Let's fill in the missing values in the table corresponding to [tex]\( x \)[/tex] and [tex]\( 3^x \)[/tex].
First, we are given the following [tex]\( x \)[/tex] values:
- [tex]\( x = -2 \)[/tex]
- [tex]\( x = \log_3 \frac{1}{3} \)[/tex]
- [tex]\( x = 0 \)[/tex]
- [tex]\( x = 1 \)[/tex]
- [tex]\( x = \log_3 3 \)[/tex]
Now, we will compute [tex]\( 3^x \)[/tex] for each of these [tex]\( x \)[/tex] values.
1. For [tex]\( x = -2 \)[/tex]:
[tex]\[
3^{-2} = \frac{1}{3^2} = \frac{1}{9} \approx 0.111
\][/tex]
2. For [tex]\( x = \log_3 \frac{1}{3} \)[/tex]:
Since the logarithm is the inverse of the exponential function, [tex]\( 3^{\log_3 \frac{1}{3}} = \frac{1}{3} \)[/tex].
3. For [tex]\( x = 0 \)[/tex]:
[tex]\[
3^0 = 1
\][/tex]
4. For [tex]\( x = 1 \)[/tex]:
[tex]\[
3^1 = 3
\][/tex]
5. For [tex]\( x = \log_3 3 \)[/tex]:
[tex]\[
\log_3 3 = 1 \quad \text{since} \quad 3^1 = 3
\][/tex]
Now, we can fill the table with the computed values:
[tex]\[
\begin{array}{|c|c|}
\hline
x & 3^x \\
\hline
-2 & 0.111 \\
-1 & 0.333 \\
0 & 1 \\
1 & 3 \\
1 & 3 \\
\hline
\end{array}
\][/tex]
The values match our computed results. Here, [tex]\(\log_3 \frac{1}{3} = -1\)[/tex] and [tex]\(\log_3 3 = 1\)[/tex].
Completed table is:
[tex]\[
\begin{array}{|c|c|}
\hline
x & 3^x \\
\hline
-2 & 0.111 \\
-1 & 0.333 \\
0 & 1 \\
1 & 3 \\
1 & 3 \\
\hline
\end{array}
\][/tex]
Therefore, the table is properly filled with the correct values of [tex]\( x \)[/tex] and [tex]\( 3^x \)[/tex].
