Chapter Four of Maths in Focus
-
Upload
rishi-maran -
Category
Documents
-
view
222 -
download
0
Transcript of Chapter Four of Maths in Focus
8/10/2019 Chapter Four of Maths in Focus
http://slidepdf.com/reader/full/chapter-four-of-maths-in-focus 1/33
Exponential equation: Equation where the pronumeral isthe index or exponent such as 3 9
x
=
Exponential function: A function in the form y a x
= wherethe variable x is a power or exponent
Logarithm: A logarithm is an index. The logarithm is thepower or exponent of a number to a certain basei.e. 2 8
x
= is the same as log 8 x 2
=
TERMINOLOGY
Exponentialand Logarithmic
Functions
4
8/10/2019 Chapter Four of Maths in Focus
http://slidepdf.com/reader/full/chapter-four-of-maths-in-focus 2/33
161Chapter 4 Exponential and Logarithmic Functions
DID YOU KNOW?
John Napier (1550–1617), a Scottish theologian and an amateur mathematician, was the first toinvent logarithms. These ‘natural’, or ‘Naperian’, logarithms were based on ‘e’. Napier was also
one of the first mathematicians to use decimals rather than fractions. He invented the notation of
the decimal, using either a comma or a point. The point was used in England, but a few European
countries still use a comma.
Henry Briggs (1561–1630), an Englishman who was a professor at Oxford, decided that
logarithms would be more useful if they were based on 10 (our decimal system). These are called
common logarithms. Briggs painstakingly produced a table of logarithms correct to 14 decimal
places. He also produced sine tables—to 15 decimal places—and tangent tables—to 10 decimal
places.
The work on logarithms was greatly appreciated by Kepler, Galileo and other astronomers
at the time, since they allowed the computation of very large numbers.
INTRODUCTION
THIS CHAPTER INTRODUCES A new irrational number, ‘e’, that has
special properties in calculus. You will learn how to differentiate
and integrate the exponential function.
( ) f x e x
=
The definition and laws of logarithms are also introduced in
this chapter, as well as differentiation and integration involving
logarithms.
Differentiation of Exponential Functions
When differentiating exponential functions ( )f x ax= from first principles, an
interesting result can be seen. The derivative of any exponential function gives
a constant which is multiplied by the original function.
EXAMPLE
Sketch the derivative (gradient) function of y 10x= .
Solution
CONTINUED
8/10/2019 Chapter Four of Maths in Focus
http://slidepdf.com/reader/full/chapter-four-of-maths-in-focus 3/33
162 Maths In Focus Mathematics Extension 1 HSC Course
The graph of y 10x= always has a positive gradient that is becoming steeper.
So the derivative function will always be positive, becoming steeper.
The derivative function of an exponential function will always have a
shape similar to the original function.
We can use differentiation from first principles to find how close this
derivative function is to the original function.
EXAMPLE
Differentiate ( )f x 10x= from first principles.
Solution
( )x
10
=
=
( ) ( )
( )
lim
lim
lim
lim
f h
f x h f x
h
h
h
10 10
10 10 1
10 1
h
h
x h x
h
x h
x
h
h
0
0
0
0
+ −
=−
=−
−
"
"
"
"
+
l
Using the 10 x key on the calculator, and finding values of
h
10 1h−
when h
is small, gives the result:
or
.2 3026 10Z( )x
( ) .dx
d 10 2 3026 10
x
x xZ
f l
Drawing the graphs of . y 2 3026 10x
= and y 10x
= together shows
how close the derivative function is to the original graph.
12
10
8
6
4
2
1 2−1−2
y
y = 2.3026 ×10 x
x
y =10 x
You can explore
limits using a
graphics package
on a computer or a
graphical calculator.
´
´
´
8/10/2019 Chapter Four of Maths in Focus
http://slidepdf.com/reader/full/chapter-four-of-maths-in-focus 4/33
163Chapter 4 Exponential and Logarithmic Functions
Similar results occur for other exponential functions. In general,
( )dx
d a ka
x x= where k is a constant.
Application
If y a x = then
dx
dy ka
ky
x =
=
This means that the rate of change of y is proportional to y itself. That is, if y is
small, its rate of change is small, but if y is large, then it is changing rapidly.
This is called exponential growth (or decay, if k is negative) and has many
applications in areas such as population growth, radioactive decay, the cooling of
objects, the spread of infectious diseases and the growth of technology.
Different exponential functions have different values of k.
EXAMPLES
1. ( ) .dx
d 2 0 6931 2x xZ .
12
10
8
6
4
2
1 2 3−3 −2 −1
y
y = 2 x
y = 0.6931 × 2 x
x
2. ( ) .dx
d 3 1 0986 3x xZ .
12
10
8
6
4
2
−3
y
y = 3 x
y = 1.0986 × 3 x
x1 32−2−1
You will study exponential
growth and decay in
Chapter 6.
´
´
8/10/2019 Chapter Four of Maths in Focus
http://slidepdf.com/reader/full/chapter-four-of-maths-in-focus 5/33
164 Maths In Focus Mathematics Extension 1 HSC Course
Notice that the derivative function of y 3x
= is very close to the original
function.
We can find a number close to 3 that gives exactly the same graph for the
derivative function. This number is approximately 2.71828, and is called e.
e x KEY
Use this key to find powers of e.
For example, to find e2:
Press SHIFT e 2 7.389056099e2 x
=
To find e:
Press SHIFT 1 2.718281828e e1 x
=
EXAMPLES
1. Sketch the curve . y ex
=
Solution
Use ex on your calculator to draw up a table of values:
x -3 -2 -1 0 1 2 3
y 0.05 0.1 0.4 1 2.7 7.4 20.1
( )dx
d e ex x=
DID YOU KNOW?
