# A colony of bacteria grows according to the law of uninhibited growth. If 100 grams of bacteria are present initially and 250 grams are present after two hours, how many will be present after 4 hours?

Oct 5, 2016

There will be a population of $625 g$

#### Explanation:

The law of uninhibited growth follows the rule

$P = {P}_{0} {e}^{k t}$, where

• $P$ is the population at time $t$
• ${P}_{0}$ is the initial population
• $k$ is the rate of exponential growing
• $t$ measures time, in your case hours.

Since ${P}_{0} = 100 g$, the equation becomes

$P = 200 {e}^{k t}$

Now we can find $k$, knowing that after two hours ($t = 2$), the population has risen to $250 g$:

$250 = 100 \cdot {e}^{k \cdot 2}$

Divide both sides by a hundred:

$2.5 = {e}^{2 k}$

Consider the natual logarithm of both sides:

$\ln \left(2.5\right) = 2 k \setminus \implies k = \ln \frac{2.5}{2}$

To find the answer, simply plug $t = 4$ in the equation, now that we know $k$:

$P = 100 {e}^{k \cdot 4} = 100 {e}^{\ln \frac{2.5}{2} \cdot 4} = 100 {e}^{2 \ln \left(2.5\right)} = 100 {e}^{\ln \left({2.5}^{2}\right)}$

And since ${e}^{\ln} \left(z\right) = z$, we finally have

$P = 100 {e}^{\ln \left({2.5}^{2}\right)} = 100 \cdot {2.5}^{2} = 100 \cdot 6.25 = 625$