Certainly! Let's work through these logarithmic equations one by one.
### Part (a)
Given: [tex]\(\log_2 11 + \log_2 9 = \log_2 \square\)[/tex]
We use the product rule for logarithms, which states:
[tex]\[
\log_b(m) + \log_b(n) = \log_b(m \cdot n)
\][/tex]
Here, [tex]\( \log_2 11 + \log_2 9 \)[/tex] can be written as:
[tex]\[
\log_2 (11 \cdot 9)
\][/tex]
Therefore, the value of the square is:
[tex]\[
11 \cdot 9 = 99
\][/tex]
So,
[tex]\[
\log_2 11 + \log_2 9 = \log_2 99
\][/tex]
### Part (b)
Given: [tex]\(\log_9 8 - \log_9 \square = \log_9 \frac{8}{11}\)[/tex]
We use the quotient rule for logarithms, which states:
[tex]\[
\log_b(m) - \log_b(n) = \log_b \left(\frac{m}{n}\right)
\][/tex]
Here, [tex]\(\log_9 8 - \log_9 \square\)[/tex] can be written as:
[tex]\[
\log_9 \left(\frac{8}{\square}\right)
\][/tex]
Matching this with the given right-hand side of the equation, [tex]\(\log_9 \frac{8}{11}\)[/tex], it follows that:
[tex]\[
\square = 11
\][/tex]
So,
[tex]\[
\log_9 8 - \log_9 11 = \log_9 \frac{8}{11}
\][/tex]
### Part (c)
Given: [tex]\(\log_3 4 = \square \log_3 2\)[/tex]
We use the change of base or properties of logarithms. Specifically, knowing that [tex]\(4 = 2^2\)[/tex], we can rewrite the expression using the power rule:
[tex]\[
\log_3 4 = \log_3 (2^2)
\][/tex]
Using the power rule for logarithms, which states:
[tex]\[
\log_b (m^k) = k \log_b m
\][/tex]
We get:
[tex]\[
\log_3 (2^2) = 2 \log_3 2
\][/tex]
So, the value of the square is:
[tex]\[
\square = 2
\][/tex]
Thus,
[tex]\[
\log_3 4 = 2 \log_3 2
\][/tex]
### Summary
Here are the missing values filled in:
(a) [tex]\(\log_2 11 + \log_2 9 = \log_2 99\)[/tex]
(b) [tex]\(\log_9 8 - \log_9 11 = \log_9 \frac{8}{11}\)[/tex]
(c) [tex]\(\log_3 4 = 2 \log_3 2\)[/tex]
