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The number of bacteria in a culture is increasing according to the law of exponential growth. The initial population is 240 bacteria, and the population after 7 hours is double the population after 1 hour. How many bacteria will be there after 4 hours? (Round your answers to the nearest whole number.)

Respuesta :

Answer:

381 bacteria

Step-by-step explanation:

[tex]\boxed{\begin{minipage}{9 cm}\underline{General form of an Exponential Function with base $e$}\\\\$y=Ae^{kt}$\\\\where:\\\phantom{ww}$\bullet$ $A$ is the initial value ($y$-intercept). \\ \phantom{ww}$\bullet$ $e$ is Euler's number. \\ \phantom{ww}$\bullet$ $k$ is some constant.\\ \phantom{ww}$\bullet$ $t$ is the independent value time.\\\end{minipage}}[/tex]

If the initial population is 240 bacteria, then A = 240:

[tex]\implies y=240e^{kt}[/tex]

Create equations for the population after 1 hour and 7 hours:

[tex]t=1\implies y=240e^{k}[/tex]

[tex]t=7\implies y=240e^{7k}[/tex]

Given the population after 7 hours is double the population after 1 hour:

[tex]\implies 240e^{7k}=2 \cdot 240e^k[/tex]

[tex]\implies e^{7k}=2e^k[/tex]

Solve the equation to find the value of k:

[tex]\implies e^{7k}=2e^k[/tex]

[tex]\implies \dfrac{e^{7k}}{e^k}=2[/tex]

[tex]\implies e^{6k}=2[/tex]

[tex]\implies \ln e^{6k}=\ln 2[/tex]

[tex]\implies 6k \ln e = \ln 2[/tex]

[tex]\implies 6k=\ln 2[/tex]

[tex]\implies k=\dfrac{1}{6}\ln 2[/tex]

Therefore, the equation that models the given parameters is:

[tex]y=240e^{\frac{1}{6}t\ln 2}[/tex]

To find how many bacteria there will be after 4 hours, substitute t = 4 into the equation:

[tex]\implies y=240e^{\frac{1}{6}(4)\ln 2}[/tex]

[tex]\implies y=240e^{\frac{2}{3}\ln 2}[/tex]

[tex]\implies y=240(1.58740105...)[/tex]

[tex]\implies y=380.9762525[/tex]

[tex]\implies y=381[/tex]

Therefore, there will be 381 bacteria after 4 hours

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