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Higher Outcome 3
Higher Unit 3Higher Unit 3
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Exponential & Log Graphs
Special “e” and Links between Log and ExpRules for Logs
Exam Type Questions
Solving Exponential Equations
Experimental & Theory
Harder Exponential & Log Graphs
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Higher Outcome 3
The Exponential & Logarithmic Functions
Exponential Graph Logarithmic Graph
y
x
y
x
(0,1)
(1,0)
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Higher Outcome 3
The letter e represents the value 2.718….. (a never ending decimal).
This number occurs often in nature
f(x) = 2.718..x = ex is called the exponential function
to the base e.
A Special Exponential Function – the “Number” e
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Higher Outcome 3
y
x
( ) 2xf x
1( )f x (0,1)
(1,0)
In Unit 1 we found that the
exponential function has an
inverse function, called the
logarithmic function.log 1 0
log 1
log
a
a
xa
a
y a x y
2log x
The log function is the inverse of the exponential function, so it ‘undoes’ the
exponential function:
Linking the Exponential and the Logarithmic Function
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Higher Outcome 3
f (x) = 2x
ask yourself :
1 2 21 = 2 so log22 = 1 “2 to what power gives 2?”
2 4 22 = 4 so log24 = “2 to what power gives 4?”
3 8 23 = 8 so log28 = “2 to what power gives 8?”
4 16 24 = 16 so log216 = “2 to what power gives 16?”
f (x) = log2x
2
34
Linking the Exponential and the Logarithmic Function
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Higher Outcome 3f (x) = 2x
ask yourself :
1 2 21 = 2 so log22 = 1 “2 to what power gives 2?”
2 4 22 = 4 so log24 = “2 to what power gives 4?”
3 8 23 = 8 so log28 = “2 to what power gives 8?”
4 16 24 = 16 so log216 = “2 to what power gives 16?”
f (x) = log2x
234
Examples(a) log381 = “ to what power gives ?”(b) log42 = “ to what power gives ?”
1
27
(c) log3 = “ to what power gives ?”
4 3 81
4 2
-3 3
Linking the Exponential and the Logarithmic Function
1 2
1
27
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Higher Outcome 3
log log loga a axy x y
log log loga a a
xx y
y
log logpa ax p x
Rules of Logarithms
Three rules to learn in this section
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Higher Outcome 3Examples
Simplify:
a) log102 + log10500 b) log363 – log37
10log (2 500)
10log 1000
3
3
63log
7
3log 9
2
Rules of Logarithms
310log (10)
103 log 10
Sincelog logna aa n a
32 log 3
23log (3)
Sincelog logna aa n a Since
log 1a a
Sincelog 1a a
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Higher Outcome 3Example
2 2
1 1log 16 log 8
2 3Simplif y
11
322 2log (16) log (8)
2 2log 4 log 2
2
4log
2
1
Sincelog 1a a
Rules of Logarithms
2log 2
Since
log log na an b b
log log log
a a a
xx y
y
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Higher Outcome 3
You have 2 logarithm buttons on your calculator:
which stands for log10
which stands for loge
log log10x
lnxe
ln
Try finding log10100 on your calculator 2
Using your Calculator
and its inverse
and its inverse
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Higher Outcome 1
Logarithms & Exponentials
We have now reached a stage where trial and error is no longer required!
Solve ex = 14 (to 2 dp)
ln(ex) = ln(14)
x = ln(14)x = 2.64
Check e2.64 = 14.013
Solve ln(x) = 3.5 (to 3 dp)
elnx = e3.5
x = e3.5
x = 33.115
Check ln33.115 = 3.499
April 18, 2023April 18, 2023 www.mathsrevision.comwww.mathsrevision.com
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Higher Outcome 1
Solve 3x = 52 ( to 5 dp )ln3x =
ln(52)xln3 = ln(52) (Rule 3)
x = ln(52) ln(3)
x = 3.59658
Check: 33.59658 = 52.0001…. April 18, 2023April 18, 2023 www.mathsrevision.comwww.mathsrevision.com
Logarithms & Exponentials
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Higher Outcome 3
Solve 5 11x 51 = 5 and 52 = 25
so we can see that x lies between 1 and 2Taking logs of both sides and applying the rules
10 10log 5 log 11x
10 10log 5 log 11x 10
10
log 111.489
log 5x
Solving Exponential Equations
Since
log logna ab n b
Example
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Higher Outcome 3
For the formula P(t) = 50e-2t:
a) Evaluate P(0)
b) For what value of t is P(t) = ½P(0)?
