EquationsininOne
OneVariable
VariableVV
Equations
This
chapter
a review
solving
logarithmic
and
exponential
equations.
This
chapter
is ais review
forfor
solving
logarithmic
and
exponential
equations.
Inequalities
Inequalities
Here
four
rules
inequalities
that
should
know.
Here
areare
four
rules
forfor
inequalities
that
youyou
should
know.
b, then:
> then:
If aIf>a b,
• a• +a+d>
d > b +b+d
d
• a• −a—d>b—d
d>b−d
*
*
*
*
•ca>cb
• ca
> cb
if cifc>O
>0
•ca<cb
• ca
< cb
if cifc<O
<0
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
Implied
domains
logarithmicfunctions
functions
Implied
domains
forforlogarithmic
The
domain
a logarithm
is the
of positive
numbers,
oc).A number
A number
The
domain
of aoflogarithm
is the
setset
of positive
numbers,
(0,(0,
∞).
positive
take
logarithm.
hashas
to to
be be
positive
to to
take
its its
logarithm.
Examples.
Examples.
• Let’s
find
implied
domain
only
makes
sense
• Let’s
find
thethe
implied
domain
of of
loglog(2x
only
makes
sense
e (2x − 3).3).It It
take
a logarithm
a number
if it’s
positive,so so
implied
domain
to to
take
a logarithm
of of
a number
if it’s
positive,
thethe
implied
domain
of of
is the
of all
numbers
x with
property
that
logloge(2x
is the
setset
of all
numbers
x with
thethe
property
that
2x2x
− 3 >3 0.
> 0.
e (2x − 3) 3)
Adding
3 to
both
sides
inequality,
have
Dividing
Adding
3 to
both
sides
of of
thethe
inequality,
wewe
have
2x2x
>>
3. 3.Dividing
byby
2 2
3
have
x >2 . That
That
implied
domain
wewe
have
x >
is, is,
thethe
implied
domain
of of
logloge(2x
is is
thethe
setsetof of
e (2x − 3) 3)
3
numbers
x with
property
that
x >
., or more succinctly, the implied
all all
numbers
x with
thethe
property
that
x>
2 , or more succinctly, the implied
3
domain
of 1og(2x
oc).
domain
of log
is (is2 ,(,
∞).
e (2x − 3) 3)
—
—
—
.
—
—
/3
p
214 195
• The implied domain of 3 log
(4 5x) is the set of all x with the
2
• The
implied
3 log
the set oftoallwriting
x with the
Writing of
that
4 2 (45x−>5x)
property that
4 5x
0 isis equivalent
> 0. domain
the
• The
domain
of domain
(that
2
4 of
5x)
is> the
set
all x).
property
that
4 −So5xthe
>
0. Writing
4−
5x
0 is5x)
equivalent
towith
writing
3 log
4> 5x
or that
> implied
x.
implied
(
2
31og
4
isof(—oc,
4
4
Writing
4 5x
to writing
property
0 is2 (4equivalent
4 > 5x orthat
that45 >5x
x. >So0.the
impliedthat
domain
of >
3 log
− 5x) is (−∞,
5 ).
4> 5x or that > x. So the implied domain of 31og
(4 5x) is (—oc, ).
2
—
—
—
—
—
—
—
—
\
‘5
\
*
‘5
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
Implied domains for logarithmic equations
domains
logarithmic
TheImplied
implied domain
of an for
equation
is the set ofequations
all numbers that are in
Implied
domains
for
logarithmic
equations
implied domain
of an equation
of all numbers that are in
each ofThe
the domains
of the functions
involvedis inthe
thesetequation.
Theofimplied
domain
of an
equationinvolved
is the set
of all
numbers that are in
each
the domains
of the
functions
in the
equation.
Examples.
each of the domains of the functions involved in the equation.
Examples.
• To find the implied domain of the equation
Examples.