The number e was linked to logarithms before this useful result in calculus was known. It is
a transcendental (irrational) number. This was proven by a French mathematician, Hermite,
in 1873. Leonhard Euler (1707–83) gave e its symbol, and he gave an approximation of e to
23 decimal places. Currently, e is known to about 100 000 decimal places.
Euler studied mathematics, theology, medicine, astronomy, physics and oriental languages.
He did extensive research into mathematics and wrote more than 500 books and papers.
Euler gave mathematics much of its important notation. He caused π to become standardnotation and used i for the square root of –1. He first used small letters to show the sides of
triangles and the corresponding capital letters for their opposite angles. Also, he introduced
the symbol for sums and f(x) notation.
A transcendental
number is a number
beyond ordinary
numbers. Another
transcendental
number is π .
e is an irrational numberlike π .
8/10/2019 Chapter Four of Maths in Focus
http://slidepdf.com/reader/full/chapter-four-of-maths-in-focus 6/33
165Chapter 4 Exponential and Logarithmic Functions
2. Differentiate .e5 x
Solution
`
( )
( ) ( )
dx
d
e e
dx
d e
dx
d e
e
5 5
5
x x
x x
x
=
=
=
3. Find the equation of the tangent to the curve y e3 x= at the point (0, 3).
Solution
dx
dy e3 x
=
`
At ( , ),dx
dy e
m
0 3
3
3
3 0=
=
=
Equation ( )
( )
y y m x x
y x
x
y x
3 0
3
3 3
31 1
− = −
− = −
=
= +
dx
dy gives the gradient of
the tangent.
CONTINUED
8/10/2019 Chapter Four of Maths in Focus
http://slidepdf.com/reader/full/chapter-four-of-maths-in-focus 7/33
166 Maths In Focus Mathematics Extension 1 HSC Course
4.1 Exercises
1. Find, correct to 2 decimal places,
the value of
(a) e .1 5
(b) e 2−
(c) e2
.0 3
(d)e
1
3
(e) e3 .3 1
−
−
2. Sketch the curve
(a) y e2 x
=
(b) y e x=
−
(c) y ex= −
3. Differentiate
(a) e9 x
(b) ex
−
(c) e xx 2+
(d) x x x e2 3 5 x3 2
− + −
(e) 3( )e 1x
+
(f) 7( )e 5x
+
(g) 2( )e2 3x
−
(h) xex
(i)x
ex
(j) x ex2
(k) ( )x e2 1 x+
(l)x
e
7 3
x
−
(m)e
x5x
4. If ( ) ,f x x x e3 x3= + − find ( )1f l
and ( )f 1m in terms of e.
5. Find the exact gradient of the
tangent to the curve y ex
= at the
point (1, e).
6. Find the exact gradient of the
normal to the curve y ex
= at the
point where .x 5=
7. Find the gradient of the tangent
to the curve y e4 x
= at the point
where . ,x 1 6= correct to 2
decimal places.
8. Find the equation of the tangent
to the curve y ex
= − at the point
(1, –e).
4. Differentiate .e
x2 3x
+
Solution
v u
.
( )
( )
( )
( )
dx
dy
v
u v
e
e e x
e
e xe e
e
e xe
e
e x
e
x
2 2 3
2 2 3
2
1 2
1 2
x
x x
x
x x x
x
x x
x
x
x
2
2
2
2
2
=
−
=
− +
=− −
=− −
=
− +
=
− +
l l
This is the quotient rule
from Chapter 8 of the
Preliminary Course book.
8/10/2019 Chapter Four of Maths in Focus
http://slidepdf.com/reader/full/chapter-four-of-maths-in-focus 8/33
167Chapter 4 Exponential and Logarithmic Functions
Function of a function rule
Remember that the function of a function rule uses the result
.
dx
dy
du
dy
dx
du=
EXAMPLE
Differentiate .ex x5 32
+ −
Solution
Let u x x5 32= + −
Then y eu=
dx
dux2 5= + and
du
dy eu
=
( )
( )
( )
dx
dy
du
dy
dx
du
e xe x
x e
2 5
2 5
2 5
u
x x
x x
5 3
5 3
2
2
=
= +
= +
= +
+ −
+ −
You studied this in Chapter 8 of
the Preliminary Course book.
Can you see a quick way to
do this?
Proof
Let ( )u f x=
Then y eu=
du
dy eu
= and ( )xdx
duf = l
( )x
( )x e
dx
dy
du
dy
dx
du
e f
f ( )
u
f x
=
=
=
l
l
If y e ( )f x= then ( )x e
dx
dy ( )f x
= f l
9. Find the equation of the normal
to the curve y ex
= at the point
where ,x 3= in exact form.
10. Find the stationary point on the
curve y xex
= and determine its
nature. Hence sketch the curve.
11. Find the first and second
derivatives of y e7 x
= . Hence show
that .dx
d y y
2
2
=
12. If , y e2 1
x= +
show that.
dx
d y
y 1
2
2
= −
´
´
´
8/10/2019 Chapter Four of Maths in Focus
http://slidepdf.com/reader/full/chapter-four-of-maths-in-focus 9/33
168 Maths In Focus Mathematics Extension 1 HSC Course
4.2 Exercises
(m)x
e x
2
3
(n) x e x3 5
(o)x
e
2 5
x2 1
+
+
2. Find the second derivative of
.7( )e 1x2+
3. If ( ) ,f x e x3 2=
− find the exact value
of ( )f 1l and ( ) .f 0m
4. Find the gradient of the tangent
to the curve y e x5
= at the point
where .x 0=
EXAMPLES
1. Differentiate e x5 2−
Solution
( )x e y f
e5
( )f x
x5 2
=
= −
l l
2. Differentiate x e x2 3 .
Solution
u +
dx
dy v v u= l l
. .2 3
( )
x e e x
xe x2 3
x x
x
3 3 2
3
= +
= +
3. Given , y e2 1x3= + show that ( ) .
dx
d y y 9 1
2
2
= −
Solution
( )
( )
( )
y e
dx
dy e
dx
d y e
e
e
y
2 1
6
18
9 2
9 2 1 1
9 1
x
x
x
x
x
3
3
2
2
3
3
3
= +
=
=
=
= + −
= −
This is the product rule from
Chapter 8 of the Preliminary
Course book.
