the exponential function e a reminder from c2. use your calculator (if you need to) to work out the...
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The exponential functione
A reminder from C2.
Use your calculator (if you need to) to work out the following.Why are these the answers? (Hint: What does logax actually mean?)
Question Answer Reason
log39 =
log5125 =
log216 =
log61 =
log648 =
log51/25 =
2
3
4
0
½
-2
32 = 9
53 = 125
24 = 16
60 = 1
641/2 (= √64) = 8
5-2 = 1/25
Follow-up question:Based on your answers to the above, what is an alternative way of writing the statement:
y = logex ?
The exponential functione
What you have in fact just done is made x the subject of y = ln x.
i.e. y = ln x x = ey
If we now interchange y and x we obtain y = ex.
What effect does this have on the graph of y = ln x?
y
x
y = ln x
× (1, 0)
× (0, 1)
The exponential functione
× (2, ln2)
y = x
× (ln2, 2)
× (3, ln3)
× (ln3, 3)
Interchanging x and y has the effect of reflecting the graph in the line y = x.
y
x
y = ln x
× (1, 0)
× (0, 1)
The exponential functione
× (2, ln2)
y = x
× (ln2, 2)
× (3, ln3)
× (ln3, 3)
y = ex
The function y = ex is called the exponential function (more on this later).
The exponential functione
You know from your work on functions that reflecting in the line y = x gives an inverse function, so it follows that y = ex and y = ln x are the inverse of each other.
What does this mean?
Investigate, using your calculator, eln x and ln(ex) for different values of x. What is happening?
These functions “undo” each other.
ln(ex) = xlne (using the laws of logarithms) = x (using the fact that ln e = logee = 1.
The exponential functione
Although the function ex is called the exponential function, in fact any function of the form ax is exponential.
y
x
y = 1x = 1
y = 0.5x
y = 1.5x
y = 2x
y = ex
y = 3xUse your graphical calculator to plotthe following:
1. y = 1.5x
2. y = 2x
3. y = ex
4. y = 3x
5. y = 1x
6. y = 0.5x
7. y = e-x
8. y = (-2)x
Discuss:Why are 6, 7 & 8 different to the rest?
The exponential function y = ex increases at an ever increasing rate. This is described as exponential growth.
By contrast, the graph of y = e-x, shown below, approaches the x axis ever more slowly as x increases: this is called exponential decay.
The exponential functione
y
x
y = e-x
This works because:
y = e-x = (1/e)x.
1/e = 1/2.718281828 = 0.3678794412 < 1.
So, multiplying this by itself an increasing number of times will result in a smaller and smaller answer, similar to y = 0.5x.
You will meet ex and ln x again later in this course where you will learn to differentiate and integrate them. In this section we will be focusing on practical applications which require you to use the ln button on your calculator.
The exponential functione
The exponential functione
Example
The number, N, of insects in a colony is given by N = 2000e0.1t where t is the number of days after observations have begun.
(i) Sketch the graph of N against t.(ii) What is the population of the colony after 20 days?(iii) How long does it take the colony to reach a population of 10 000?
The exponential functione
Solution
The number, N, of insects in a colony is given by N = 2000e0.1t where t is the number of days after observations have begun.
(i) Sketch the graph of N against t.
N
tO
N = 2000e0.1t
2000When t = 0, N = 2000e0 = 2000
The exponential functione
Solution
The number, N, of insects in a colony is given by N = 2000e0.1t where t is the number of days after observations have begun.
(ii) What is the population of the colony after 20 days?
When t = 20, N = 2000e0.1×20 = 14 778.
The population is 14 778 insects.
The exponential functione
Solution
The number, N, of insects in a colony is given by N = 2000e0.1t where t is the numberof days after observations have begun.
(iii) How long does it take the colony to reach a population of 10 000?
When N = 10 000, 10 000 = 2000e0.1t
5 = e0.1t.
Taking natural logarithms of both sides,
ln 5 = ln(e0.1t) = 0.1tand so t = 10ln5 = 16.09…
It takes just over 16 days for the population to reach 10 000.
Rememberln(ex) = x.
The exponential functione
Example
The radioactive mass, M grams of a lump of material is given by M = 25e-0.0012t where t is the time in seconds since the first observation.
(i) Sketch the graph of M against t.(ii) What is the initial size of the mass?(iii) What is the mass after 1 hour?(iv) The half-life of a radioactive substance is the time it takes to decay to half of its
mass. What is the half-life of this material?
The exponential functione
Solution
The radioactive mass, M grams of a lump of material is given by M = 25e-0.0012t where t is the time in seconds since the first observation.
(i) Sketch the graph of M against t.
M
tO
25
The exponential functione
Solution
The radioactive mass, M grams of a lump of material is given by M = 25e-0.0012t where t is the time in seconds since the first observation.
(ii) What is the initial size of the mass?
When t = 0, M = 25e0
= 25.
The initial mass is 25g.
The exponential functione
Solution
The radioactive mass, M grams of a lump of material is given by M = 25e-0.0012t where t is the time in seconds since the first observation.
(iii) What is the mass after 1 hour?
After 1 hour, t = 3600. M = 25e-0.0012×3600.
The mass after 1 hour is 0.33g (to 2 decimal places).
The exponential functione
Solution
The radioactive mass, M grams of a lump of material is given by M = 25e-0.0012t where t is the time in seconds since the first observation.
(iv) The half-life of a radioactive substance is the time it takes to decay to half of its mass. What is the half-life of this material?
The initial mass is 25g, so after one half-life,
M = ½ × 25 = 12.5g.
At this point the value of t is given by 12.5 = 25e-0.0012t.
Dividing both sides by 25 gives 0.5 = e-0.0012t.
Taking logarithms of both sides: ln 0.5 = ln e-0.0012t
= -0.0012t.
t = = 577.6 (to 1 decimal place).
The half-life is 577.6 seconds.
ln 0.5-0.0012
The exponential functione
Example
Make p the subject of ln(p) – ln(1 – p) = t.
Solution
The exponential functione
p
1 - pln = t
Using log a – log b = log ab
Writing both sides as powers of e gives
ep
1 - pln = et
Remember elnx = x
p1 - p
= et
p = et(1 – p)
p = et – pet
p + pet = et
p(1 + et) = et
p =et
1 + et
Example
Make p the subject of ln(p) – ln(1 – p) = t.
The exponential functione
Example
Solve the equation e1-3x = 5.
The exponential functione
Example
Solve the equation e1-3x = 5.
Solution
Take natural logarithms of both sides: ln(e1-3x) = ln 5
Using the laws of logarithms: (1 – 3x)ln e = ln 5
Since ln e = 1: 1 – 3x = ln 5
x =1 – ln 53
= -0.203
The exponential functione
Example
Solve the equation ln(2x + 1) = 3
The exponential functione
Example
Solve the equation ln(2x + 1) = 3
Solution
ln(2x + 1) = 3 2x + 1 = e3
x =
e3 – 12
= 9.54