Alright, let's carefully go through each part step by step to solve for the missing values using basic properties of logarithms.
### Part (a)
[tex]\[
\log_4 3 + \log_4 5 = \log_4 \rrbracket
\][/tex]
We use the property of logarithms that states:
[tex]\[
\log_b a + \log_b c = \log_b (a \cdot c)
\][/tex]
Applying this property to the given expression:
[tex]\[
\log_4 3 + \log_4 5 = \log_4 (3 \cdot 5) = \log_4 15
\][/tex]
Hence, the missing value is:
[tex]\[
\boxed{15}
\][/tex]
### Part (b)
[tex]\[
\log_3 4 - \log_3 \square = \log_3 \frac{4}{11}
\][/tex]
We use the property of logarithms that states:
[tex]\[
\log_b a - \log_b c = \log_b \left(\frac{a}{c}\right)
\][/tex]
Applying this property to the given expression:
[tex]\[
\log_3 4 - \log_3 x = \log_3 \frac{4}{11}
\][/tex]
Therefore, we need to find [tex]\( x \)[/tex] such that:
[tex]\[
\log_3 4 - \log_3 x = \log_3 \frac{4}{11}
\][/tex]
Setting the arguments equal:
[tex]\[
\frac{4}{x} = \frac{4}{11} \implies x = 11
\][/tex]
Thus, the missing value is:
[tex]\[
\boxed{11}
\][/tex]
### Part (c)
[tex]\[
\log_6 32 = \square \log_6 2
\][/tex]
We use the property of logarithms that states:
[tex]\[
\log_b a^c = c \log_b a
\][/tex]
Here, we need to express 32 as a power of 2:
[tex]\[
32 = 2^5
\][/tex]
Therefore:
[tex]\[
\log_6 32 = \log_6 (2^5) = 5 \log_6 2
\][/tex]
Thus, the missing value is:
[tex]\[
\boxed{5}
\][/tex]
### Summary:
1. For part (a): [tex]\(\boxed{15}\)[/tex]
2. For part (b): [tex]\(\boxed{11}\)[/tex]
3. For part (c): [tex]\(\boxed{5}\)[/tex]
These are the values that complete the given logarithmic equations.