1. Differentiate
(a) e x7
(b) e x−
(c) e x6 2−
(d) ex 1
2+
(e) ex x5 73
+ +
(f) e x5
(g) e x2−
(h) e x10
(i) e xx2+
(j) x x e2 x2 1
+ + −
(k) 5( )x e x4
+
(l) xe
x2
8/10/2019 Chapter Four of Maths in Focus
http://slidepdf.com/reader/full/chapter-four-of-maths-in-focus 10/33
169Chapter 4 Exponential and Logarithmic Functions
Integration of Exponential Functions
Since ( ) ,dx
d e e
x x= then the reverse must be true.
5. Find the equation of the tangent
to the curve y e x3x2
= − at the
point (0, 1).
6. Find the exact gradient of the
normal to the curve y e x3
= at the
point where .x 1=
7. Find the equation of the tangent to
the curve y ex2
= at the point (1, e).
8. If ( ) ,f x x x e4 3 x3 2 2= + −
− find
( )f 1−m in terms of e.
9. Find any stationary points on
the curve y x e x2 2= and sketch the
curve.
10. If , y e ex x4 4= + − show that
.dx
d y y 16
2
2
=
11. Prove ,dx
d y
dx
dy y 3 2 0
2
2
− + = given
. y e3 x2=
12. Showdx
d y b y
2
2
2= for . y aebx
=
13. Find the value of n if y e x3=
satisfies the equation
y .
dx
d
dx
dy ny 2 0
2
2
+ + =
14. Sketch the curve , y ex x 22
= + −
showing any stationary points
and inflexions.
15. Sketch the curve
e
x y
1x
2
=+
,
showing any stationary points
and inflexions.
e dx e Cx x= +
#
Integration is the inverse of
differentiation.
To find the indefinite integral (primitive function) when the function of
a function rule is involved, look at the derivative first.
EXAMPLE
Differentiate .e x2 1+
Hence find .e dx2 x2 1+ #
Find .e dxx2 1+
Solution
( )dx
d e e
e dx e C
e dx e dx
e C
2
2
2
12
2
1
x x
x x
x x
x
2 1 2 1
2 1 2 1
2 1 2 1
2 1
`
=
= +
=
= +
+ +
+ +
+ +
+
# # #
8/10/2019 Chapter Four of Maths in Focus
http://slidepdf.com/reader/full/chapter-four-of-maths-in-focus 11/33
170 Maths In Focus Mathematics Extension 1 HSC Course
EXAMPLES
1. Find ( ) .e e dxx x2−
− #
Solution
( )( )
e e dx e e C
e e C
2
1
1
1
2
1
x x x x
x x
2 2
2
− = −
−
+
= + +
− −
−
#
2. Find the exact area enclosed between the curve , y e x3
=
the x-axis andthe lines x 0= and .x 2=
Solution
Area
( )
( ) units
e dx
e
e e
e e
e
3
1
3
1
3
1
3
1
31 1
x
x
3
0
3
0
2
6 0
6 0
6 2
=
=
= −
= −
= −
2
; E #
3. Find the volume of the solid of revolution formed when the curve
y ex= is rotated about the x-axis from x 0= to .x 2=
Solution
( )
y e
y e
e
x
x
x
2 2
2
`
=
=
=
In general
Use index laws to simplify ( ) .e 2 x
e dxa
e C1ax b ax b
= ++ +
Proof
( )dx
d e ae
ae dx e C
e dxa
ae dx
a e C
1
1
ax b ax b
ax b ax b
ax b ax b
ax b
`
=
= +
=
= +
+ +
+ +
+ +
+
# # #
8/10/2019 Chapter Four of Maths in Focus
http://slidepdf.com/reader/full/chapter-four-of-maths-in-focus 12/33
171Chapter 4 Exponential and Logarithmic Functions
( ) units
π
π
π
π
π
π
V y dx
e dx
e
e e
e
e
2
1
2
1
2
1
2
1
2
1
2 1
a
x
x
2
2
0
2
0
2
4 0
4
4 3
=
=
=
= −
= −
= −
b
2
cc
mm
; E
#
#
4.3 Exercises
(c) ( )e x dx2 3x 5
5
6+ −
+
(d) ( )e t dt t 3 4
0−
+1
(e) ( )e e dxx x4 2
1+
2
4. Find the exact area enclosed by
the curve , y e2 x2= the x-axis and
the lines x 1= and .x 2=
5. Find the exact area bounded by
the curve , y e x4 3
=− the x-axis and
the lines x 0= and .x 1=
6. Find the area enclosed by the
curve , y x e x
= + − the x-axis and
the lines x 0= and ,x 2= correct
to 2 decimal places.
7. Find the area bounded by the
curve , y e x5
= the x-axis and the
lines x 0= and ,x 1= correct to
3 significant figures.
8. Find the exact volume of the
solid of revolution formed when
the curve y ex
= is rotated about
the x-axis from x 0= to .x 3=
9. Find the volume of the solid formed
when the curve y e 1x
= +− is rotated
about the x-axis from x 1= to ,x 2=
correct to 1 decimal place.
