1
Exponential functions and
logarithms
2
Overview
A. Exponential functions
1. Example: the function y=2x
2. Exponential function versus power function
B. Exponential growth
C. Exponential decrease
D. Logarithms
E. Some rules for calculations with logarithms
F. Simple exponential equations
G. More complicated exponential equations
3
Example: the function y=2x
Table
Graph
x
y
-4
2-4=1/24=1/16=0.0625
-3
2-3=1/23=1/8=0.125
-2
2-2=1/22=1/4=0.25
-1
2-1=1/2=0.5
0
20=1
0.25
0.5
0.75
20.25 = 21/4 =
4
2=1.1892…
20.5 = 21/2 = 2 =1.4142…
20.75 = 23/4 =
4
23 =1.6817…
1
21=2
2
22=4
3
23=8
4
24=16
4
Exponential function versus power function
x is the exponent
y=2x describes an
exponential function
x is the base
y=x2 describes a
(quadratic function),
power function
A power function is a function having an equation of
the form y=xr (where r is a real number), i.e. x
serves as the base.
An exponential function is a function having an
equation of the form y=bx (where b is a positive
number distinct from 1), i.e. x is the exponent.
5
Overview
A. Exponential functions
B. Exponential growth
1. Example: a growing capital
2. Exponential growth
3. Exercise: growth percentage and growth factor
C. Exponential decrease
D. Logarithms
E. Some rules for calculations with logarithms
F. Simple exponential equations
G. More complicated exponential equations
6
Example: a growing capital
An amount of 1000 EUR is invested in a savings account yielding
3% of compound interest each year. Express the amount A in the
savings account in terms of the time t (in years, starting from the
time of the investment).
in the beginning: 1000 EUR
each year: + 3% (of the preceding value)
general formula???
t=1: A=1000+0.031000=1000+30=1030
t=2: A=1030+0.031030=1030+30.9=1060.9
t=3: A=1060.9+0.031060.9=1060.9+31.82…=1092.72…
t=4: A=1092.72…+0.031092.72…=1092.72…+32.78…=1125.50…
t=5: A=1125.50…+0.03 1125.50…=1125.50…+33.76…=1159.27…
7
Example: a growing capital
An amount of 1000 EUR is invested in a savings account yielding 3% of
compound interest each year. Express the amount A in the savings
account in terms of the time t (in years, starting from the time of the
investment).
t=1: A=1000+0.031000=1000+30=1030
A=1000+0.031000=1000(1+0.03)=10001.03=1030
t=2: A=1030+0.031030=1030+30.9=1060.9
A=1030+0.031030=1030(1+0.03)=10301.03
=10001.031.03=10001.032(=1060.9)
t=3: A=1060.9+0.031060.9=1060.9+31.82…=1092.72…
A=1060.9+0.031060.9=1060.9(1+0.03)=1060.91.03
=10001.031.031.03 =10001.033(=1092.72…)
each year ×1.03
A=10001.03t
8
Example: a growing capital
An amount of 1000 EUR is invested in a savings account yielding
3% of compound interest each year. Express the amount A in the
savings account in terms of the time t (in years, starting from the
time of the investment).
‘each year: +3%’ corresponds to
‘each year ×1.03’ (1.03=1+3/100)
we will use this formula
also if t is not an integer
A=10001.03t=
3 

1000  1 

 100 
multiple of an
exponential function!
t
graph has J-form
9
Example: a growing capital
An amount of 1000 EUR is invested in a savings account yielding
3% of compound interest each year. Express the amount A in the
savings account in terms of the time t (in years, starting from the
time of the investment).
yearly growth percentage=3%
initial value=1000
A=10001.03
3 

1000  1 

 100 
t
t=
growth
factor
= 1.03
growth
factor
graph has J-form
10
Exponential growth
cf. examples in parts A and B
• A variable y grows exponentially iff y=y0bt
(y0: initial value; b growth factor (b>0, b≠1))
• If y increases by p% every time unit
(p: growth percentage), then
cf. example in part B
♦ y grows exponentially
♦ growth factor is
♦ the equation is
p
b 1
100
p 