2 0(0) 50 50P e
Solving Exponential Equations
(a)
Remember
a0 always equals 1
Example
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Higher Outcome 3
For the formula P(t) = 50e-2t:
b) For what value of t is P(t) = ½P(0)?
1 1(0) 50 25
2 2P
225 50 te21
2te
21ln ln
2te
Solving Exponential Equations
0.693 2 lnt e
0.693 2 1t
0.693
2
t
0.346t
ln = loge e
logee = 1Example
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Higher Outcome 3
The formula A = A0e-kt gives the amount of a radioactive substance after time t minutes. After 4 minutes 50g is reduced to 45g.
(a) Find the value of k to two significant figures.
(b) How long does it take for the substance to
reduce to half it original weight?
Example
(a)
4t (0) 50A (4) 45A
Solving Exponential Equations
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Higher Outcome 3
(a) 4t (0) 50A (4) 45A
445 50 ke
4ln(45) ln 50 ke
4ln(45) ln 50 ln ke4ln 45 ln 50 ln ke
Solving Exponential Equations
Example log log log a a axy x y
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Higher Outcome 3
4ln 45 ln 50 ln ke
45ln 4 ln
50k e
0.1054 4k 0.1054
4
k
0.0263k
Solving Exponential Equations
log log log
a a a
xx y
y
ln = loge e
logee = 1
Example
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Higher Outcome 3
(b) How long does it take for the substance to
reduce to half it original weight?0.02631
(0) (0)2
tA A e 0.02631
2 te
0.02631ln ln
2
te
0.693 0.0263 lnt e
0.693 0.0263 t
Solving Exponential Equations
ln = loge e
logee = 1
Example
26.35 minutes t
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Higher Outcome 3
When conducting an experiment scientists may analyse the data to find if a formula connecting the
variables exists.
Data from an experiment may result in a graph of the form shown in the
diagram, indicating exponential growth. A
graph such as this implies a formula of the type y =
kxn
Experiment and Theory
y
x
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Higher Outcome 3
log k
We can find this formula by using logarithms: ny kxIf
Then
log log ny kx
So log klog nx
log y log n x
Compare this to
Y mX c
log y Yn m
log k c
SoIs the equation
of a straight line
Experiment and Theory
log log log y n x k
log y
log y
log x(0,log k)
log x X
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Higher Outcome 3
Experiment and Theory
From ny kxWe see by taking logs that
we can reduce this problem to a straight line
problem where:
And
log y
log x
(0,log k)
log log log y n x k
Y m X c= +Y X cm
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Higher Outcome 3
ln(y)
ln(x)
m = 50.69
Express y in terms of x.
NB: straight line with
gradient 5 and intercept 0.69Using Y = mX + cln(y) = 5ln(x) + 0.69
ln(y) = 5ln(x) + ln(e0.69)
ln(y) = 5ln(x) + ln(2)ln(y) = ln(x5) + ln(2) ln(y) = ln(2x5)
y = 2x5
Experiment and Theory
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Higher Outcome 1
log10y
log10
x1
0.3
Find the formula connecting x and y.
straight line with intercept 0.3
Using Y = mX + c
Taking logs
log10y = -0.3log10x + 0.3
log10y = -0.3log10x + log10100.3
log10y = -0.3log10x + log102 log10y = log10x-0.3 + log102 log10y = log102x-
0.3
y = 2x-0.3
m = -0.3/1 = -0.3
Experiment and Theory
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Higher Outcome 1
Experimental Data
When scientists & engineers try to find relationships between variables in experimental data the figures are often very large or very small and drawing meaningful graphs can be difficult. The graphs often take exponential form so this adds to the difficulty.
By plotting log values instead we often convert from
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Higher Outcome 3
The variables Q and T are known
to be related by a formula in the form
The following data is obtained from experimenting
Q 5 10 15 20 25
T 300 5000 25300 80000 195300
Plotting a meaningful graph is too difficult so taking log values instead we get ….
log10Q 0.7 1 1.2 1.3 1.4
log10T 2.5 3.7 4.4 4.9 5.3
T = kQn
log10Q
log10T
m = 5.3 - 2.5 1.4 - 0.7
= 4 Point on line (a,b) = (1,3.7)
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Higher Outcome 3
Since the graph does not cut the y-axis use
Y – b = m(X – a)
where X = log10Q and Y = log10T ,
log10T – 3.7 = 4(log10Q – 1)
log10T – 3.7 = 4log10Q – 4
log10T = 4log10Q – 0.3log10T = log10Q4 – log10100.3
log10T = log10Q4 – log102log10T = log10(Q4/2) T = 1/2Q4
Experiment and Theory
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Higher Outcome 3Example
The following data was collected during an experiment:
a) Show that y and x are related by the formula y = kxn
.
b) Find the values of k and n and state the formula that
connects x and y.