1. Find these indefinite integrals.
(a) e dxx2 #
(b) e dxx4 #
(c) e dxx− #
(d) e dxx5 #
(e) e dxx2− #
(f) e dxx4 1+ #
(g) e dx3 x5
− # (h) e dt
t 2 # (i) ( )e dx2x7
− # (j) ( )e x dx
x 3+
− #
2. Evaluate in exact form.
(a) e dxx5
0
1
#
(b) e dxx
0
− −
2
#
(c) e dx2 x3 4
1
+4 #
(d) ( )x e dx3 x2 2
2−
3 #
(e) ( )e dx1x2
0+
2 #
(f) ( )e x dxx
1−
2 #
(g) ( )e e dxx x2
0−
−
3 #
3. Evaluate correct to 2 decimal places.
(a) e dxx
1
−
3 #
(b) e dy 2 y 3
0
2 #
8/10/2019 Chapter Four of Maths in Focus
http://slidepdf.com/reader/full/chapter-four-of-maths-in-focus 13/33
172 Maths In Focus Mathematics Extension 1 HSC Course
10. Use Simpson’s rule with 3
function values to find an
approximation to ,xe dxx
1
2
correct to 1 decimal place.
11. (a) Differentiate.x e
x2
(b) Hence find ( ) .x x e dx2 x+
12. Find x e dxx2 1
3+ # using the
substitution .u x 13= +
13. Use the substitution u x2
= to
evaluate xe dxx
0
22
(give exact
value).
14. The curve y e 1x
= + is rotated
about the x-axis from x 0= to
.x 1= Find the exact volume of
the solid formed.
15. Find the exact area enclosed
between the curve y e x2
= and the
lines y 1= and .x 2=
Application
The exponential function occurs in many fields, such as science and economics.
P P e0
kt = is a general formula that describes exponential growth.
P P e0
kt =
- is a general formula that describes exponential decay.
Logarithms
‘Logarithm’ is another name for the index or power of a number. Logarithms
are related to exponential functions, and allow us to solve equations like
.2 5x
= They also allow us to change the subject of exponential equations such
as y ex
= to x.
Definition
If ,y a x
= then x is called the logarithm of y to the base a.
You will study these formulae in
Chapter 6.
If , y ax= then logx y
a=
Logarithm keys
log is used for log x 10
In is used for log x e
8/10/2019 Chapter Four of Maths in Focus
http://slidepdf.com/reader/full/chapter-four-of-maths-in-focus 14/33
173Chapter 4 Exponential and Logarithmic Functions
EXAMPLES
1. Find .log 5 310
correct to 1 decimal place.
Solution
log 5.3 0.724275869
0.7 correct to 1 decimal place10
=
=
2. Evaluate log 80e
correct to 3 significant figures.
Solution
.
. correct to 3 significant figures
log 80 4 382026634
4 38e
=
=
3. Evaluate .log 813
Solution
Let
Then (by definition)
i.e.
So .
log
log
x
x
81
3 81
3 3
4
81 4
x
x
3
4
3
`
=
=
=
=
=
4. Find the value of .log 4
12
Solution
Let
Then
So .
log
log
x
x
41
241
2
1
2
2
41 2
x
2
2
2
2
`
=
=
=
=
= −
= −
−
Use the log key.
Use the In key.
8/10/2019 Chapter Four of Maths in Focus
http://slidepdf.com/reader/full/chapter-four-of-maths-in-focus 15/33
174 Maths In Focus Mathematics Extension 1 HSC Course
1. Evaluate
log(a)2 16
log(b)4 16
log(c)5 125
log(d)3 3
log(e)7 49
log(f)7 7
log(g)5 1
log(h)2 128
2. Evaluate3 log(a)
2 8
log(b)5 25 + 1
3 – log(c)3 81
4 log(d)3 27
2 log(e)10
10 000
1 + log(f)4 64
3 log(g)4 64 + 5
2 + 4 log(h)6 216
(i)log
2
93
(j)log
log
8
64 4
2
8 +
3. Evaluate
(a) log2
12
(b) log 33
log(c)4 2
(d) log25
15
(e) log 77
4
(f) log3
13 3
(g) log2
14
log(h)8 2
(i) log 6 66
(j) log 4
22
4. Evaluate correct to 2 decimal
places.
log(a)10
1200
log(b)10
875
log(c)e 25
ln 140(d)
5 ln 8(e)
log(f)10
350 + 4.5
(g)
log
2
1510
ln 9.8 + log(h)10
17
(i)log
log
30
30
e
10
4 ln 10 – 7(j)
4.4 Exercises
Class Investigation
Sketch the graph of1. log y x2
= .
There is no calculator key for logarithms to the base 2. Use the
definition of a logarithm to change the equation into index form,and the table of values:
x
y –3 –2 –1 0 1 2 3
On the same set of axes, sketch the curve2. 2 y x= and the line y x= .
What do you notice?
8/10/2019 Chapter Four of Maths in Focus
http://slidepdf.com/reader/full/chapter-four-of-maths-in-focus 16/33
175Chapter 4 Exponential and Logarithmic Functions
5. Write in logarithmic form.
(a) y 3x=
(b) z5x
=
(c) x y 2=
(d) a2b=
(e)b d 3=
(f) y 8x
=
(g) y 6x
=
(h) y ex
=
(i) y ax
=
(j) e=Q x
6. Write in index form.
(a) log x53
=
(b) log x7a
=
(c) log a b3
=
(d) 9log y x=
(e) log b y a
=
(f) log y 62
=
(g) log y x3
=
(h) log y 910
=
(i) 4ln y =
(j) log y x7
=
7. Solve for x, correct to 1 decimal
place where necessary.
(a) log x 610
=
(b) log x 53 =
(c) log 343 3x
=
(d) log 64 6x
=
(e) log x
51
5 =
(f) log 32
1x
=
(g) 3.8ln x =
(h) log x3 2 1010
− =
(i) log x
2
34
=
(j) log 431
x=
8. Evaluate y given that .log 125 3 y
=
9. If . ,log x 1 6510
= evaluate x correct
to 1 decimal place.
