Sub: Biology Topic: Microbiology
Question:
Application of exponential and logarithmic functions
Q(t) = Qe^kt
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The count in a bacterial culture was 400 after 2 hours and 25,600 after 6 hours.
We are assuming exponential growth.
1).What is the rate of growth of the population of bacteria?
2).What was the initial population at time t = 0 hours?
3).Write the function that models the population n(t) after t hours.
4).When will the number of bacteria exceed 100,000?
Solution:
General Information:
Exponential Functions:
There are many things that grow exponentially, for example population, compound
interest and charge in a capacitator.
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Sub: Biology Topic: Microbiology
Exponential Growth
The special thing about exponential growth is that the rate of growth increases as time
increases. The exponential growth is represented in the graph above. The curve gets steeper
and steeper as time goes on.
Exponential functions have the form:
f(x) = bx
where b is the base and x is the exponent (or power).
If b is positive, the function continuously increases in value. A special property of
exponential functions is that the slope of the function also continuously increases as x
increases.
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Sub: Biology Topic: Microbiology
It is common to write exponential functions using the carat (^), which means "raised to the
power". Computer programing uses the ^ sign, as do some calculators.
Other calculators have a button labeled xy which is equivalent to the ^ symbol.
Example for exponential function:
Consider the function f(x) = 2x.
In this case, we have an exponential function with base 2. Some typical values for this function
would be:
x -2 -1 0 1 2 3
f(x) 1/4 1/2 1 2 4 8
Here is the graph of y = 2x.
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Sub: Biology Topic: Microbiology
Notice:
That as x increases, y also increases.
That as x increases; the slope of the graph also increases.
That the curve passes through (0, 1). All exponential curves of the form f(x) = bx pass
through (0, 1), if b > 0.
The curve does not pass through the x-axis. It just gets closer and closer to the x-axis as
we take smaller and smaller x-values.
Logarithmic functions:
Logarithms were developed in the 17th century by the Scottish mathematician, John
Napier. He developed a clever method of reducing long multiplications into much simpler
additions (and reducing divisions into subtractions). The use of logarithms made trigonometry
and many other fields of mathematics much simpler to calculate.
The logarithmic function is defined as:
f(x) = logbx
The base of the logarithm is b.
The 2 most common bases that we use are base 10 and base e.
The logarithmic function has many real-life applications, in acoustics, electronics,
earthquake analysis and population prediction.
A logarithm is simply an exponent that is written in a special way.
For example, we know that the following exponential equation is true:
32 = 9
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Sub: Biology Topic: Microbiology
In this case, the base is 3 and the exponent is 2. We can write this equation in logarithm form
(with identical meaning) as follows:
log39 = 2
We say this as "the logarithm of 9 to the base 3 is 2". What we have effectively done is to move
the exponent down on to the main line. This was done historically to make multiplications and
divisions easier, but logarithms are still very handy in mathematics.
1) What is the rate of growth of the population of bacteria?
Solution:
We are provided with a formula for the exponential growth in our query which is
Q(t) = Qe^kt
Where,
Q(t) = Number of bacteria after time‘t’
Q = Initial number of bacterial population at time t = 0
K = non-zero constant.
If k is positive then the equation will grow without bound and is called the
exponential growth equation. Likewise, if k is negative the equation will die down to
zero and is called the exponential decay equation.
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Sub: Biology Topic: Microbiology
Provided Data’s:
We are provided with two data’s in our query i.e.,
When Q(t) = 400, t = 2 and
When Q(t) = 25, 600, t = 6
Let us now substitute each pair of Q(t) and t in the given formula,
Q(t) = Qe^kt
In the above equation the rate of growth of the bacteria could be found only by finding the
value of k by substituting the given values.
