www.mathsrevision.com Higher Outcome 3 Higher Unit 3 Higher Unit 3 www.mathsrevision.com www.mathsrevision.com Exponential & Log Graphs Special “e” and Links between Log and Exp Rules for Logs Exam Type Questions Solving Exponential Equations Experimental & Theory Harder Exponential & Log Graphs
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Higher Outcome 3 Higher Unit 3 Exponential & Log Graphs Special “e” and Links between Log and Exp Rules for.
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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
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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
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 ….