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### Basic Properties of Logarithms

Fill in the missing values to make the equations true.

(a) [tex]\(\log_4 3 + \log_4 5 = \log_4 \boxed{15}\)[/tex]

(b) [tex]\(\log_3 4 - \log_3 \boxed{11} = \log_3 \frac{4}{11}\)[/tex]

(c) [tex]\(\log_6 32 = \boxed{5} \log_6 2\)[/tex]


Sagot :

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.