X 50.1 194.9 501.2 707.9
y 20.9 46.8 83.2 102.3
Experiment and Theory
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Higher Outcome 3
a) Taking logs of x and y and plotting points we get:
0.5
0.5
1
1
1.5
1.5
2
2
2.5
2.5
3
3
0.5
0.5
1
1
1.5
1.5
2
2
2.5
2.5
3
3
log10 Y
log10 X
Since we get a straight line the
formula connecting X and Y is of the
formY mX c
X 50.1 194.9 501.2 707.9
y 20.9 46.8 83.2 102.3
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Higher Outcome 3
log log log y n x k
b) Since the points lie on a straight line, formula is of the form:
ny kx
Graph has equation
Compare this to
Y mX c
Experiment and Theory
Selecting 2 points on the graph and substituting them into the straight line equation we get:
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Higher Outcome 3
1.69,1.32 2.84,2.00
1.32 1.69 (B)m c 2.00 2.84 (A)m c
0.68 1.15m
0.68
1.15m
0.6m
1.32 1.69 0.6 c Sub in B to find value of
c
0.3c
Experiment and Theory
Sim. Equations Solving we
get
The two points picked are and
( any will do ! )
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Higher Outcome 3
So we have
0.6 0.3Y X
Compare this to log log log y n x k
n and log k 0.6 0.3so
Experiment and Theory
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Higher Outcome 3
solving
0.310k
log 0.3k
1.99k
so 0.61.99ny kx y x
You can always check this on your
graphics calculato
r
Experiment and Theory
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Higher Outcome 3
Transformations of Log & Exp Graphs
In this section we use the rules from Unit 1 Outcome 2
Here is the graph of y = log10x.
Make sketches of
y = log101000x and y = log10(1/x) .
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Higher Outcome 3
log101000x = log101000 + log10x
= log10103 + log10x = 3 +
log10xIf f(x) = log10x
then this is f(x) + 3
y = log101000x
Graph Sketching
(10,1)
(1,0)
(1,3)(10,4)
y = log10x
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Higher Outcome 3
Graph Sketching
log10(1/x) = log10x-1= -log10x
If f(x) = log10x -f(x) ( reflect in x - axis )
(1,0)(10,1)
(10,-1)
y = log10x
y = -log10x
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Higher Outcome 3
Here is the graph of y = ex
y = ex
Sketch the graph of
y = -e(x+1)
Graph Sketching
(0,1)
(1,e)
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Higher Outcome 3
If f(x) = ex
reflect in x-axis move 1 left
y = -e(x+1)
Graph Sketching
(-1,1)
(0,-e)
-e(x+1) = -f(x+1)
Revision
Logarithms & ExponentialsHigher Mathematics
www.maths4scotland.co.uk
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Logarithms Revision
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Reminder
All the questions on this topic will depend
upon you knowing and being able to use,
some very basic rules and facts.
Click to show
When you see this button
click for more information
Logarithms Revision
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Three Rules of logs
log log loga a ax y xy
log log loga a a
xx y
y
log logpa ax p x
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Two special logarithms
log 1a a
log 1 0a
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Relationship between log and exponential
log ya x y a x
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Graph of the exponential function
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Graph of the logarithmic function
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Related functions of
( )
( )
( )
( )
( )
( )
y f x a
y f x a
y f x
y f x
y f x a
y f x a
Move graph left a units
Move graph right a units
Reflect in x axis
Reflect in y axis
Move graph up a unitsMove graph down a units
( )y f x
Click to show
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Calculator keys
ln = loge
log = 10log
Logarithms Revision
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Calculator keys
ln=log 2.5e 2 . 5 = = 0.916…
log=10log 7.6 7 . 6 = = 0.8808…
Click to show
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Solving exponential equations
2.4 3.1 xe
Show
log 2.4 log 3.1 xe e eTake loge both sides
log 2.4 log 3.1 log xe e e e
log 2.4 log 3.1 loge e ex e Use log ab = log a + log b
log 2.4 log 3.1e ex
Use log ax = x log a
Use loga a = 1 log 2.4 log 3.1e e x
0.25593... (2dp)0.26
Logarithms Revision
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60 80 keSolving exponential equations
log 60 log 80 ke e eTake loge both sides
log 60 log 80 log ke e e e
log 60 log 80 loge e ek e
Use log ab = log a + log b
log 80 log 60e ek
Use log ax = x log a
Use loga a = 1 log 60 log 80e e k
0.2876... (2dp)0.29
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Solving logarithmic equations
3log 0.5y 0.53y Change to exponential form
Change to exponential form1
21
2
1 13
33
y
(2dp)0.577.... 0.58y
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3log (2 ) 2 log (3 )e ee elog loge eA B C
Simplify
expressing your answer in the form where A, B and C are whole numbers.
