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 nth day exceeds one million, then the lowest possible value of $n$ is

We start with $2$ organisms on day $1$. Each day, the number of organisms $= 2 \times (\text{previous day's count}) + 3$.

We need to find the smallest day number $(n)$ when organisms exceed $1,000,000$.

---

Identify the recurrence relation:

$T(n) = 2 \times T(n-1) + 3$, with $T(1) = 2$

This is a first-order linear recurrence relation of the form $T(n) = a \times T(n-1) + b$.

---

For any recurrence $T(n) = a \times T(n-1) + b$, the general solution is:

$T(n) = a^n \times C + \text{particular solution}$

To find the particular solution, set $T(n) = A$ (constant):

$A = 2A + 3$

$A - 2A = 3$

$A = -3$

So: $T(n) = 2^n \times C - 3$

---

Using initial condition $T(1) = 2$:

$2 = 2^1 \times C - 3$

$2 = 2C - 3$

$5 = 2C$

$C = \dfrac{5}{2}$

Therefore: $T(n) = \dfrac{5}{2} \times 2^n - 3 = 5 \times 2^{n-1} - 3$

---

We need: $5 \times 2^{n-1} - 3 > 1,000,000$

$5 \times 2^{n-1} > 1,000,003$

$2^{n-1} > \dfrac{1,000,003}{5}$

$2^{n-1} > 200,000.6$

$n - 1 > \log_2(200,000.6) \approx 17.61$ (Use calculator)

$n > 18.61$

Since $n$ must be an integer, $n \geq 19$.

---

Verification:

$T(18) = 5 \times 2^{17} - 3 = 5 \times 131,072 - 3 = 655,357 < 1,000,000$

$T(19) = 5 \times 2^{18} - 3 = 5 \times 262,144 - 3 = 1,310,717 > 1,000,000$

Answer: $n = 19$