To complete the Powers of 2 table, we populate the values of [tex]\(2^2\)[/tex], [tex]\(2^3\)[/tex], [tex]\(2^4\)[/tex], [tex]\(2^5\)[/tex], [tex]\(2^6\)[/tex], [tex]\(2^7\)[/tex], [tex]\(2^8\)[/tex], and [tex]\(2^9\)[/tex].
Here are the completed values:
[tex]\[
\begin{tabular}{|c|c|c|c|c|c|}
\hline
\multirow{2}{*}{\begin{tabular}{c}
Powers \\
of 2
\end{tabular}} & $2^2=4$ & $2^3=8$ & $2^a=16$ & $2^4=16$ & $2^5=32$ \\
\cline { 2 - 6 }
& $2^6=64$ & $2^7=128$ & $2^8=256$ & $2^9=512$ \\
\hline
\end{tabular}
\][/tex]
Now, let's solve Example A: [tex]\(3(2)^x = 96\)[/tex] for [tex]\(x\)[/tex].
1. Original Equation:
[tex]\[
3(2)^x = 96
\][/tex]
2. Divide both sides by 3 to isolate the power:
[tex]\[
\frac{3(2)^x}{3} = \frac{96}{3} \implies 2^x = 32
\][/tex]
3. Simplify:
[tex]\[
2^x = 32
\][/tex]
4. Rewrite 32 as a power of 2:
[tex]\[
32 = 2^5 \implies 2^x = 2^5
\][/tex]
5. Solve:
Since the bases are the same, we equate the exponents:
[tex]\[
x = 5
\][/tex]
Therefore, the solution for [tex]\(3(2)^x = 96\)[/tex] is [tex]\(x = 5\)[/tex].