10. Evaluate b to 3 significant figures
if . .log b 0 894e =
11. Find the value of log2 1. What is
the value of loga 1?
12. Evaluate log5 5. What is the value
of loga a?
13. (a) Evaluate ln e without a
calculator.
Using a calculator, evaluate(b)
(i) loge e3
(ii) loge e2
(iii) loge e5
(iv) loge e
(v) loge 1
e
(vi) eln 2
(vii) eln 3
(viii) eln 5
(ix) eln 7
(x) eln 1
(xi) eln e
14. Sketch the graph of .log y xe
=
What is its domain and range?
15. Sketch , log y y x10x
10= = and
y x= on the same number plane.
What do you notice about the
relationship of the curves to
the line?
16. Change the subject of log y xe
=
to x.
8/10/2019 Chapter Four of Maths in Focus
http://slidepdf.com/reader/full/chapter-four-of-maths-in-focus 17/33
8/10/2019 Chapter Four of Maths in Focus
http://slidepdf.com/reader/full/chapter-four-of-maths-in-focus 18/33
177Chapter 4 Exponential and Logarithmic Functions
This corresponds to the
law ( )a a .m n mn
=
,log 36 26
= since .6 362=
Proof
Let x am
= and y an
=
Then logm xa
= and logn y a
=
` (by definition)
÷
log
log log
y
xa a
a
y
xm n
x y
m n
m n
a
a a
=
=
= −
= −
−
b l
log logx n xa
n
a=
Proof
Let x am=
Then logm xa
=
`
( )
(by )log
log
x a
a
x mn
n x
definition
n m n
mn
a
n
a
=
=
=
=
EXAMPLES
1. Evaluate .log log3 126 6
+
Solution
( )log log log
log
3 12 3 12
36
2
6 6 6
6
+ =
=
=
2. Given .log 3 0 685
= and . ,log 4 0 865
= find
(a) log 125
(b) .log 0 755
(c) log 95
(d) log 205
CONTINUED
´
8/10/2019 Chapter Four of Maths in Focus
http://slidepdf.com/reader/full/chapter-four-of-maths-in-focus 19/33
178 Maths In Focus Mathematics Extension 1 HSC Course
Solution
(a) ( )
. .
.
log log
log log
12 3 4
3 4
0 68 0 86
1 54
5 5
5 5
=
= +
= +
=
(b) .
. .
.
log log
log log
0 754
3
3 4
0 68 0 86
0 18
5 5
5 5
=
= −
= −
= −
(c)
.
.
log log
log
9 3
2 3
2 0 68
1 36
5 52
5
=
=
=
=
(d) ( )
.
.
log log
log log
20 5 4
5 4
1 0 86
1 86
5 5
5 5
=
= +
= +
=
3. Solve log2 12 = log
2 3 + log
2 x.
Solution
log log log
log
x
x
12 3
32 2 2
2
= +
=
So 12 = 3x
4 = x
,log 5 15
= since .5 5 1=
1. Use the logarithm laws to
simplify
log(a)a 4 + log
a y
log(b) a 4 + loga 5log(c)
a 12 – log
a 3
log(d)a b – log
a 5
3 log(e)x y + log
x z
2 log(f)k 3 + 3 log
k y
5 log(g)a x – 2 log
a y
log(h)a x + log
a y – log
a z
log(i)10
a + 4 log10
b + 3 log10
c
3 log(j)3 p + log
3 q – 2 log
3 r
2. Given log7 2 = 0.36 and
log7 5 = 0.83, find
log(a)7 10
log(b) 7 0.4log(c)
7 20
log(d)7 25
log(e)7 8
log(f)7 14
log(g)7 50
log(h)7 35
log(i)7 98
log(j)7 70
4.5 Exercises
´
´
´
8/10/2019 Chapter Four of Maths in Focus
http://slidepdf.com/reader/full/chapter-four-of-maths-in-focus 20/33
179Chapter 4 Exponential and Logarithmic Functions
3. Use the logarithm laws to
evaluate
log(a)5 50 – log
5 2
log(b)2 16 + log
2 4
log(c)4 2 + log
4 8
log(d) 5 500 – log5 4log(e)
9 117 – log
9 13
log(f)8 32 + log
8 16
3 log(g)2 2 + 2 log
2 4
2 log(h)4 6 – (2 log
4 3 + log
4 2)
log(i)6 4 – 2 log
6 12
2 log(j)3 6 + log
3 18 – 3 log
3 2
4. If loga 3 = x and log
a 5 = y , find an
expression in terms of x and y for
log(a)a 15
log(b) a 0.6log(c)
a 27
log(d)a 25
log(e)a 9
log(f)a 75
log(g)a 3a
log(h)a 5
a
log(i)a 9a
log(j)a a
125
5. If loga x = p and loga y = q, find, interms of p and q.
log(a)a xy
log(b)a y 3
log(c)a x
y
log(d)a x2
log(e)a xy 5
log(f)a y
x2
log(g)a ax
log(h) a y
a
2
log(i)a a3 y
log(j)a ay
x
6. If loga b = 3.4 and log
a c = 4.7,
evaluate
log(a)a b
c
log(b)a bc 2
log(c)a (bc )2
log(d)a abc
log(e) a a2
c log(f)
a b7
log(g)a c
a
log(h)a a3
log(i)a bc 4
log(j)a b4c 2
7. Solve
log(a)4 12 = log
4 x + log
4 3
log(b)3 4 = log
3 y – log
3 7
log(c)a 6 = log
a x – 3 log
a 2
log(d) 2 81 = 4 log2 xlog(e)
x 54 = log
x k + 2 log
x 3
Change of base
Sometimes we need to evaluate logarithms such as log2 7. We use a change of
base formula.
loglog
logx
a
x
a
b
b
=
8/10/2019 Chapter Four of Maths in Focus
http://slidepdf.com/reader/full/chapter-four-of-maths-in-focus 21/33
180 Maths In Focus Mathematics Extension 1 HSC Course
EXAMPLE
Find the value of ,log 25
correct to 2 decimal places.
