Question

I would like help solving question number two. Please be patient when answering as I will be taking notes from the answer screen. Thank you for your help.

Answer 1

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:

$\begin{gathered} A(t)=12e^{0.4t} \\ B(t)=24e^{kt} \end{gathered}$

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:

$A(4)=12\cdot e^{0.4\cdot4}=12\cdot e^{1.6}$

Using a calculator, we get the approximation:

$A(4)\approx59\text{ bacteria}$

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:

$A(t)=400$

Hence plugging the exponential function, we get

$12e^{0.4t}=400$

Divide both sides of the equation by a factor of 12

$e^{0.4t}=\dfrac{100}{3}$

Take the natural logarithm on both sides of the equation

$\ln(e^{0.4t})=\ln\left(\dfrac{100}{3}\right)$

Apply the following property:

$\log_a(a^b)=b\cdot\log_a(a)=b\cdot1=b$

Knowing that:

$\log_e(x)=\ln(x)$

Hence we get:

$0.4t=\ln\left(\dfrac{100}{3}\right)$

Multiply both sides of the equation by a factor of 2.5

$\begin{gathered} 2.5\cdot0.4t=2.5\cdot\ln\left(\dfrac{100}{3}\right) \\ \\ t=2.5\ln\left(\dfrac{100}{3}\right) \end{gathered}$

Using a calculator, we get the following approximation:

$t\approx8.76\text{ hours}$

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

$\begin{gathered} B(4)=60 \\ \\ 24e^{k\cdot4}=24e^{4k}=60 \end{gathered}$

Divide both sides of the equation by a factor of 24

$e^{4k}=\frac{5}{2}$

Taking the natural logarithm of both sides of the equation

$\ln(e^{4k})=\ln\left(\dfrac{5}{2}\right)$

Applying the property presented before, we get

$4k=\ln\left(\dfrac{5}{2}\right)$

Divide both sides by a factor of 4

$k=\dfrac{1}{4}\cdot\ln\left(\dfrac{5}{2}\right)$

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

$B(t)=24e^{\frac{1}{4}\ln\left(\frac{5}{2}\right)t}=24\cdot e^{\ln\left(\frac{5}{2}\right)^{\frac{t}{4}}}=24\cdot\left(\dfrac{5}{2}\right)^{\frac{t}{4}}$

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:

$\begin{gathered} A(n)>B(n) \\ \\ 12e^{0.4n}>24e^{^{\frac{1}{4}\ln\left(\frac{5}{2}\right)n}} \end{gathered}$

Since both exponential functions are powers of e and

$e\approx2.7182818\cdots>1$

We can solve the inequality without having to swap its order.

Divide both sides of the inequality by a factor of

$24e^{0.4n}$

Hence we get

$e^{\frac{1}{4}\ln\left(\frac{5}{2}\right)n-0.4n}<\dfrac{1}{2}$

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:

$\ln(e^{^{\left[\frac{1}{4}\ln\left(\frac{5}{2}\right)-0.4\right]n}})<\ln(\dfrac{1}{2})$

Applying the following property:

$\begin{gathered} \ln\left(\dfrac{a}{b}\right)=\ln(a)-\ln(b),\text{ b not equal to zero;} \\ \\ \ln(1)=0 \end{gathered}$

We get that

$\ln\left(\dfrac{1}{2}\right)=-\ln(2)$

Hence we get by applying the very first property that

$\left[\dfrac{1}{4}\ln\left(\dfrac{5}{2}\right)-0.4\right]n<-\ln(2)$

Divide both sides by a factor of

$\dfrac{1}{4}\ln\left(\dfrac{5}{2}\right)-0.4$

Notice it is a negative number, hence we swap the inequality sign as follows

$n\gt-\dfrac{\operatorname{\ln}(2)}{\dfrac{1}{4}\operatorname{\ln}\left(\dfrac{5}{2}\right)-0.4}$

Which evaluates to

$n>4.055$

Since n is an integer, then we say

$n=5$

Is the first hour for which the number of bacteria on colony A exceeds the number of bacteria on colony B.