Sure, let's solve each equation step by step:
1. For the first equation:
[tex]\[
\log 11 - \log 5 = \log \left( \frac{11}{5} \right)
\][/tex]
Therefore,
[tex]\[
\log 11 - \log 5 = \log \left( 2.2 \right)
\][/tex]
2. For the second equation:
[tex]\[
\log \square + \log 3 = \log 15
\][/tex]
Since [tex]\(\log a + \log b = \log (a \cdot b)\)[/tex], it implies:
[tex]\[
\log \left( \square \cdot 3 \right) = \log 15
\][/tex]
Therefore,
[tex]\[
\square \cdot 3 = 15
\][/tex]
Solving for the unknown:
[tex]\[
\square = \frac{15}{3} = 5
\][/tex]
3. For the third equation:
[tex]\[
\log 25 = 2 \log \square
\][/tex]
This implies:
[tex]\[
\log 25 = \log (\square^2)
\][/tex]
Therefore from the properties of logarithms, if [tex]\(\log 25 = \log (\square^2)\)[/tex], it follows that:
[tex]\[
\square^2 = 25
\][/tex]
Solving for the unknown:
[tex]\[
\square = \sqrt{25} = 5
\][/tex]
Putting it all together, the values that make the equations true are:
[tex]\[
\begin{array}{l}
\log 11 - \log 5 = \log \left(2.2\right) \\
\log 5 + \log 3 = \log 15 \\
\log 25 = 2 \log 5 \\
\end{array}
\][/tex]
