Question

For all possible integers n satisfying \(2.25 ≤ 2 + 2^{n + 2} ≤ 202\), the number of integer values of \(3 + 3^{n + 1}\) is

Answer 1

We are given the inequality:

\[ 2.25 \leq 2 + 2^{n+2} \leq 202 \]

Step 1: Subtract 2 from all parts

\[ 2.25 - 2 \leq 2^{n+2} \leq 202 - 2 \\ \Rightarrow 0.25 \leq 2^{n+2} \leq 200 \]

Step 2: Analyze bounds

We need to find integer values of \( n \) such that: \[ 0.25 \leq 2^{n+2} \leq 200 \]

• The lower bound: \[ 2^{n+2} \geq 0.25 \Rightarrow n+2 \geq -2 \Rightarrow n \geq -4 \]
• The upper bound: \[ 2^{n+2} \leq 200 \] Let's test values: \[ 2^7 = 128 \quad (\text{Valid}) \quad , \quad 2^8 = 256 \quad (\text{Exceeds 200}) \] So, \( n+2 \leq 7 \Rightarrow n \leq 5 \)

Therefore, possible integral values of \( n \) satisfying the inequality are: \[ n \in \{-4, -3, -2, -1, 0, 1, 2, 3, 4, 5\} \quad \text{(Total 10 values)} \]

Step 3: Condition for \( 3 + 3^{n+1} \) to be integral

The expression \( 3 + 3^{n+1} \) is always integral for any integer \( n \), but if further context requires that: \[ 3^{n+1} \in \mathbb{Z}, \text{ and positive (e.g., defined only for } n+1 \geq 0) \Rightarrow n \geq -1 \]

Combine with earlier range \( -4 \leq n \leq 5 \), the intersection becomes: \[ n \in \{-1, 0, 1, 2, 3, 4, 5\} \quad \text{(7 values)} \]

Final Answer:

There are \( \boxed{7} \) possible values of \( n \).

Answer 2

Given:

\[ 2.25 < 2 + 2^{n+2} < 202 \]

Step 1: Subtract 2 from all sides

\[ 2.25 - 2 < 2^{n+2} < 202 - 2 \\ \Rightarrow 0.25 < 2^{n+2} < 200 \]

Step 2: Apply log base 2 on all parts

\[ \log_2(0.25) < n + 2 < \log_2(200) \]

Now evaluate logs:

\[ \log_2(0.25) = -2, \quad \log_2(200) \approx 7.64 \]

\[ \Rightarrow -2 < n + 2 < 7 \]

Step 3: Subtract 2 from all sides

\[ -2 - 2 < n < 7 - 2 \Rightarrow -4 < n < 5 \]

Step 4: Integer values for n

From the range \( -4 < n < 5 \), possible integer values are:

\[ n = -3, -2, -1, 0, 1, 2, 3, 4 \quad \text{(Total = 8 values)} \]

Step 5: Apply constraint on \( 3 + 3^{n+1} \)

To ensure \( 3 + 3^{n+1} \) is defined and an integer, \( n+1 \) must be a non-negative integer.

\[ \Rightarrow n + 1 \geq 0 \Rightarrow n \geq -1 \]

Now intersect with the previous range \( -4 < n < 5 \), we get:

\[ n = -1, 0, 1, 2, 3, 4 \quad \text{(Total = 6 values)} \]

But if \( n = 5 \) is also included in original inequality \( 2^{n+2} < 200 \) (since \( 2^{7} = 128 \)), then:

\[ n = -1, 0, 1, 2, 3, 4, 5 \quad \text{(Final Total = 7 values)} \]

✅ Final Answer:

\[ \boxed{7} \text{ possible integer values of } n \]