• To find the implied domain of the equation
x = loge(x + 5) + loge(2
domain
equation
• To find the implied
of the
x = log
+ log
(2 − x)
e (x + 5) in
the eequation. Those functions
first list the implied domains of the functions
x = loge(x
5) + loge(2
+
list the implied
domains
of
thetheir
functions
in domains
the equation.
are x,first
log(x+5),
and log(2—x),
and
implied
are R,Those
(—5, functions
oc),
are
x,
log
(x
+
5),
and
log
(2
−
x),
and
their
implied
domains
are
R,
(−5, ∞),
functions
in
the
equation.
Those
functions
first list
of the
2), the
respectively.
and (—oc,
e implied domains
e
2), respectively.
are
log(x+5),
their
R, (—5, oc),
x,(−∞,
contained
Theand
implied
domain
ofand
thelog(2—x),
equation is and
the set
of implied
numbersdomains
that areare
The
implied
domain
the and
equation
theand
set of
numbers
are contained
2), respectively.
and
that
set is that
precisely
the
in each
of(—oc,
the
sets
R, (—5,ofcc),
(—cc,is2),
in The
each
of the domain
sets R, of(−5,
and (−∞,
2), of
and
that set
is are
precisely
the
contained
implied
the ∞),
equation
is the set
numbers
that
(—5,2).
interval
interval
in each (−5,
of the2).sets R, (—5, cc), and (—cc, 2), and that set is precisely the
interval (—5,2).
(-oo,z’)
—
—
(-oo,z’)
(-5, z)
Thus, the implied domain of x = loge(x +(-5,
5) +z)loge(2 x) is (—5,2).
196
Thus, the implied domain of x = loge (x + 5) + loge (2 − x) is (−5, 2).
215
Thus, the implied domain of x = loge(x
+ 5) + loge(2 x) is (—5,2).
—
—
196
• To find the implied domain of the equation
log4 (x) domain
= log4 (xof+the
4) equatioll
• To find the 7implied
first find the implied domains of the71og
and log4 (x + 4). They
(functions
4
x) = 7log
(log
4
x
+ 4)
4 (x)
are (0, ∞)
respectively.
find(−4,
the ∞)
implied
domains of the functions 71og
(x) and log
4
firstand
(x+4). They
4
The implied
domain
of
the
equation
is
the
set
of
numbers
that
are
in both
are (0, oo) and (—4, oc) respectively.
(0, ∞) and
(−4,
∞), which
is the
set equatioll
(0, ∞). is the set of numbers that are in both
implied
domaill
of the
The
(0, oc) and (—4, oc), which is the set (>4oo).
(L.j-,oo)
(o,co)
0-
p
(o,oo)
Therefore, (0, ∞) is the implied domain of 7 log4 (x) = log4 (x + 4).
Therefore,
*
*
*
*
(— oc)
is the implied domain of Tlog
(x)
4
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
=
log
(
4
x + 4).
*
*
*
*
Algebraic rules for exponentials and logarithms
TheseAlgebraic
are the most important
that you shouldand
knowlogarithms
for the functions
rules forrules
exponentials
x
a and logThese
a is a positive number, not equal to 1.)
a (x). (Here
are the most important rules that you should know for the functions
ax and log(x). (Here a is a positive number, not equal to 1.)
logarithms (x, y > 0)
exponentials
exponentials
• ax ay = ax+y
•
ax
ay
• a’a’
=
= ax−y
logarithms
• loga (x) + loga (y) = loga (xy)
• 1og(x) + log(y)
• loga (x) − loga (y) = loga
a’
• log(x) — loga(y)
• loga (xz ) = z loga (x)
•
1
a
• (ax )y = axy
*
*
*
*
*
*
*
*
• log(xY)
a’
• (ax)Y
*
*
*
*
216
*
*
*
*
197
*
*
=
*
*
=
logo
()
ylog(x)
*
*
loga(xy)
x
y
*
*
*
*
Solving
exponential
Solving
exponentialand
andlogarithmic
logarithmicequations
equations
Follow
these three steps to solve an exponential or logarithmic equation.
