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

The inequality \(2.25 ≤ 2 + 2^{n + 2} ≤ 202\) can be simplified to ensure that \(2 + 2^{n + 2}\) is less than or equal to 202.

If we take 27 = 128, this satisfies the inequality, but 28 does not.

So, the maximum value for n is 5.

Additionally, when subtracting 2 from both sides of the inequality, we get \(0.25 ≤ 2^{n + 2}\), which can be further simplified to \(2 - 2 ≤ 2^{n + 2}\). This implies that \(n ≥ -4.\)

Therefore, the integral values of n can range from -4 to 5.

For the expression \(3 + 3^{n + 1}\) to have integral values, n must be greater than or equal to -1 and go up to 5. Hence, the possible values of n are -1, 0, 1, 2, 3, 4, and 5.

So, there are 7 possible values for n.

Answer 2

Given that:
\(2.25 < 2 + 2^{n+2} < 202\)
\(2.25 - 2 < 2 + 2^{n+2} - 2 < 202 - 2\)
\(0.25 < 2^{n+2} < 200\)
Taking log both side,
\(log_2 0.25 < n + 2 < log_2 200\)
\(-2 < n + 2 < 7\)
\(-2-2 < n < 7-2\)
\(-4< n < 5\)
Now, the possible integers  are \(-4, -3, -2, -1, 0, 1, 2, 3, 4, 5\)
For \(3 + 3^{n+1}\), it is clear that n should be at least \(-1\).
So, \(n = -1, 0, 1, 2, 3, 4, 5\)

Hence, there are 7 values.