Let's fill in the missing values to make the given logarithmic equations true. We'll use properties of logarithms to solve these equations.
### (a)
Equation:
[tex]$\log _4 5+\log _4 11=\log _4 \; \square$[/tex]
To combine the logarithms on the left side, we use the property of logarithms:
[tex]$\log_b(a) + \log_b(c) = \log_b(ac)$[/tex]
Applying this property:
[tex]$\log _4 5+\log _4 11 = \log _4 (5 \times 11) = \log _4 55$[/tex]
So, the missing value is:
[tex]$\square = 55$[/tex]
### (b)
Equation:
[tex]$\log _3 \; \square - \log _3 5 = \log _3 \frac{7}{5}$[/tex]
To combine the logarithms on the left side, we use the property of logarithms:
[tex]$\log_b(a) - \log_b(c) = \log_b\left(\frac{a}{c}\right)$[/tex]
Considering the right side of the equation:
[tex]$\log _3 \frac{7}{5} = \log _3 \left(\frac{\square}{5}\right)$[/tex]
To make the equations equal, the numerator of the fraction on the right side must match the argument of the logarithm on the left:
[tex]$\frac{\square}{5} = \frac{7}{5}$[/tex]
Thus:
[tex]$\square = 7$[/tex]
### (c)
Equation:
[tex]$-2 \log _8 5 = \log _8 \; \square$[/tex]
To simplify the left side, use the property of logarithms:
[tex]$-a \log_b(c) = \log_b(c^{-a})$[/tex]
Applying this property:
[tex]$-2 \log _8 5 = \log _8 (5^{-2}) = \log _8 \left(\frac{1}{5^2}\right) = \log _8 \left(\frac{1}{25}\right) $[/tex]
So, the missing value is:
[tex]$\square = \frac{1}{25}$[/tex]
Therefore, the filled-out equations are:
(a) [tex]\(\log _4 5+ \log _4 11=\log _4 55\)[/tex]
(b) [tex]\(\log _3 7 - \log _3 5=\log _3 \frac{7}{5}\)[/tex]
(c) [tex]\(-2 \log _8 5=\log _8 \frac{1}{25}\)[/tex]