Solution
.0 430676558Z
(by change of base)
.
loglog
log2
5
2
0 43
5 =
=
Proof
Let log y xa
=
Then x a y
=
Take logarithms to the base b of both sides of the equation:
`
log log
log
log
log
log
x a
y a
a
x y
x
b b
y
b
b
b
a
=
=
=
=
You can use the change of base formula to find the logarithm of any number,
such as .log 25
You change it to either log x10
or ,log xe
and use a calculator.
Exponential equations
You can also use the change of base formula to solve exponential equations
such as .5 7x
=
You studied exponential equations such as 2 8x
= in the Preliminary
Course. Exponential equations such as 2 9x
= can be solved by taking
logarithms of both sides, or by using the definition of a logarithm and the
change of base formula.
You can use either log or In
8/10/2019 Chapter Four of Maths in Focus
http://slidepdf.com/reader/full/chapter-four-of-maths-in-focus 22/33
181Chapter 4 Exponential and Logarithmic Functions
EXAMPLES
1. Solve 5 7x
= correct to 1 decimal place.
Solution
5 7x
=
Using the definition of a logarithm, this means:
(using change of base formula)
.
log
log
log
x
x
x
7
5
7
1 2
5 =
=
=
If you do not like to solve the equation this way, you can use the
logarithm laws instead.
Taking logs of both sides:
.
log log
log log
log
log
x
x
5 7
5 7
5
7
1 2 1correct to decimal place
x
`
=
=
=
=
2. Solve 4 9 y 3
=− correct to 2 decimal places.
Solution
y 3−
4 9=
Using the logarithm definition and change of base:
.
log
log
log
log
log
y
y
y
y
9 3
4
93
4
93
4 58
4 = −
= −
+ =
=
Using the logarithm laws:
Taking logs of both sides:
( )
.
log loglog log
log
log
log
log
y
y
y
4 93 4 9
34
9
4
93
4 58 2correct to decimal place
y 3 =
− =
− =
= +
=
−
You can use either
log or ln.
Use log x n log x a
n
a=
8/10/2019 Chapter Four of Maths in Focus
http://slidepdf.com/reader/full/chapter-four-of-maths-in-focus 23/33
182 Maths In Focus Mathematics Extension 1 HSC Course
1. Use the change of base formula
to evaluate to 2 decimal places.
log(a)4 9
log(b) 6 25log(c)
9 200
log(d)2 12
log(e)3 23
log(f)8 250
log(g)5 9.5
2 log(h)4 23.4
7 – log(i)7 108
3 log(j)11
340
2. By writing each equation as a
logarithm and changing the base,solve the equation correct to
2 significant figures.
(a) 4 9x
=
(b) 3 5x
=
(c) 7 14x
=
(d) 2 15x
=
(e) 5 34x
=
(f) 6 60x
=
(g) 2 76x
=
(h) 4 50x
=
(i) 3 23x
=
(j) 9 210x
=
3. Solve, correct to 2 decimal places.
(a) 2 6x
=
(b) 5 15 y =
(c) 3 20x
=
(d) 7 32m
=
(e) 4 50k=
(f) 3 4t =
(g) 8 11x
=
(h) 2 57 p=
(i) .4 81 3x
=
(j) .6 102 6n
=
4. Solve, to 1 decimal place.
(a) 3 8x 1
=+
(b) n35 71=
(c) 2 12x 3
=−
(d) 4 7n2 1
=−
(e) 7 11x5 2
=+
(f) .8 5 7n3=
−
(g) .2 18 3x 2
=+
(h) k7 3−
.3 32 9=
(i) 29 50=
x
(j) .6 61 3 y 2 1
=+
5. Solve each equation correct to
3 significant figures.
(a) e 200x
=
(b) e 5t 3=
(c) e2 75t =
(d) e45 x
=
(e) e3000 100 n
=
(f) e100 20 t 3
=
(g) e2000 50 . t 0 15
=
(h) e15 000 2000 . k0 03
=
(i) Q Qe3 . t 0 02
=
(j) . M Me0 5 . k0 016
=
4.6 Exercises
8/10/2019 Chapter Four of Maths in Focus
http://slidepdf.com/reader/full/chapter-four-of-maths-in-focus 24/33
183Chapter 4 Exponential and Logarithmic Functions
Derivative of the Logarithmic Function
Drawing the derivative (gradient) function of a logarithm function gives a
hyperbola.
EXAMPLE
Sketch the derivative function of .log y x2
=
Solution
The gradient is always positive but is decreasing.
If log y xe
= then1
dx
dy
x=
Proof
dx
dy
dy
dx
1=
Given log y xe
=
Then y x e=
`
y
dy
dxe
dx
dy
e
x
1
1
y
=
=
=
dx
dy
dy
dx
1= is a special result that
can be proved by differentiating
from first principles.
8/10/2019 Chapter Four of Maths in Focus
http://slidepdf.com/reader/full/chapter-four-of-maths-in-focus 25/33
184 Maths In Focus Mathematics Extension 1 HSC Course
Function of a function rule
If .( ), then ( )( ) ( )
( )log y f x
dx
dy f x
f x f x
f x1e
= = =ll
Proof
Let ( )u f x=
Then log y ue
=
`
Also
.
.
.