y  y0  1 

 100 
♦ the graph has J-form
t
11
Exercise: growth percentage/factor
growth percentage
(+ …% each time unit)
growth factor
(×… each time unit)
+5%
×1.05
+50%
×1.5
+0.5%
×1.005
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Overview
A. Exponential functions
B. Exponential growth
C. Exponential decrease
1. Example: decreasing population
2. Exponential increase/decrease
3. Exercise: growth percentage/factor (continued)
D. Logarithms
E. Some rules for calculations with logarithms
F. Simple exponential equations
G. More complicated exponential equations
13
Example: decreasing population
A small village had 1000 inhabitants on 1 Jan. 1950, but since
then its population decreased by 3% each year. Express the
population N in terms of the time t (in years, starting from 1 Jan.
1950).
t=1: N=1000-0.031000=1000(1-0.03)=10000.97=970
t=2: N=970-0.03970=970(1-0.03)
=10000.970.97=10000.972
t=3: N=940.9-0.03940.9=940.9(1-0.03)
=10000.973
N=10000.97t
graph has
reflected
J-form
14
Exponential increase/decrease
• If y decreases by p% every time unit
(negative growth percentage), then
♦ y grows exponentially
♦
♦
p
growth factor is <1: b  1 
100
t
p 

the equation is y  y0  1 

 100 
♦ the graph has reflected J-form
• An exponential function y=bx is
• increasing if b>1
• decreasing if b<1
cf. example
15
Exercise: growth percentage/growth (ctd.)
growth percentage
(+ …% each time unit)
growth factor
(×… each time unit)
+5%
×1.05
+50%
×1.5
+0.5%
×1.005
–5%
×0.95
–50%
×0.5
–0.5%
×0.995
+100%
×2
+1000%
×11
16
Overview
A. Exponential functions
B. Exponential growth
C. Exponential decrease
D. Logarithms
1. Example
2. Logarithms
3. Logarithms using the calculator
E. Some rules for calculations with logarithms
F. Simple exponential equations
G. More complicated exponential equations
17
Example
Find x such that …
10  x  1000
x  1000  10  990
10  x  1000
1000
x
 100
10
10 x  1000
x
in words: which exponent
do you need to obtain 1000
when the base of the
power is 10?
log1000  3
3 is the (common) logarithm (or
logarithm base 10) of 1000
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Logarithms
in words: log x is the exponent needed to
make a power with base 10 equal to x
(common) logarithm (logarithm base 10) of x:
log x = y iff 10y = x
Calculate the following logarithms (without calculator)
log100
log1 000 000
log 0.001
log10
log1
log 100
log 0
10  100
?
!
102 100
log100  2
log1 000 000  6
log 0.001  3
log10  1
log1  0
undefined
undefined
19
Logarithms using the calculator
Calculate the following logarithms
and verify the result
log 2
log 3
log 4
log 5
log 6
log8
log 9
log 20
log 3000
0.301 029 ...
0.477 121 ...
0.602 059 ...
0.698 970 ...
0.778 151 ...
0.903 089 ...
0.954 242 ...
1.301 029 ...
3.477 121 ...
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Overview
A. Exponential functions
B. Exponential growth
C. Exponential decrease
D. Logarithms
E. Some rules for calculations with logarithms
1. Logarithm of a product
2. Logarithm of a quotient
3. Logarithm of a power
F. Simple exponential equations
G. More complicated exponential equations
21
Logarithm of a product
log 2
log 3
log 4
log 5
log 6
log8
log 9
log 20
log 3000
0.301 029 ...
0.477 121 ...
0.602 059 ...
0.698 970 ...
0.778 151 ...
0.903 089 ...
0.954 242 ...
1.301 029 ...
3.477 121 ...
log 6  log 2  log 3
log10  log 2  log 5
log 20  log 2  log10
log 3000  log1000  log 3
log 3000
3000
||
3 0.477...
10
 1000 
||
!

