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Logarithms and Exponentials
8.1 Definition of a Logarithm
2
8.2 Laws of Logarithms
3
8.3 The ExponentialFunction
5
8.4 FurtherExamples
7
FurtherExercises
10
Answersto Exercises
11
@ J.Carr/HWU/1996
.
Logarithmsand Exponentials
Definition of a Logarithm
Page2
"
8.1 Definition of a Logarithm
If a and b arepositive numbers,thenwe canalwaysfind x suchthat ax = b.
For example,if a = 10 and b = 1000, then x = 3, since 103= 1000. The number
x is calledthe logarithmof b to base a, andwe write x = loga(b).
If b
~
~
a' then x
~
log.(b)
I
As examples: 100= 102 so loglo(100) = 2
0.1 = 10-:1 so 10glo(0.1)= -I
If the baseof the logarithmof b is 10 thenwe write this as log(b).
Example 8.1 Use a calculator to find (i)
log(2) (ii) log(21.6) (iii) log(0.3).
Solution:
Using a calculator we obtain:
(i) log(2) = 0.301 (ii) log(21.6)= 1.334 (iii) log(0.3)= --0.523 .
Useyour calculatorto checktheseresults.
Example8.2 Find log(O.O1) withoutusinga calculator.
Solution:
0.01 = 10-2, log(O.OI)= -2.
Since
Most calculators have two different logarithm functions; base 10 logarithms and
logarithms to base e. The letter e standsfor the number 2.7182... We write In(b)
for loge(b). On a calculator, logarithms to base e are usually denoted by In or loge'
Example 8.3 Use your calculator to find (i)
.
Solution:
In(IO) (ii) In(7t) (iii) In(5).
Usingthe In functionon the calculator,
In( 10)
In(5)
=
Example 8.4 Show that
Solution:
If
c~;
..'"-:~:"
1.609.
loga(a) = I.
loga(a) = x, then ax = a.
Sinceal = a, x = I .
Exercises 8.1
I.
2.
;:'~~
:~~::!
= 2.303
In(7t) = 1.145
(i)
(ii)
(iii)
Without using a calculator, find log(b) when:
,
"C~
c'
(i) b = 10-7
(ii) b = 10.
Use a calculator to find correct to two decimal places: (i) log(5)
(ii) In(7).
'2-,
,.
\
Logarithmsand Exponentials
Lawsof Logarithms
Page3
8.2 Laws of Logarithms
In a previous unit, we found a set of rules for indices. There is a similar set of laws
for logarithms. Below we state these rules and show how to use them. The
explanation of the rules is given at the end of this section.
Let B and C be positive numbers. Then for any base a:
1
loga(BC) = loga(B) + loga(C)
2
lOga(~) = loga(B) -loga(C)
3
loga(B)n = nloga(B)
4
loga(l)
=
0
Example8.5 Simplifyto a singlelog term
(i)
(ii) log(6) - log(3) + 3log(2)
log(7) + log(4)
Solution:
log(7) + log(4) = log(7 x 4)
= log(28)
(i) Using rule 1,
;;?
log(6) - log(3) + 3log(2) = log(6/3) + 310g(2)
(ii) Using rule 2,
= log(2) + 3log(2)
= 4log(2)
= log(24)
By rule 3
= log(16).
Example 8.6 Express 10g(A2B4) in terms of
:
log(A) and log(B).
"'"
Solution:
Using rule 1:
Using rule 3:
10g(A2B4)
=
log(A 2) + log(B4)
= 2log(A) + 410g(B)
Example8,7 Find the valueof x whichsatisfies (l.l)X = 100
Solution:
(l.l)X
.
100 so 10g(1.1)X= log(lOO)
Using rule 3:
xlog(l.l)
= log(lOO)
so x
=
=
log(lOO)
log(l.l)
2
=
= 48.32
using a calculator.
