Certainly! Let's solve the given problem step by step.
### Problem
We are tasked with solving the expressions
[tex]\[2^{2+1}\][/tex]
and
[tex]\[e^{1-e}\][/tex]
### Step-by-Step Solution
1. Evaluating [tex]\(2^{2+1}\)[/tex]:
- First, let's simplify the exponent [tex]\(2 + 1\)[/tex]:
[tex]\[2 + 1 = 3\][/tex]
- Next, we raise 2 to the power of 3:
[tex]\[2^3 = 2 \times 2 \times 2\][/tex]
- Performing the multiplication:
[tex]\[2 \times 2 = 4\][/tex]
[tex]\[4 \times 2 = 8\][/tex]
- Thus,
[tex]\[2^{2+1} = 8\][/tex]
2. Evaluating [tex]\(e^{1-e}\)[/tex]:
- Here, [tex]\(e\)[/tex] is the base of the natural logarithm, approximately equal to [tex]\(2.71828\)[/tex].
- The expression we need to evaluate is [tex]\(e\)[/tex] raised to the power of [tex]\((1 - e)\)[/tex].
- Numerically, this evaluates to:
[tex]\[e^{1-e} \approx 0.17937407873401723\][/tex]
### Final Answer
- The value of [tex]\(2^{2+1}\)[/tex] is [tex]\(8\)[/tex].
- The value of [tex]\(e^{1-e}\)[/tex] is approximately [tex]\(0.17937407873401723\)[/tex].
Therefore, the results are:
[tex]\[2^{2+1} = 8\][/tex]
[tex]\[e^{1-e} \approx 0.17937407873401723\][/tex]