( )
( )
( ) ( )
du
dy
u
dx
duf x
dx
dy
du
dy
dx
du
u f x
f x f x
1
1
1
=
=
=
=
=
l
l
l
EXAMPLES
1. Differentiate ( 3 1) .log x x2
e− +
Solution
[ ( )]logdx
d x x
x x
x3 1
3 1
2 3e
2
2− + =
− +
−
2. Differentiate3 4
1.log
x
x
e−
+
Solution
Let
( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( )
log
log log
y x
x
x x
dx
dy
x x
x x
x x
x x
x x
x x
3 4
1
1 3 4
1
1
3 4
3
1 3 4
1 3 4 3 1
1 3 4
3 4 3 3
1 3 4
7
e
e e
=−
+
= + − −
=+
−−
=
+ −
− − +
=
+ −
− − −
=
+ −
−
8/10/2019 Chapter Four of Maths in Focus
http://slidepdf.com/reader/full/chapter-four-of-maths-in-focus 26/33
185Chapter 4 Exponential and Logarithmic Functions
3. Find the gradient of the normal to the curve ( )log y x 5e
3= − at the
point where 2.x =
Solution
dxdy
xx5
33
2
=
−
When 2,x =
( )
dx
dy
m
2 5
3 2
4
3
2
1
=
−
=
The normal is perpendicular to the tangent
m m
m
m
1
4 1
4
1
i.e.1 2
2
2`
= −
= −
= −
4. Differentiate .log y x2
=
Solution
log
log
log
log log
log
log
y x
x
x
dxdy
x
x
2
2
1
21 1
2
1
e
e
ee
e
e
2=
=
=
=
=
5. Find the derivative of 2x.
Solution
( )
ln
ln
ln
e
e
e
dx
dy e
2
2
2
2 2
2 2
2
ln
ln
ln
ln
x x
x
x
x
x
2
2
2
`
=
=
=
=
=
=
This result comes from the
Preliminary Course.
´
´
´
8/10/2019 Chapter Four of Maths in Focus
http://slidepdf.com/reader/full/chapter-four-of-maths-in-focus 27/33
186 Maths In Focus Mathematics Extension 1 HSC Course
1. Differentiate
(a) logx xe
+
(b) log x1 3e
−
(c) ( )ln x3 1+
(d) ( )log x 4e
2−
(e) ( )ln x x5 3 93+ −
(f) ( )log x x5 1e
2+ +
(g) lnx x x3 5 5 42
+ − +
(h) ( )log x8 9 2e
− +
(i) (2 4) (3 1)log x xe
+ −
(j)2 7
4 1log
x
x
e−
+
(k) ( )log x
1 e
5+
(l) ( )ln x x 9
−
(m) ( )log xe
4
(n) ( )logx xe
2 6+
(o) logx xe
(p)log
x
xe
(q) ( ) logx x2 1e
+
(r) ( )logx x 1e
3+
(s) ( )log log xe e
(t) lnx
x
2−
(u)log x
e
e
x2
(v) lne xx
(w) ( )log x5e
2
2. If ( ) 2 ,logf x xe
= − find ( )f 1l .
3. Find the derivative of .log x10
4. Find the equation of the tangentto the curve log y xe
= at the
point ( , )log2 2e
.
5. Find the equation of the tangent
to the curve ( 1)log y xe
= − at the
point where 2x = .
6. Find the gradient of the normal
to the curve ( )log y x xe
4= + at the
point ( , )log1 2e
.
7. Find the exact equation of the
normal to the curve log y xe
= at
the point where 5x = .
8. Find the equation of the tangent
to the curve ( )log y x5 4e
= + at
the point where 3x = .
9. Find the point of inflexion on the
curve .log y x x xe
2= −
10. Find the stationary point on the
curveln
x
x y = and determine its
nature.
11. Sketch, showing any stationary
points and inflexions.
(a) log y x xe
= −
(b) ( )log y x 1e
3= −
(c) ln y x x=
12. Find the derivative of ( )log x2 53
+ .
13. Differentiate
(a) 3x
(b) 10x
(c) 2 x3 4−
14. Find the equation of the tangent
to the curve y 4x 1
= + at the point
(0, 4).
15. Find the equation of the normal
to the curve log y x3
= at the
point where 3x = .
4.7 Exercises
8/10/2019 Chapter Four of Maths in Focus
http://slidepdf.com/reader/full/chapter-four-of-maths-in-focus 28/33
187Chapter 4 Exponential and Logarithmic Functions
logx
dx
x dx x C
1e
= = + #
Integration and the Logarithmic Function
EXAMPLES
1. Find the area enclosed between the hyperbola , y
x
1= the x-axis and
the lines 1x = and 2x = , giving the exact value.
Solution
log
log log
log
Ax
dx
x
1
2 1
2
e
e e
e
1
1
2
=
=
= −
=
2
7 A
#
So area is log 2e
units2.
2. Find .x
xdx
73
2
+
#
Solution
( )log
x
xdx
x
xdx
x C
7 3
1
7
3
31
7e
3
2
3
2
3
+
=
+
= + +
#
( )
( )( )log
f x
f xdx f x C
e= +
l
CONTINUED
Integration is the inverse of
differentiation.
8/10/2019 Chapter Four of Maths in Focus
http://slidepdf.com/reader/full/chapter-four-of-maths-in-focus 29/33
188 Maths In Focus Mathematics Extension 1 HSC Course
1. Find the indefinite integral
(primitive function) of
(a)2 5
2
x +
(b)x
x
2 1
4
2+
(c)2
5
x
x
5
4
−
(d)2
1
x
(e)
2
x
(f)3
5
x
(g)3
2 3
x x
x
2−
−
(h)x
x
22+
(i)x
x
7
3
2+
(j)x x
x
2 5
1
2+ −
+
2. Find
(a)x
dx4 1
4
−
#
(b)x
dx
3+ #
(c)x
xdx
2 73
2
−
#
(d)x
xdx
2 56
5
+
#
(e)x x
xdx
6 2
3
2 + +
+
#
3. Evaluate correct to 1 decimal
place.