0.04l4
Note that we could also have taken logarithms to base e :
. . the sameresult as before.
x = In(lOO) = 46.052
0
3 = 48.32 giVIng
In(l.l)
.095
.
,
r
!
.
Logarithms
and
Exponentials
Laws
of
Logarithms
Paae
=-
4
Example8.8 If a,b and c arepositivenumbersand bax= c , showthat:
x=
Solution:
In(c / b)
.
InCa) ,
.
bax= c so ax= c/b and In(aX)= In(c/b)
,.,
"!
Thus x InCa) = In(c/b) so x = In(c / b)
InCa)
"""~4;
;
We now showthat rules 1 to 4 hold:
Let p
= 10ga(A) and
q = 10ga(B). Then A = aP and B = aq.
law 1: AB = aPaq= aP+q so 10ga(AB) = p + q = 10ga(A)
+ 10ga(B).
law 2:
A = ;q
aP = aP-q
B
so loga( A ) = p - q = 10ga(A)-loga(B)
B
law 3: An = (aP)"= apn so 10ga(A") = pn = nloga(A) .
law 4: From rulesof indices aD= 1 so loga(1)= 0 .
Exercises8.2
1.
Expressthe following in termsof 10g(A) and 10g(B):
(i) log(AB3)
(ii)
IO~
(i) 3x= 5
~) .
2.
Solvefor x:
(ii) 4x = 7 .
3.
Showthat InCA)+ In(I/A) = 0 .
4.
Find the smallestinteger, n suchthat (1.05)"> 2.
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Logarithmsand Exponentials
The ExponentialFunction
Page5
,8.3 The Exponential
Function
We begin by looking at the graphof y = 2x
x
-2
1/4
2x
-1
1/2
2x
0
1
1
2
2
4
3
8
4
16
16
14
12
10
j
-2
-100
1
2
3
4
x
Figure1 Thegraphof y = 2x .
More generally,if a > 1 thenthe graphof y = aX is similar
a
!
x
'.
,
. f;~~
1
x
Figure2 Thegraphof y = ax .
The function ex where e = 2.7183... is calledthe exponentialfunction. The graphof
y = ex has the same form as shown in figure 2.
On a calculator, the exponential function is usually written as ex or exp. However
the abbreviation exp is sometimes used for other purposes,so you should check
before using a particular calculator.
e3 = 20.0855, e-l = 0.3679
You should use your calculator to check that:
~~:fj,
-".
~'
Logarithms and Exponentials
The ExponentialFunction
Page6
"
,~,t;;::l",:';:"'q;
,..j;~
;:;q,.. ';"fc;;
The laws of indices apply to ex so:
;
x
exeY = ex + Y
=
eo =' 1
ex - Y
(eX)n= enx
e
Example8.9 Simplify
(i) e3x(e2X
- e-2\)
(ii)
(eX+ e-;x)2- (eX- e-x)2
..
Solution:
(i)
e3x(e2x-
e-2x) = e3xe2x
- e3xe-2x= e5x- ex
(ii)
(eX+ e-x)2 = (ex)2 + 2exe-x+ (e-x)2
= e2x+ 2 + e-2x
similarly (eX- e-x)2 = e2x - 2 + e-2x
Usinglawsof indices
so (eX+ e-x)2- (eX- e-X)2 :;: 4..
An alternative notation for ex is exp(x)
i..e.. ex = exp(x).. This is useful in
complicated expressions such as exp(- 4x2 + 3x) ..
Example8.10 Let f(x)
2 exp(- ax/b),
:;:
where a:;: 10 and b:;: 2..5.. Find the value of
f(O) , f(O.l) , f(I).
Solution:
-a :;: -10 :;: 4
b
2.5
so f(x):;: 2e- 4x.. Then:
f(O) = 2eo :;: 2.
Usinga calculator: f(O.I):;: 2e-O.4= 1.34,
f(I):;: 2e-4:;: 0.04.
Exercises 8.3
I.
