Question

The number of bacteria in a culture is given by the function n(t)=990e^0.45t where t is measured in hours (a) What is the exponential rate of growth of this bacterium population?Your answer is % (b) What is the initial population of the culture (at t=0)?Your answer is (c) How many bacteria will the culture contain at time t=7?Your answer is

Answer 1

.

Answer:

The function is given below as

$n(t)=990e^{0.45t}$

The exponential function is given below as

$\begin{gathered} A(t)=A_0e^{rt} \\ t=time \\ A_0=initial\text{ number } \\ A(t)=number\text{ after time t} \\ r=rate \end{gathered}$

By comparing coefficients, we will have

$\begin{gathered} A_0=990 \\ r=0.45 \end{gathered}$

Hence,

the rate will be

$r=0.45\times100=45\%$

Therefore,

The rate is

$\Rightarrow45\%$

Step 2:

To figure out the initial population, we will substitute t=0 in the function above

$\begin{gathered} n(t)=990e^{0.45t} \\ n(0)=990e^{0.45\times0} \\ n(0)=990e^0 \\ n(0)=990 \end{gathered}$

Hence,

The initial population of the culture is

$\Rightarrow990$

Step 3:

To figure out the number of bacteria at t=7, we will have

$\begin{gathered} n(t)=990e^{0.45t} \\ n(7)=990e^{0.45\times7} \\ n(7)=990e^{3.15} \\ n(7)=990\times23.3361 \\ n(7)=23102.7 \\ n(7)=23103 \end{gathered}$

Hence,

The number of bacteria at time t=7 will be

$\Rightarrow23103$