Question 15.

$2.25 ≤ 2 + 2^{n+2} ≤ 202$

$2.25 − 2 ≤ 2 + 2^{n+2} − 2 ≤ 202 − 2$

$0.25 ≤ 2^{n+2} ≤ 200$

$\log_2 0.25 ≤ n + 2 ≤ \log_2 200$

$−2 ≤ n + 2 ≤ 7.xx$

$−4 ≤ n ≤ 7.xx − 2$

$−4 ≤ n ≤ 5.xx$

Possible integers = −4, −3, −2, −1, 0, 1, 2, 3, 4, 5

If we see the second expression provided, i.e.

$3 + 3^{n+1}$, it can be implied that n should be at least −1 for this expression to be an integer.

So, n = −1, 0, 1, 2, 3, 4, 5

Hence, there are a total of 7 values.