A lab experiment measures the number of organisms at 8 am every day. Starting with 2 organisms on the first day, the number of organisms on any day is equal to 3 more than twice the number on the previous day. If the number of organisms on the $$ n $$th day exceeds one million, then the lowest possible value of $$ n $$ is:

Question

Accepted Answer

The population growth follows a recurrence relation:  
$$ P_n = 2P_{n-1} + 3, $$  
where $$ P_1 = 2 $$.

#### Step 1: Express general formula for population
The recurrence relation can be expanded step by step:  
- Day 1: $$ P_1 = 2 $$.  
- Day 2: $$ P_2 = 2P_1 + 3 = 2 \cdot 2 + 3 = 7 $$.  
- Day 3: $$ P_3 = 2P_2 + 3 = 2 \cdot 7 + 3 = 17 $$.  
- Continue expanding to find a general formula.

The pattern reveals that the population on the $$ n $$th day is given by:  
$$ P_n = 2^{n+1} + (2^n - 1) - 3. $$

#### Step 2: Solve for $$ n $$ when population exceeds one million
The condition is:  
$$ P_n > 10^6. $$  
Substitute the general formula:  
$$ 2^{n+1} + (2^n - 1) - 3 > 10^6. $$

Simplify and approximate:  
$$ P_n \approx 2^{n+1} \quad 	ext{(dominant term)}. $$  
Solve:  
$$ 2^{n+1} > 10^6. $$  
Take logarithms (base 2):  
$$ n+1 > \log_2(10^6). $$

Since $$ \log_2(10^6) \approx 19.93 $$, we find that:  
$$ n+1 = \lceil \log_2(10^6) ceil = 20. $$  
Thus, the smallest value of $$ n $$ is:  
$$ n = 19. $$

#### Final Answer:
The lowest possible value of $$ n $$ is **19**.