Follow these three steps to solve an exponential or logarithmic equation.
First:
find its domain.
First: find i domain.
Second: use the algebraic rules for exponentials and logarithms to make
Second: use the algebraic rules for exponentials and logarithms to make
the equation look like either
the equation look like either
af (x) = c or
loga (f (x)) = c
a = c
or
loga(f(x)) = c
Third: use the techniques we’ve reviewed throughout the semester to solve
Third: use the techniques we’ve reviewed throughout the semester to solve
the equation fully.
the equation fully.
Problem. Find the set of solutions of the equation 5ex = e3xe
.
T.
Problem. Find the set of solutions of the equation 5e’ 3
Solution. First, the implied domain of this equation is R.
Solution. First, the implied domain of this equation is IR.
Second, divide both sides by ex :eX:
Second, divide both sides by
e3x
5 = x 3x
= e3x−x = e2x
5 =e
= e’ = e
’
2
cx
Third, we can erase the exponential base e from the right side of the equaThird, we can erase the expollential base e from the right side of the equa
tion if we apply a logarithm base e to the left side:
tion if we apply a logarithm base e to the left side:
loge (5) = 2x
loge(5) = 2x
We can divide by 2 to see that the only solution is
We can divide by 2 to see that the only solution is
√
1
1
2 ) = log (
loge(5)
x = log
(5)
=
log
(5
5)
e
e
e
=1og(5)
= 1og()
2
2
5X
217
198
roblem. Find the set of solutions of the equation (31)5 = 81.
Problem. Find the set of solutions of the equation (3x )5 = 81.
olution. First, the implied domain of this equation is IR.
(31)5
351 the implied domain of this equation is R.
Solution.
Second, rewrite
asFirst,
so that we have the equation
x 5
Second, rewrite (3351
) as 35x so that we have the equation
= 81
35x = 81
Third, we can erase the exponential base 3 from the left side of the equation
we can
from the left side of the equation
y applying the Third,
logarithm
baseerase
3 tothe
theexponential
right side ofbase
the 3equation:
by applying the logarithm base 3 to the right side of the equation:
5x = log(81) = 4
5x = log3 (81) = 4
hus, the solution is x =
Thus, the solution is x = 54 .
()5
199
218
Problem. Find the
of solutions
the equation
1og(x)
log(1
Problem. Find the set ofset
solutions
of theofequation
loge (x)
− loge (1
− x) =x)0. = 0.
—
—
Solution. First, we need to find the implied domain of the given equation.
Solution. First, we need
to find the implied domain of the given equation.
The implied domain of log(x) is (0, cc). The implied domain of 1og(1
The implied domain of loge (x) is (0, ∞). The implied domain of loge (1 − x)
is (—cc, 1). Thus, the implied domain of the equation is the set of numbers
is (−∞, 1). Thus, the implied
domain of the equation is the set of numbers
that are in both the interval (0, cc) and the interval (—cc, 1). Therefore, the
that are in both the interval (0, ∞) and the interval (−∞, 1). Therefore, the
implied domain of the equation is (0, 1).
implied domain
of the equation is (0, 1).
—
(oo,i)
-o
I
0
(o,oo)
I
0
Second,
we use
can the
us algebraic
e algebraic
for logarithms
to rewrite
loge(x)
Second,
we can
rulesrules
for logarithms
to rewrite
loge (x)
−
loge(1
x)
loge(as
hus,
we
are
trying
to
find
the
solutions
of
x
loge (1 − x) as loge 1−x . Thus, we are trying to find the solutions of
x x
loge loge
=0 = 0
1−x
Third,
we
can
erase
the
logarithm
on left
the side
left side
of equation
the equation
Third, we can erase the logarithm base base
e on ethe
of the
by applying
an exponential
e to right
the right
sidethe
of equation:
the equation:
by applying
an exponential
base base
e to the
side of
x
= e0 = 1
1—x
1−x
Multiply
both
sides
by
1
to xget
= 1
and then
to 2x
get=2x1 = 1
Multiply both sides by 1 − x to xget
= 1x −
x andx then
add xadd
to xget
so that
Notice
is in the domain of our original equation, the
1
so that
x = x21 . =Notice
that that
2 is in the domain of our original equation, the
set1).(0,Therefore,
1). Therefore,
is indeed a solution.