103
3
||
 100.477...
log1000  log 3
Logarithm of a product: loga  b   log a  log b
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Logarithm of a quotient
log 2
log 3
log 4
log 5
log 6
log8
log 9
log 20
log 3000
0.301 029 ...
0.477 121 ...
0.602 059 ...
0.698 970 ...
0.778 151 ...
0.903 089 ...
0.954 242 ...
1.301 029 ...
3.477 121 ...
log
log636log
log62log
log33
log
 log10  log 5
log10
10
5  log 2  log 5
20
log 20
 log 20  log10
10  log 2  log10
log 3000  log1000  log 3
log 3000  log 3  log1000
log1000  log 3000  log 3
log 3000
 log 3000  log 3
3
 log
 alog
Logarithm of a product:
quotient: log
logbaa blog
a blog b
23
Logarithm of a power
log 2
log 3
log 4
log 5
log 6
log8
log 9
log 20
log 3000
0.301 029 ...
0.477 121 ...
0.602 059 ...
0.698 970 ...
0.778 151 ...
0.903 089 ...
0.954 242 ...
1.301 029 ...
3.477 121 ...
log22  log 4  2 log 2
log23  log8  3 log 2
log32  log9  2 log3
!
log3  3  log 3  log 3
Logarithm of a power: log a r  r log a
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Overview
A. Exponential functions
B. Exponential growth
C. Exponential decrease
D. Logarithms
E. Some rules for calculations with logarithms
F. Simple exponential equations
1. Example: a growing capital
2. Solution procedure
G. More complicated exponential equations
25
Example: a growing capital
An amount of 1000 EUR is invested in a savings account yielding
3% of compound interest each year. Express the amount A in the
savings account in terms of the time t (in years, starting from the
time of the investment).
When will the amount in the savings account be equal to 1500 EUR?
A=10001.03t
t? such that A=1500
exponential equation:
unknown is in the exponent
1000 1.03t  1500 (divide by 1000)
1.03t  1.5
(take logarithm of both sides)
t
log(1.03 )log(1.5)
t  log1.03  log1.5
log1.5
t
 13.7...
log1.03
(apply log a r  r log a )
Answer: After about 13.7…
years, the amount is equal to
1500 EUR.
26
Solution procedure
• An exponential equation is an equation in which
the unknown appears at least once in an
exponent.
• To solve an exponential equation of the form 𝑎 ∙
𝑏 𝑥 = 𝑐 (where a, b and c are positive numbers)
♦ divide both sides by a: 𝑏 𝑥 =
𝑐
𝑎
♦ take the logarithm of both sides:
log 𝑏 𝑥
=
𝑐
log
𝑎
♦ apply rule for logarithm of a power: 𝑥 ∙ log 𝑏 = log
𝑐
♦ divide both sides by log 𝑏 : 𝑥 =
log𝑎
log 𝑏
𝑐
𝑎
memorize
procedure,
not formulas!!!
27
Overview
A. Exponential functions
B. Exponential growth
C. Exponential decrease
D. Logarithms
E. Some rules for calculations with logarithms
F. Simple exponential equations
G. More complicated exponential equations
1. Example: two growing capitals
2. Solution procedure
28
Example: two growing capitals
Ann invests an amount of 1000 EUR in a savings account yielding
3% of compound interest each year. John invests 900 EUR in a
savings account yielding 3.5% of compound interest each year.
When will they have the same amount in their savings account?
A=10001.03t
J=9001.035t
t? such that A=J
1000 1.03t  900 1.035t
1.03t
900

t
1.035 1000
t
900
 1.03 

 
 1.035  1000
t
900
 1.03 
  log
log 
1000
 1.035 
1.03
900
t  log
 log
1.035
1000
900
log
1000
t
1.03
log
1.035
(
)
(
)
29
Example: two growing capitals
Ann invests an amount of 1000 EUR in a savings account yielding
3% of compound interest each year. John invests 900 EUR in a
savings account yielding 3.5% of compound interest each year.
When will they have the same amount in their savings account?
A=10001.03t
J=9001.035t
900
log
1000
t
1.03
log
1.035
t  21.7...
Answer: It takes nearly 22 years before the two amounts
are equal.
30
Solution procedure
• To solve an exponential equation of the form
𝑥
𝑥
𝑎1 ∙ 𝑏1 = 𝑎2 ∙ 𝑏2 (where 𝑎1 , 𝑎2 , 𝑏1 and 𝑏2 are
positive numbers)
♦ divide both sides by 𝑎1 : 𝑏1
♦ divide both sides by 𝑏2
𝑥
:
𝑥
𝑏1
𝑏2
=
𝑎2
𝑎1
𝑥
𝑥
=
∙ 𝑏2
𝑥
𝑎2
𝑎1
♦ apply rule for quotient of powers:
𝑏1 𝑥
𝑏2
=
𝑎2
𝑎1
♦ finally, apply the procedure for simple exponential
equations
memorize the procedure, not the
formulas!!!