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: 7. Solution: Given: 2.25 ≤ 2+2^(n+2) ≤ 202 We isolate the exponential term: 0.25 ≤ 2^(n+2) ≤ 200 Key insight: 0.25 = 1/4 = 1/2^2 = 2^(-2) So our inequality becomes: 2^(-2) ≤ 2^(n+2) ≤ 200 For the left inequality: 2^(-2) ≤ 2^(n+2) Since the exponential function 2^x is increasing, we can compare exponents directly: -2 ≤ n+2 Therefore: n ≥ -4 For the right inequality: 2^(n+2) ≤ 200 We need: n+2 ≤ _2(200) Since 2^7 = 128 We calculate 3+3^(n+1) for each value of n:, n, n+1, 3^(n+1), 3+3^(n+1), Integer?, -4, -3, 1/27, 3 + 1/27 = 82/27, No, -3, -2, 1/9, 3 + 1/9 = 28/9, No, -2, -1, 1/3, 3 + 1/3 = 10/3, No, -1, 0, 1, 3 + 1 = 4, Yes, 0, 1, 3, 3 + 3 = 6, Yes, 1, 2, 9, 3 + 9 = 12, Yes, 2, 3, 27, 3 + 27 = 30, Yes, 3, 4, 81, 3 + 81 = 84, Yes, 4, 5, 243, 3 + 243 = 246, Yes, 5, 6, 729, 3 + 729 = 732, Yes, The integer values of 3+3^(n+1) are: \4, 6, 12, 30, 84, 246, 732\

Modern Math

Logarithms

Hard

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

Entered answer:

✅ Correct Answer: 7

Given: $2.25 \leq 2+2^{n+2} \leq 202$

We isolate the exponential term:

$0.25 \leq 2^{n+2} \leq 200$

Key insight: $0.25 = \tfrac{1}{4} = \tfrac{1}{2^2} = 2^{-2}$

So our inequality becomes:

$2^{-2} \leq 2^{n+2} \leq 200$

For the left inequality: $2^{-2} \leq 2^{n+2}$

Since the exponential function $2^x$ is increasing, we can compare exponents directly:

$-2 \leq n+2$

Therefore: $n \geq -4$

For the right inequality: $2^{n+2} \leq 200$

We need: $n+2 \leq \log_2(200)$

Since $2^7 = 128 < 200 < 256 = 2^8$ , we have $7 < \log_2(200) < 8$

More precisely, $\log_2(200) \approx 7.64$

Since we need integers, $n+2 \leq 7$ , so $n \leq 5$

Therefore: $-4 \leq n \leq 5$

This means $n \in \{-4, -3, -2, -1, 0, 1, 2, 3, 4, 5\}$

We calculate $3+3^{n+1}$ for each value of $n$ :

$n$	$n+1$	$3^{n+1}$	$3+3^{n+1}$	Integer?
-4	-3	$\tfrac{1}{27}$	$3 + \tfrac{1}{27} = \tfrac{82}{27}$	No
-3	-2	$\tfrac{1}{9}$	$3 + \tfrac{1}{9} = \tfrac{28}{9}$	No
-2	-1	$\tfrac{1}{3}$	$3 + \tfrac{1}{3} = \tfrac{10}{3}$	No
-1	0	$1$	$3 + 1 = 4$	Yes
0	1	$3$	$3 + 3 = 6$	Yes
1	2	$9$	$3 + 9 = 12$	Yes
2	3	$27$	$3 + 27 = 30$	Yes
3	4	$81$	$3 + 81 = 84$	Yes
4	5	$243$	$3 + 243 = 246$	Yes
5	6	$729$	$3 + 729 = 732$	Yes

The integer values of $3+3^{n+1}$ are: $\{4, 6, 12, 30, 84, 246, 732\}$

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