1
set (0,
2 is indeed a solution.
—
—
—
)
( )
—
0
0
Ioe( eO)
0
0
.
219
200
Problem. Find the set of solutions of log
(x 2) + log
2
(x + 1) = 2.
2
set of solutions
m. Find the Problem.
(set
2
xof solutions
2) + log
(x
2
Find of
thelog
of log
+ 1)2 (x=−2.2) + log2 (x + 1) = 2.
Solution. First, the implied domain of this equation is the set of all x with
implied domain
all x with
n. First, theSolution.
of this
equation
is the
First, the
implied
domain
of set
thisofequation
is the set of all x with
x > 2 and x> —1. That is, the implied domain is (2, cc).
nd x> —1. That
is (2,implied
cc). domain is (2, ∞).
thex implied
domain
x > 2is,and
> −1. That
is, the
—
—
p
p
I
2.
I
I -1
2.
I
-1
a,oo
a,oo
Second. we know that
d. we know that
Second, we know that
x x —2)
(x —2) + log
9
log
(x + 1)
9
(
9
log
9 ((x 2)(x + 1)) = 2
log
x+
—2)
9
x
g —2) + log
(x log
9
= 2
((x
9
x
9
2)(x
− 2)(x
1) = log2 (x2 − x − 2)
2) +((x
log2 (x
+ 1)
= log
+ 1))
+ 1)
2 (x −log
2log
so we are trying to solve the equation
e trying to solve
so wethe
areequation
trying to solve the equation
(x x 2) = 2
log
2
(x x 2) = 2log2 (x2 − x − 2) = 2
log
2
Third, the previous equation is equivalent to
equation
equivalent
he previous Third,
to
theis previous
equation
is equivalent to
x
—2 22 = 4
x —2 22 = 4 x2 − x − 2 = 22 = 4
or
or
x —6 = 0
x —6 = 0
x2 − x − 6 = 0
—2 or x
3. Only
Now the quadratic formula tells us that either x
—2that
or xeither3. x Only
either tells
x
formula
e quadratic Now
us that
the tells
quadratic
formula
us
= −2 or x = 3. Only
one of these solutions, 3, is in the domain of our original equation, (2, cc).
equation,
cc). equation, (2, ∞).
hese solutions,
of isourin original
in the
domain 3,
one 3,of isthese
solutions,
the domain
of our(2,original
Therefore, the only solution is x 3.
solution is the
re, the only Therefore,
x only
3. solution is x = 3.
—
—
—
—
—
—
—
—
—
—
—
—
z;f
I%f,
201
I
3
z;f
I%f,
I
3
iô2.Z -2’) + Io(x+I) = 2
iô2.Z -2’) + Io(x+I) = 2
201
220
Exercises
Find the set of solutions of the equations given in #1-10. (Remember that
equations of the form ef (x) = c have no solution if c ≤ 0.)
2
1.) e7 ex = e3x
2.) (e2x )3 = 7
3.)
e4x
ex
=2
4.) e3x−4 = −1
2
5.) ex (ex )2 = e3
6.) loge (3x + 1) = 2
7.) loge (x − 3) − loge (4x + 2) = 0
8.) log3 (x) + log3 (x + 2) = 1
9.) loge (x + 1) + loge (x) = 0
10.) loge (2x − 7) = loge (x + 1)
For #11-12, find the inverse of the given matrix.
2 −1
11.)
0 3
−3 4
12.)
−1 2
221