\log_5 (s^5 \cdot b)$
Using the Product Rule:
\[
\log_5 (s^5 \cdot b) = \log_5 (s^5) + \log_5 b
\]
Then, apply the Power Rule to $\log_5 (s^5)$:
\[
\log_5 (s^5) = 5 \log_5 s
\]
So, the expanded form is:
\[
5 \log_5 s + \log_5 b
\]
\[
\boxed{5 \log_5 s + \log_5 b}
\]

Problem: Condense each expression to one logarithm.

We will use the properties of logarithms to condense each expression. The key properties are:
1. Product Rule: $\log_b M + \log_b N = \log_b (MN)$
2. Quotient Rule: $\log_b M - \log_b N = \log_b \left(\frac{M}{N}\right)$
3. Power Rule: $p \log_b M = \log_b (M^p)$

#### 7) $3 \log_{11} z - 3 \log_{11} w$
Factor out the common coefficient 3:
\[
3 \log_{11} z - 3 \log_{11} w = 3 (\log_{11} z - \log_{11} w)
\]
Using the Quotient Rule:
\[
\log_{11} z - \log_{11} w = \log_{11} \left(\frac{z}{w}\right)
\]
So, the condensed form is:
\[
3 \log_{11} \left(\frac{z}{w}\right) = \log_{11} \left(\left(\frac{z}{w}\right)^3\right)
\]
\[
\boxed{\log_{11} \left(\frac{z^3}{w^3}\right)}
\]

#### 8) $6 \log_8 9 + 3 \log_8 7$
Using the Power Rule for both terms:
\[
6 \log_8 9 = \log_8 (9^6) \quad \text{and} \quad 3 \log_8 7 = \log_8 (7^3)
\]
So the expression becomes:
\[
\log_8 (9^6) + \log_8 (7^3)
\]
Using the Product Rule:
\[
\log_8 (9^6) + \log_8 (7^3) = \log_8 (9^6 \cdot 7^3)
\]
\[
\boxed{\log_8 (9^6 \cdot 7^3)}
\]

#### 9) $\log 4 + \log 3$
Using the Product Rule:
\[
\log 4 + \log 3 = \log (4 \cdot 3) = \log 12
\]
\[
\boxed{\log 12}
\]

#### 10) $\log_6 7 - 2 \log_6 9$
Using the Power Rule for the second term:
\[
2 \log_6 9 = \log_6 (9^2) = \log_6 81
\]
So the expression becomes:
\[
\log_6 7 - \log_6 81
\]
Using the Quotient Rule:
\[
\log_6 7 - \log_6 81 = \log_6 \left(\frac{7}{81}\right)
\]
\[
\boxed{\log_6 \left(\frac{7}{81}\right)}
\]

#### 11) $6 \log_3 4 + 4 \log_3 6 + \log_3 3$
Using the Power Rule for the first two terms:
\[
6 \log_3 4 = \log_3 (4^6) \quad \text{and} \quad 4 \log_3 6 = \log_3 (6^4)
\]
So the expression becomes:
\[
\log_3 (4^6) + \log_3 (6^4) + \log_3 3
\]
Using the Product Rule:
\[
\log_3 (4^6) + \log_3 (6^4) + \log_3 3 = \log_3 (4^6 \cdot 6^4 \cdot 3)
\]
\[
\boxed{\log_3 (4^6 \cdot 6^4 \cdot 3)}
\]

#### 12) $2 \log_{15} 6 + 6 \log_{15} 9 + 4 \log_{15} 2$
Using the Power Rule for all terms:
\[
2 \log_{15} 6 = \log_{15} (6^2) = \log_{15} 36
\]
\[
6 \log_{15} 9 = \log_{15} (9^6)
\]
\[
4 \log_{15} 2 = \log_{15} (2^4) = \log_{15} 16
\]
So the expression becomes:
\[
\log_{15} 36 + \log_{15} (9^6) + \log_{15} 16
\]
Using the Product Rule:
\[
\log_{15} 36 + \log_{15} (9^6) + \log_{15} 16 = \log_{15} (36 \cdot 9^6 \cdot 16)
\]
\[
\boxed{\log_{15} (36 \cdot 9^6 \cdot 16)}
\]

Final Answers:

\[
\boxed{
\begin{array}{ll}
1) & \log_4 7 + \log_4 9 + \log_4 3 \\
2) & \frac{1}{4} \log_{16} p + \frac{1}{4} \log_{16} s + \frac{1}{4} \log_{16} d \\
3) & \log_6 m - \log_6 s \\
4) & 6 \log_5 s - \log_5 m \\
5) & 2 \log_{12} d + 2 \log_{12} m \\
6) & 5 \log_5 s + \log_5 b \\
7) & \log_{11} \left(\frac{z^3}{w^3}\right) \\
8) & \log_8 (9^6 \cdot 7^3) \\
9) & \log 12 \\
10) & \log_6 \left(\frac{7}{81}\right) \\
11) & \log_3 (4^6 \cdot 6^4 \cdot 3) \\
12) & \log_{15} (36 \cdot 9^6 \cdot 16) \\
\end{array}
}
\]