Find eO.4. Give your answer to 2 decimal places.
2.
Simplify
3.
Show that
4.
If f(x)
places.
eX(ex- e-X).
:;:
(eX - 1)2
=
e2x
-
2ex + 1 . Hence simplify
(e2x- 2ex + 1)1/2.
4[2 - exp(-5x)] find f(O) and f(I). Give your answerto 2 decimal
.
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Logarithms and Exponentials
Further Examples.
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8.4
Further
.'c.-'-"
Page 7
~-~_:,-""',::~~.;
Examples
The function y = ax (a> 1) has the inverse function x = 10gaUr) defined for y> 0 .
Thus
In(eX) = x and exp[ln(y)] = y
log(10X) = x
and
10Iog(y)= y
Example8.11 Find x when: (i) ex = 4 (ii)
log(x) = c.
Solution:
(i) Takinglogarithmsto base e gives x = In(4)
(ii) Since 10Iog(x)= x, X = 10c .
=
1.386 .
Example8.12 If Y = a exp(-bt) , express t in termsof the otherletters.
Solution:
exp(-bt)
In[ exp(-bt)]
-bt
t
= y/a
= In(y/a)
= In(y/a)
= -b-1ln(y/a)
Example8.13 Simplify: (i) In[exp(x1]
Solution:
(i) In[exp(x2)]
(ii)
=
X2
(ii) exp[-3 In(x)]
(iii) exp[2 In(x)] .
.
exp[-3 In(x)] = exp[ln(x-3)] = x-3 .
(iij) 2 Inx = In(x2) so exp[2 In(x)] = exp[ln(x2)] =
X2
.
Example8.14 x and yare relatedby the equation:
y = a( 1 - e- bX) where a and b are constants.
(i) What is the valueof y when x = 0 ?
(ii) Express x in termsof the other letters.
(iii) Use your answerto part (ii) to find x, when a = 10, b = 0.1 and y = 5.
Solution:
(i) When x = 0
y = a( 1 - eo) = a( 1 - 1) = O.
(ii) ya-1 = l-e-bx
so e-bx = l-ya-l
.
Hence -bx = In(1 - ya-l) so x = -b-1ln(1
(iii) When
a=10, b=O.1 and y=5.
Example 8.15
If
4 log(y) = 2 log( x)
- ya-l) .
x = -10In(I/2)
= 6.931.
express y In termsof x .
Solution:
log(y) = -1 1og(x ) --
1og(x 1/2)
so y -- x.1/2
~'Ji'i
2
[I' ~,
,
Logarithmsand Exponentials
FurtherExamples
:..
Page8
'0
--
-.~-
,~
i'
2000
Example8.16 The numberof fish, f(t) in a loch is givenby t1t) =
-kt
1+ 2e
I
w
'-'"-
.
i
where t is the time in years after the first census and k is a positive constant
(i) How many fish were there at the first census?
(ii) If there were 1500 fish after 5 years,find k.
(iii) How many fish will there be after 10 years?
(iv) How many fish will there be long term?
Solution:
(i) First census was taken at t = O. Since eO= 1 :
f(O) =
2000
~
(ii)
=
667
to the nearest significant fish.
2000
1500 =
1+ 2 exp( -5k)
. 1ymg
.
cross mu 1tIp
I + 2exp(-5k) = -2000 = -4
1500
3
2exp(-5k) = -4 - 1 = -1
3
so
exp(-5k)
=
3
I
-
6
-5k = In(I/6)
so
(iii) If t
= 10, then
so
k = 0.358.
kt = 3.58,
f( I 0) =
-
2000
1+ 2 exp(-3.58)
.
2000
1+ (2 x 0.028)
= 1894.
(iv) For large t, exp(-kt) is very small, so f(t) will be closeto -2000
1
= 2000.
+0
To draw the graphof y = loglo(x) we notethe following:
(i) log(x) is definedfor x> 0
(ii) 10g(I)=0
(iii) 10g(10)= 1 , 10g(100)= 2, log(1000)= 3 and in general log(x) increases
as x Increases.
