Sure, let's work through each part carefully:
### Part (a):
Equation: [tex]\(\log_7 2 - \log_7 9 = \log_7 \square\)[/tex]
We use the logarithm subtraction rule: [tex]\(\log_b(a) - \log_b(c) = \log_b\left(\frac{a}{c}\right)\)[/tex].
So:
[tex]\[
\log_7 2 - \log_7 9 = \log_7\left(\frac{2}{9}\right)
\][/tex]
Thus, the missing value is:
[tex]\[
\square = \frac{2}{9}
\][/tex]
### Part (b):
Equation: [tex]\(\log_4 3 + \log_4 \square = \log_4 15\)[/tex]
We use the logarithm addition rule: [tex]\(\log_b(a) + \log_b(c) = \log_b(ac)\)[/tex].
So:
[tex]\[
\log_4 3 + \log_4 \square = \log_4 (3 \cdot \square) = \log_4 15
\][/tex]
This means:
[tex]\[
3 \cdot \square = 15 \implies \square = \frac{15}{3}
\][/tex]
Thus, the missing value is:
[tex]\[
\square = 5
\][/tex]
### Part (c):
Equation: [tex]\(\log_6 25 = 2 \log_6 \square\)[/tex]
We use the logarithm power rule: [tex]\(\log_b(a^c) = c \log_b(a)\)[/tex].
Given the equation, we can rewrite it as:
[tex]\[
\log_6 25 = 2 \log_6 \square
\][/tex]
This implies:
[tex]\[
\log_6 25 = \log_6 (\square^2)
\][/tex]
Thus, comparing the arguments of the logarithms:
[tex]\[
25 = \square^2 \implies \square = \sqrt{25}
\][/tex]
Thus, the missing value is:
[tex]\[
\square = 5
\][/tex]
### Summary
The solutions for the missing values in each part are as follows:
- (a) [tex]\(\log_7 2 - \log_7 9 = \log_7 \left( \frac{2}{9} \right)\)[/tex]
- (b) [tex]\(\log_4 3 + \log_4 5 = \log_4 15\)[/tex]
- (c) [tex]\(\log_6 25 = 2 \log_6 5\)[/tex]
