# A bacteria culture has an initial population of 10,000. If it's initial population declines to 6,000 in 8 hours, what will it be at the end of 10 hours? Assume that population decreases according to the exponential model.

May 29, 2017

There will be $5281$ bacteria after $10$ hours.

#### Explanation:

Initial population of bacteria is p_0 =10000 ; p_8=6000 ; t= 8 hrs.

Exponential model is ${p}_{t} = {p}_{0} \cdot {e}^{k t} \therefore 6000 = 10000 \cdot {e}^{k \cdot 8}$ or

${e}^{8 \cdot k} = \frac{6000}{10000} = \frac{3}{5} = 0.6$

Taking natural log on both sides we get 8*k ln(e) = ln (0.6) ; or 8*k ~~ -0.5108 [ln(e)=1] or k ~~ -0.5108/8 = -0.06385

Population of bacteria after 10 hours is ${p}_{10} = {p}_{0} \cdot {e}^{k \cdot 10}$ or

${p}_{10} = 1000 \cdot {e}^{- 0.06385 \cdot 10} = 10000 \cdot {e}^{- 0.6385} = \frac{10000}{e} ^ \left(0.6385\right) \approx 5281$

There will be $5281$ bacteria after $10$ hours. [Ans]