So the graphis shownin figure 3 below.
.
.
Logarithmsand Exponentials
FurtherExamples
Page9
:
\
1
5
10
15
20
x
-1
Figure3 Thegraphof y = loglo(x).
Exercises 8.4
I.
If Y = 1 - exp(-t/a),
express t in terms of the o~er letters.
2.
The instantaneouscharge q in a capacitative circuit is given by:
,
q = Q[ 1- exp(~)]
.
(i) Express t in termsof the otherletters.
(ii) Find the valueof t when q = 0.01,Q = 0.015,c = 0.0001and R = 7000.
3.
The area A (measuredin cm2)occupiedby bacteriaat any time t (measuredin
daysfrom the presenttime) is givenby:
A =.
(i)
(ii)
(iii)
(iv)
4.
h
k IS
. a constant.
49.5' kt ' were
1+197e-
What is the initial areaof bacteria?
If the colony occupies49.21cm2 after 5 days,find the valueof k.
How long will it takefor the colonyto occupy 36.68cm2?
What will the areaof bacteriabe in the long term?
Sketchthe graphof y = log(x- I) .
,,""
.
Logarithms
andExponentials MoreExercises
Page10
'
.:
Further Exercises
1.
Simplify the following to a singlelog term:
(i) logx+310gy
(ii) 210gA-510gB.
2.
Solve (i) lOx= 250
(ii) ex= 15 .
3.
Find the smallestinteger n suchthat (1.03)" > 37 .
4.
If f(x) = 7e2x find
5.
The atmosphericpressure,p(h) (rnrnof mercury)at height h (krn) is given by
the formula
p(h) = 760e-kh where k is a constant.
(i) f(O)
The pressureis 590 at height 2.
v = Eexp( ~)
(ii) f(I.5)
Find (i) k
(ii) The pressureat height 1 .
6.
If
, expresst in termsof the otherletters.
7.
Acertainradioactivematerialdecaysaccordingto m(t) = moe-kt where t is
the time (in years), mo is the massinitially (in grammes)and m(t) is the mass
at time t.
A block of materialwith a massof 100g is reducedto 80g after 20 years.
(i) Find the valueof k.
(ii) The half life of the materialis the time takenfor half of its massto
decay. Find the half life of the material.
8.
Let
f(t) =
50-kt
where a and k are constants.
l+ae
(i) If f(O) = 10 find a.
W\.f}
(ii) Find k if f( 10) = 25 .
11.( (li)
"" )
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Logarithmsand Exponentials
Answersto Exertises
Page11
.
,
it
Answers to Exercises
Exercises8.1
1. (i) -7 (ii) 1
2. (i) 0.70 (ii) 1.95
Exercises8.2
1.(i) log(A) + 3 log(B) (ii) 2 log(A) - 3 log(B)
2. (i) 1.46 to 2 d.p. (ii) 1.40 to 2 d.p.
4. 15
Exercises8.3
1. 1.49
2. e2x-1
3. ex-1
4. f(O)1 =4
f(1)=7.97
i
Exercises 8.4
1.t=-aln(1-y)
2.(i) t = - cRIn(1-~)
(ii) 0.769
3. (i) 0.25 cm2 (ii) k = 2.08
-
i
.
(iii) 3.05days (iv) 49.5cm2
4.
log(.-1)
. =1
"I
~
1
4
6
B
10
.
.1
.
Further Exercises
1.
(i) logxy3 (ii)
log
~
2.
3.
B
(i) x = 2.4 (ii) x = 2.7
123
4.
(i) 7
5.
6.
(i) 0.127 (ii) 670.,
t=-Tln -v )
7.
(i) k=0.0112
8.
(i) 4
i
(ii) 140.6
1
..
(
E
(ii) 62 years
f
II
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i
(ii) 0.14.1
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