Sure, let's go through each equation step-by-step.
1. [tex]\(\log 3 - \log 5 = \log \square\)[/tex]:
- Using the property of logarithms, [tex]\(\log a - \log b = \log \left( \frac{a}{b} \right)\)[/tex]:
- So, [tex]\(\log 3 - \log 5 = \log \left( \frac{3}{5} \right)\)[/tex].
- Therefore, [tex]\(\log 3 - \log 5 = \log \left( \frac{3}{5} \right) = \log 0.6\)[/tex].
So, [tex]\(\log 3 - \log 5 = \log 0.6\)[/tex].
2. [tex]\(\log 11 + \log \square = \log 77\)[/tex]:
- Using the property of logarithms, [tex]\(\log a + \log b = \log (a \times b)\)[/tex]:
- So, [tex]\(\log 11 + \log x = \log (11 \times x) = \log 77\)[/tex].
- Solving for [tex]\( x\)[/tex], we get [tex]\( 11x = 77\)[/tex].
- Therefore, [tex]\( x = \frac{77}{11} = 7\)[/tex].
So, [tex]\(\log 11 + \log 7 = \log 77\)[/tex].
3. [tex]\(\log 32 = \square \log 2\)[/tex]:
- Using the property of logarithms, [tex]\(\log a^b = b \log a \)[/tex]:
- We know that [tex]\(32 = 2^5\)[/tex].
- So, [tex]\(\log 32 = \log (2^5) = 5 \log 2\)[/tex].
Therefore, [tex]\(\log 32 = 5 \log 2\)[/tex].
In summary, the missing values are:
[tex]\[
\begin{array}{r}
\log 3-\log 5=\log 0.6 \\
\log 11+\log 7=\log 77 \\
\log 32=5 \log 2
\end{array}
\][/tex]
