Answer :
Let's solve each part of the question step-by-step.
### Part (a)
We have the expression:
[tex]\[ \log_8 7 - \log_8 3 \][/tex]
Using the properties of logarithms, specifically, the difference rule:
[tex]\[ \log_b a - \log_b c = \log_b \left( \frac{a}{c} \right) \][/tex]
So,
[tex]\[ \log_8 7 - \log_8 3 = \log_8 \left( \frac{7}{3} \right) \][/tex]
Thus, the missing value is:
[tex]\[ \boxed{\frac{7}{3}} \][/tex]
### Part (b)
We are given:
[tex]\[ \log_4 9 + \log_4 \square = \log_4 45 \][/tex]
Using the properties of logarithms, specifically, the addition rule:
[tex]\[ \log_b a + \log_b c = \log_b (a \cdot c) \][/tex]
So,
[tex]\[ \log_4 9 + \log_4 x = \log_4 (9 \cdot x) \][/tex]
For the equation to be true,
[tex]\[ 9 \cdot x = 45 \][/tex]
Solving for [tex]\(x\)[/tex],
[tex]\[ x = \frac{45}{9} = 5 \][/tex]
Thus, the missing value is:
[tex]\[ \boxed{5} \][/tex]
### Part (c)
We are given:
[tex]\[ -3 \log_3 2 = \log_3 \square \][/tex]
Using the properties of logarithms, specifically, the power rule:
[tex]\[ a \log_b c = \log_b (c^a) \][/tex]
So,
[tex]\[ -3 \log_3 2 = \log_3 (2^{-3}) \][/tex]
Simplifying the expression:
[tex]\[ 2^{-3} = \frac{1}{2^3} = \frac{1}{8} \][/tex]
Thus, the missing value is:
[tex]\[ \boxed{\frac{1}{8}} \][/tex]
Therefore, the filled-in equations are:
(a) [tex]\(\log_8 7 - \log_8 3 = \log_8 \boxed{\frac{7}{3}}\)[/tex]
(b) [tex]\(\log_4 9 + \log_4 \boxed{5} = \log_4 45\)[/tex]
(c) [tex]\(-3 \log_3 2 = \log_3 \boxed{\frac{1}{8}}\)[/tex]
### Part (a)
We have the expression:
[tex]\[ \log_8 7 - \log_8 3 \][/tex]
Using the properties of logarithms, specifically, the difference rule:
[tex]\[ \log_b a - \log_b c = \log_b \left( \frac{a}{c} \right) \][/tex]
So,
[tex]\[ \log_8 7 - \log_8 3 = \log_8 \left( \frac{7}{3} \right) \][/tex]
Thus, the missing value is:
[tex]\[ \boxed{\frac{7}{3}} \][/tex]
### Part (b)
We are given:
[tex]\[ \log_4 9 + \log_4 \square = \log_4 45 \][/tex]
Using the properties of logarithms, specifically, the addition rule:
[tex]\[ \log_b a + \log_b c = \log_b (a \cdot c) \][/tex]
So,
[tex]\[ \log_4 9 + \log_4 x = \log_4 (9 \cdot x) \][/tex]
For the equation to be true,
[tex]\[ 9 \cdot x = 45 \][/tex]
Solving for [tex]\(x\)[/tex],
[tex]\[ x = \frac{45}{9} = 5 \][/tex]
Thus, the missing value is:
[tex]\[ \boxed{5} \][/tex]
### Part (c)
We are given:
[tex]\[ -3 \log_3 2 = \log_3 \square \][/tex]
Using the properties of logarithms, specifically, the power rule:
[tex]\[ a \log_b c = \log_b (c^a) \][/tex]
So,
[tex]\[ -3 \log_3 2 = \log_3 (2^{-3}) \][/tex]
Simplifying the expression:
[tex]\[ 2^{-3} = \frac{1}{2^3} = \frac{1}{8} \][/tex]
Thus, the missing value is:
[tex]\[ \boxed{\frac{1}{8}} \][/tex]
Therefore, the filled-in equations are:
(a) [tex]\(\log_8 7 - \log_8 3 = \log_8 \boxed{\frac{7}{3}}\)[/tex]
(b) [tex]\(\log_4 9 + \log_4 \boxed{5} = \log_4 45\)[/tex]
(c) [tex]\(-3 \log_3 2 = \log_3 \boxed{\frac{1}{8}}\)[/tex]