3 2log (2 ) log (3 )e ee e 3 2log 8 log 9e ee e
3
2
8log
9ee
e
8log
9e
e
log 8 log log 9e e ee 1 log 8 log 9e e
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5 5 5log 2 log 50 log 4 Simplify
5
2 50log
4
5log 25
25log 5 52 log 5 2
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Find x if 4 log 6 2 log 4 1x x
4 2log 6 log 4 1x x 4
2
6log 1
4x
36log x
936
9
41
41 1 log 81 1x
1 81x 81x
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5 5log 3 log 4x Given find algebraically the value of x.
5log 3 4x 5log 12x
5 12x
10 10log 5 log 12x 10 10log 5 log 12x
10
10
log 12
log 5x 1.5439..x (2dp)1.54x
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Find the x co-ordinate of the point where the graph of the curve
with equation 3log ( 2) 1y x intersects the x-axis.
When y = 0 30 log ( 2) 1x
31 log ( 2)x
Exponential form
Re-arrange
13 2x
12 3x Re-arrange1
32x
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The graph illustrates the law ny kx
If the straight line passes through A(0.5, 0)
and B(0, 1). Find the values of k and n.
5 5log log ny kx
5 5 5log log logy k n x
5logY k nX
Gradient1
0.5 y-intercept 1
5log 1k 5k
(gradient)2n
Show
is the area covered by the fire when it was first detected
and A is the area covered by the fire t hours later.
If it takes one and a half hours for the area of the forest fire to double,
find the value of the constant k.
Logarithms Revision
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Before a forest fire was brought under control, the spread of fire was described by a law of the form
0 ktA A e where
0A
1.50 02 kA A e 1.52 ke log 2 1.5 loge ek e
log 2 1.5e k log 2
1.5ek (2 dp)0.462.. 0.46k
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The results of an experiment give rise to the graph shown.
a) Write down the equation of the line in terms of P and Q.
It is given that logeP p and logeQ q
bp aq stating the values of a and b.
b) Show that p and q satisfy a relationship of the form
log log be ep aq
log log loge e ep a b q
logeP a bQ Gradient 0.6 y-intercept 1.8
0.6 1.8P Q 1.8 6.0o .8 5l g 1e a ea a
0.6b
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The diagram shows part of the graph of
log ( )by x a
.Determine the values of a and b.
...(1)1 log (7 )b a Use (7, 1)
...(2)0 log (3 )b a Use (3, 0)
Hence, from (2) 3 1a 2a
and from (1) 5b1 log 5b
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The diagram shows a sketch of
part of the graph of 2log ( )y x
a) State the values of a and b.
b) Sketch the graph of 2log ( 1) 3y x
1a 3b
Graph moves
1 unit to the left and 3 units down
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a) i) Sketch the graph of 1, 2xy a a
ii) On the same diagram, sketch the graph of 1, 2xy a a
b) Prove that the graphs intersect at a point where the x-coordinate is 1
log1a a
11x xa a 11 x xa a 1 1xa a
log 1 log log 1xa a aa a 0 log 1ax a
log 1ax a 1log 1ax a
1
log1ax
a
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Part of the graph of
105 log (2 10)y x
is shown in the diagram.
This graph crosses the x-axis at
the point A and the straight line 8y at the point B. Find algebraically the x co-ordinates of A and B.
108 5 log (2 10)x 10
8log (2 10)
5x 101.6 log (2 10)x
1.610 2 10x 1.610 10 2x 14.9x B (14.9, 8)
100 5 log (2 10)x 2 10 1x 4.5x A ( 4.5, 0)
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The diagram is a sketch of part of the graph of , 1xy a a
a) If (1, t) and (u, 1) lie on this curve, write down the values of t and u.
b) Make a copy of this diagram and on it sketch the graph of
2xy ac) Find the co-ordinates of the point of intersection of
2xy a
with the line 1x
a) t a 0u b)
c)2 1y a 2y a 21, a
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The diagram shows part of the graph with equation 3xy and the straight line with equation 42y
These graphs intersect at P.
Solve algebraically the equation 3 42x and hence write down, correct to 3 decimal places, the co-ordinates of P.
10 10log 3 log 42x 10 10log 3 log 42x
10
10
log 42
log 3x 3.40217...x
P (3.402, 42)
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Higher Outcome 3
Are you on Target !
• Update you log book
• Make sure you complete and correct
ALL of the Logs and Exponentials questions in the past paper
booklet.
