Sure, let's work through each equation step-by-step to fill in the missing values:
### Part (a)
The equation is given as:
[tex]\[ \log_3 2 + \log_3 7 = \log_3 \square \][/tex]
We can use the property of logarithms that states [tex]\(\log_a b + \log_a c = \log_a (bc)\)[/tex].
So,
[tex]\[ \log_3 2 + \log_3 7 = \log_3 (2 \cdot 7) \][/tex]
[tex]\[ 2 \cdot 7 = 14 \][/tex]
Therefore, the missing value is:
[tex]\[ \log_3 2 + \log_3 7 = \log_3 14 \][/tex]
### Part (b)
The equation is given as:
[tex]\[ \log_2 5 - \log_2 \square = \log_2 \frac{5}{3} \][/tex]
We can use the property of logarithms that states [tex]\(\log_a b - \log_a c = \log_a \left(\frac{b}{c}\right) \)[/tex].
So,
[tex]\[ \log_2 5 - \log_2 x = \log_2 \frac{5}{3} \][/tex]
For these to be equal, [tex]\( x \)[/tex] must be equal to [tex]\( 3 \)[/tex].
Therefore, the missing value is:
[tex]\[ \log_2 5 - \log_2 3 = \log_2 \frac{5}{3} \][/tex]
### Part (c)
The equation is given as:
[tex]\[ \log_6 \frac{1}{32} = -5 \log_6 \square \][/tex]
We can use the property of logarithms that states [tex]\(\log_a b^c = c \log_a b \)[/tex].
First, let's express [tex]\(\frac{1}{32}\)[/tex] as a power of 2:
[tex]\[ \frac{1}{32} = 32^{-1} = (2^5)^{-1} = 2^{-5} \][/tex]
So,
[tex]\[ \log_6 (2^{-5}) = -5 \log_6 2 \][/tex]
Therefore, the missing value is:
[tex]\[ \log_6 \frac{1}{32} = -5 \log_6 2 \][/tex]
In summary, the missing values are:
[tex]\[
\begin{aligned}
\text{(a)} & \quad \log_3 2 + \log_3 7 = \log_3 14 \\
\text{(b)} & \quad \log_2 5 - \log_2 3 = \log_2 \frac{5}{3} \\
\text{(c)} & \quad \log_6 \frac{1}{32} = -5 \log_6 2 \\
\end{aligned}
\][/tex]