Substituting the data’s to find K:
Step 1:
(i) 400 = Qek(2) and
Step 2:
(ii) 25, 600 = Qek(6)
Step 3:
Let us now divide the equation (ii) with equation (i),
25, 600 Qek(6)
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Sub: Biology Topic: Microbiology
400 = Qek(2)
We will get,
64 = e6k / e2k
Step 4:
(i) 25600/400 gives us 64
(ii) Q and Q gets cancelled
Subtracting the exponent of K will give the following equation,
64 = e4k
Step 5:
In order to find out k we have take ln on both sides,
ln64 = ln e4k
ln 64 = 4k ln e / 1
k = ¼ ln 64
k = 1.03972
Final Answer:
The rate of growth of the population of bacteria was found to be
Q(t) = Qe1.03972t.
2) What was the initial population at time t = 0 hours?
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Sub: Biology Topic: Microbiology
Solution:
The initial population at time t = 0 can be also calculated from the same formula,
We are provided with a formula for the exponential growth in our query which is
Q(t) = Qe^kt
Where,
Q(t) = Number of bacteria after time ‘t’
Q = Initial number of bacterial population at time t = 0
K = non-zero constant.
As per the formula we have to find out Q, let us now substitute the value of k,
When Q(t) = 400, t = 2 and k = 1.03972
Lets us now substitute the values in the formula to find out the value of Q,
Q(t) = Qe^kt
Step 1:
400 = Qe1.03972(2)
Step 2:
Q = 400 / e1.03972(2)
Step 3:
Q = 400 / e2.07944
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Sub: Biology Topic: Microbiology
Step 4:
Q = 400 / 7.9999
Step 5:
Q = 50
When Q(t) = 25, 600, t = 6, then the value of Q will be as follows,
Step 1:
25600 = Qe1.03972(6)
Step 2:
Q = 25600 / e1.03972(6)
Step 3:
Q = 25600 / e6.23832
Step 4:
Q = 25600 / 511.997
Step 5:
Q = 50
Final answer:
From the above calculations it is clearly evident that the initial population at time t = 0 hours
was 50.
3) Write the function that models the population n(t) after t hours.
Solution:
In this query we are asked provided with n(t) which is nothing but the rate of growth of
bacteria after ‘t’ hours. In the formula it is shown as Q(t), in order to avoid confusion we can
use the same variable Q(t) instead of n(t) as used in other calculations. The function that
models the population n(t) after t hours will be as follows,
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Sub: Biology Topic: Microbiology
Q(t) = Qe^kt
We have the values Q = 50 and k = 1.03972, let us now substitute the values to get the
population n(t) after t hours in the above formula, so finally we get
Q(t) = 50e e1.03972t.
Final answer:
Q(t) = 50e e1.03972t which is the function that models the population n(t) after t hours.
4) When will the number of bacteria exceed 100,000?
Solution:
We have found the values of Q and k, which is nothing but the initial number of
bacterial population at time t = 0 and non-zero constant.
In this query we are provided rate of growth of bacteria which is
Q(t) = 100,000, we are asked to find out the time at which this growth rate is reached.
In simple words we are provided with Q(t) and we need to find out the t.
Q(t) = Qe^kt
Where the following were deduced,
Q(t) = 100,000
Q = 50
k = 1.03972
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Sub: Biology Topic: Microbiology
Let us now substitute the values and find out the time t,
Step 1:
100,000 = 50e1.03972t
Step 2:
100,000 / 50 = e1.03972t
Step 3:
2000 = e1.03972t
Step 4:
Taking ln on both sides
ln 2000 = 1n [e1.03972t]
Step 5:
ln 2000 = 1.03972t lne
Here lne will become 1, so finally we get
ln 2000 = 1.03972t
Step 6:
ln 2000 = 7.60090246
Let us now substitute the value of ln 2000, we get
7.60090246 = 1.03972t
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Sub: Biology Topic: Microbiology
Step 7:
t = 7.601 / 1.03972
Step 8:
t = 7.3105 hrs
Final answer:
The time when the number of bacteria exceeds 100,000 is 7 hours and 19 minutes.
** End of the Solution **
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