Q.In a culture, the bacteria count is 1,00,000. The number is increased by $10\%$ in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?

The key idea is Exponential Growth Rate: when the rate of growth is proportional to the number present, the population follows $N = N_0 e^{kt}$.

Step 1 — Find $k$.

Given $N_0 = 1,00,000$ and after 2 hours $N = 1,10,000$ (a $10\%$ increase).

Step 2 — Set up for target.

We want $N = 2,00,000$:

Step 3 — Solve for $t$.

Substitute $k$: …

The key idea here is exponential growth. When a quantity grows at a rate proportional to its own size — more bacteria means more new bacteria per hour — the growth follows an exponential curve. This is the classic “compound interest” model of nature, seen in populations, radioactive decay (with a minus sign), and even investments.

Why does proportionality lead to exponentials? Because the differential equation $\frac{dN}{dt} = k N$ says: the bigger $N$ gets, the faster it grows. The solution is $N(t) = N_0 e^{kt}$, where $N_0$ is the starting count and $k$ is the growth constant. Our job is to find $k$ from the given data, then solve for the time to double.

1. Set up the model. Let $N(t)$ be the number of bacteria after $t$ hours. The statement “rate of growth is proportional to the number present” translates to:

where $k$ is a positive constant. Solving this gives:

Here $N_0 = 1,00,000$ (the initial count).

2. Use the 2-hour data to find $k$. After 2 hours, the count increases by $10\%$, so it becomes $1,00,000 \times 1.1 = 1,10,000$. Plug into the model:

Divide both sides by $1,00,000$:

Take the natural logarithm:

3. Find the time to reach 2,00,000. We want $t$ such that $N(t) = 2,00,000$. Using the model:

Divide:

Take $\ln$:

Substitute $k = \frac{\ln(1.1)}{2}$:

Solve for $t$:

$t = \frac{2 \ln 2}{\ln(1.1)}$ …

Method: Proportional growth — solving for the time

Use this when population/bacteria grow at a rate proportional to their number, one data point sets the growth constant, and you must find the time to reach a target.

Steps

Step 1: Model and integrate

$\dfrac{dN}{dt}=kN$ integrates to $N=N_0\,e^{kt}$.

Step 2: Find $k$ from the given growth

Use the stated change (here $+10\%$ in 2 hours: $1.1=e^{2k}$) to get $k=\dfrac{1}{2}\log_e1.1$.

Step 3: Solve for the target time …

Common Mistakes

Mistake 1: Finding $k$ as a decimal too early and losing accuracy

Why it's wrong: $k=\frac12\log_e1.1$; converting prematurely muddies the final time. Correct approach: keep $k$ symbolic and combine with the target log.

Mistake 2: Confusing the two ratios (1.1 and 2)

Why it's wrong: $1.1$ sets the growth constant over 2 hours; $2$ is the doubling target — swapping them gives nonsense. Correct approach: use $1.1=e^{2k}$ for $k$, then $2=e^{kt}$ for $t$. …