(a)x
dx2 5
2
1
3
+
(b)x
dx
12 +
5
(c)x
xdx
23
2
1
7
+
(d)x x
xdx
2 1
4 1
20
3
+ +
+
(e)x x
xdx
2
1
23
4
−
−
4. Find the exact area between the
curve , y x
1= the x-axis and the
lines 2x = and 3.x =
5. Find the exact area bounded by
the curve , y x 1
1=
−
the x-axis and
the lines 4x = and 7.x =
6. Find the exact area between the
curve , y x1
= the x-axis and the
lines y x= and 2x = in the first
quadrant.
7. Find the area bounded by the
curve , y x
x
12
=
+
the x-axis and
the lines 2x = and 4x = , correct
to 2 decimal places.
4.8 Exercises
3. Find .x x
xdx
4
1
2+ +
+ #
Solution
( )
( )log
x x
x
dx x x
x
dx
x x
xdx
x x C
2 4
1
2
1
2 4
2 1
2
1
2 4
2 2
2
12 4
e
2 2
2
2
+ +
+
=+ +
+
=
+ +
+
= + + +
# # #
8/10/2019 Chapter Four of Maths in Focus
http://slidepdf.com/reader/full/chapter-four-of-maths-in-focus 30/33
189Chapter 4 Exponential and Logarithmic Functions
8. Find the exact volume of the
solid formed when the curve
y x
1= is rotated about the x-axis
from 1x = to 3x = .
9. Find the volume of the solid
formed when the curve
y x2 1
2=
−
is rotated about the
x-axis from 1x = to 5x = , giving
an exact answer.
10. Find the area between the curve
ln y x= , the y -axis and the lines
2 y = and 4 y = , correct to 3
significant figures.
11. Find the exact volume of the
solid formed when the curve
log y xe
= is rotated about the
y -axis from 1 y = to 3 y = .
12. (a) Show that
9
3 3
3
1
3
2
x
x
x x2−
+=
++
−
.
(b) Hence findx
xdx
9
3 3
2−
+.
13. (a) Show thatx
x
x1
6
1
51
−
−
−
−
= .
(b) Hence findx
xdx
1
6
−
−
.
14. Find the indefinite integral
(primitive function) of 3 x2 1− .
15. Find, correct to 2 decimal places,
the area enclosed by the curve
log y x2
= , the x-axis and the
lines 1x = and 3x = by using
Simpson’s rule with 3 functionvalues.
8/10/2019 Chapter Four of Maths in Focus
http://slidepdf.com/reader/full/chapter-four-of-maths-in-focus 31/33
190 Maths In Focus Mathematics Extension 1 HSC Course
Test Yourself 41. Evaluate to 3 significant figures.
(a) e 12−
(b) log 9510
(c) log 26e
(d) 7log4
(e) 3log4
(f) ln50
(g) 3e +
(h) ln
e
4
5 3
(i) eln 6
(j) eln 2
2. Differentiate
(a) e x5
(b) e2 x1−
(c) 4log xe
(d) ( )ln x4 5+
(e) xex
(f)ln
x
x
(g) 10( )e 1x
+
3. Find the indefinite integral (primitive
function) of
(a) e x4
(b)x
x
92−
(c) e x−
(d)4
1
x +
4. Find the equation of the tangent to the
curve y e2 x3
= + at the point where 0x = .
5. Find the exact gradient of the normal to
the curve y x e x
= −− at the point where
2x = .
6. Find the exact area bounded by the curve
y e x2
= , the x-axis and the lines 2x = and
5x = .
7. Find the volume of the solid formed if
the area bounded by y e x3
= , the x-axis
and the lines 1x = and 2x = is rotated
about the x-axis.
8. If .log 2 0 367
= and .log 3 0 567
= , find the
value of
(a) log 67
(b) log 87
(c) .log 1 57
(d) log 147
(e) .log 3 57
9. Find the area enclosed between the curve
ln y x= , the y -axis and the lines 1 y = and
3 y = .
10. (a) Use Simpson’s rule with 3 function
values to find the area bounded by the
curve ln y x= , the x-axis and the lines
2x = and 4x = .
(b) Change the subject of ln y x= to x.
(c) Hence find the exact area in part (a).
11. Solve
(a) 3 8x
=
(b) 2 3x3 4
=−
(c) log 81 4x
=
(d) log x 26
=
(e) e12 10 . t 0 01
=
12. Evaluate
(a) e dx3 x2
0
1
(b)x
dx
3 21 −
4
(c)x
x x xdx
2 5 33 2
1
2 − + + #
13. Find the equation of the tangent to the
curve y ex
= at the point ( , )e4 4 .
14. Evaluate log 89
to 1 decimal place.
8/10/2019 Chapter Four of Maths in Focus
http://slidepdf.com/reader/full/chapter-four-of-maths-in-focus 32/33
8/10/2019 Chapter Four of Maths in Focus
http://slidepdf.com/reader/full/chapter-four-of-maths-in-focus 33/33
192 Maths In Focus Mathematics Extension 1 HSC Course
14. Use Simpson’s rule with 3 function
values to find the area enclosed by thecurve , y e
x2= the y -axis and the line 3 y = ,
correct to 3 significant figures.
15. Find the derivative of .log
e
x x
x
e
16. Use the substitution u x3 12
= + to find
xe dxx3 1
2+ .
17. If y e ex x
= + − , show
dx
d y y
2
2
= .
18. Use the substitution u x 23
= − to
evaluate x e dxx2 2
0
3−
2 # .
19. Provedx
d y
dx
dy y 4 5 10 0
2
2
− − − = , given
y e3 2x5
= − .
20. Find the equation of the curve that has
( )f x e12 x2=m and a stationary point at
( , )0 3 .
21. Sketch ( )log y x xe
2= − .
22. A curve has xedx
dy x2
= and passes through
the point ,021c m . Use the substitution
u x2
= to find the equation of the curve.