Respuesta :
STak
Hello there. To solve this question, we'll have to remember some properties about exponential growth and solving exponential equations, inequalities.
Given the colonies A and B, for which the number of bacteria are, respectivelly, modelled by the equations:
[tex]\begin{gathered} A(t)=12e^{0.4t} \\ B(t)=24e^{kt} \end{gathered}[/tex]For t > 0 in hours. We have to determine:
a) The number of bacteria in colony A after 4 hours.
For this, we take t = 4 and calculate the following number:
[tex]A(4)=12\cdot e^{0.4\cdot4}=12\cdot e^{1.6}[/tex]Using a calculator, we get the approximation:
[tex]A(4)\approx59\text{ bacteria}[/tex]b) How long does it take for the number of bacteria in colony A to reach 400?
For this, we have to determine t such that:
[tex]A(t)=400[/tex]Hence plugging the exponential function, we get
[tex]12e^{0.4t}=400[/tex]Divide both sides of the equation by a factor of 12
[tex]e^{0.4t}=\dfrac{100}{3}[/tex]Take the natural logarithm on both sides of the equation
[tex]\ln(e^{0.4t})=\ln\left(\dfrac{100}{3}\right)[/tex]Apply the following property:
[tex]\log_a(a^b)=b\cdot\log_a(a)=b\cdot1=b[/tex]Knowing that:
[tex]\log_e(x)=\ln(x)[/tex]Hence we get:
[tex]0.4t=\ln\left(\dfrac{100}{3}\right)[/tex]Multiply both sides of the equation by a factor of 2.5
[tex]\begin{gathered} 2.5\cdot0.4t=2.5\cdot\ln\left(\dfrac{100}{3}\right) \\ \\ t=2.5\ln\left(\dfrac{100}{3}\right) \end{gathered}[/tex]Using a calculator, we get the following approximation:
[tex]t\approx8.76\text{ hours}[/tex]c) Find the value of k
For this, we have to use the fact that there are 60 bacteria in colony B after four hours;
Hence we get
[tex]\begin{gathered} B(4)=60 \\ \\ 24e^{k\cdot4}=24e^{4k}=60 \end{gathered}[/tex]Divide both sides of the equation by a factor of 24
[tex]e^{4k}=\frac{5}{2}[/tex]Taking the natural logarithm of both sides of the equation
[tex]\ln(e^{4k})=\ln\left(\dfrac{5}{2}\right)[/tex]Applying the property presented before, we get
[tex]4k=\ln\left(\dfrac{5}{2}\right)[/tex]Divide both sides by a factor of 4
[tex]k=\dfrac{1}{4}\cdot\ln\left(\dfrac{5}{2}\right)[/tex]In this case it is better to keep the answer like this instead of using an approximaton.
With this, we have that B(t) will be
[tex]B(t)=24e^{\frac{1}{4}\ln\left(\frac{5}{2}\right)t}=24\cdot e^{\ln\left(\frac{5}{2}\right)^{\frac{t}{4}}}=24\cdot\left(\dfrac{5}{2}\right)^{\frac{t}{4}}[/tex]But we can keep it in the first form in order to solve part d).
d) The number of bacteria in colony A first exceeds the number of bacteria in colony B after n hours, where n is in integers. Find the value of n.
For this, we have to solve the following inequality:
[tex]\begin{gathered} A(n)>B(n) \\ \\ 12e^{0.4n}>24e^{^{\frac{1}{4}\ln\left(\frac{5}{2}\right)n}} \end{gathered}[/tex]Since both exponential functions are powers of e and
[tex]e\approx2.7182818\cdots>1[/tex]We can solve the inequality without having to swap its order.
Divide both sides of the inequality by a factor of
[tex]24e^{0.4n}[/tex]Hence we get
[tex]e^{\frac{1}{4}\ln\left(\frac{5}{2}\right)n-0.4n}<\dfrac{1}{2}[/tex]Since the logarithm is an one-to-one function, we take the natural logarithm on both sides of the inequality, preserving the order, therefore we get:
[tex]\ln(e^{^{\left[\frac{1}{4}\ln\left(\frac{5}{2}\right)-0.4\right]n}})<\ln(\dfrac{1}{2})[/tex]Applying the following property:
[tex]\begin{gathered} \ln\left(\dfrac{a}{b}\right)=\ln(a)-\ln(b),\text{ b not equal to zero;} \\ \\ \ln(1)=0 \end{gathered}[/tex]We get that
[tex]\ln\left(\dfrac{1}{2}\right)=-\ln(2)[/tex]Hence we get by applying the very first property that
[tex]\left[\dfrac{1}{4}\ln\left(\dfrac{5}{2}\right)-0.4\right]n<-\ln(2)[/tex]Divide both sides by a factor of
[tex]\dfrac{1}{4}\ln\left(\dfrac{5}{2}\right)-0.4[/tex]Notice it is a negative number, hence we swap the inequality sign as follows
[tex]n\gt-\dfrac{\operatorname{\ln}(2)}{\dfrac{1}{4}\operatorname{\ln}\left(\dfrac{5}{2}\right)-0.4}[/tex]Which evaluates to
[tex]n>4.055[/tex]Since n is an integer, then we say
[tex]n=5[/tex]Is the first hour for which the number of bacteria on colony A exceeds the number of bacteria on colony B.
