Equations in One Variable VI
Equations
OneVariable
Variable
VI
Equations
in
One Variable
VI
Equations
ininOne
VI
o of a function f(x) is a number z E R such that f(z) = 0.
zero
a function
fof
(x)
a number
∈
R
such
that
= 0.that f(z) = 0.
A zero
is such
z Ef (z)
f(x)
a number
R= such
zero
AA
of of
a function
isa aisfunction
number
z Ez R
f(x)
that
f(z)
0.
ples.
Examples. Examples.
Examples.
The number 1 is a zero of the function loge(x). That’s because
• The
number
is
a zero
That’s
becauseThat’s because
•1 The
number
1function
isfunction
a zerologe(x).
oflog
the
function
loge(x).
e (x).
loge(1) = 0.
• The
number
1 is
a zero
of of
thethe
That’s
because
loge (1)
= 0. loge(1) = 0.
loge(1)
= 0.
JOSe
()
JOSe
()
JOSe
()
6 is a zero of the function x 6. That’s because [6] 6 = 0.
a zero
− 6.
That’s
because
[6] 6−=6because
= 0. [6]
isfunction
•the
6function
a zerox ofx the
function
x 6. [6]That’s
• 6• is6 is
a zero
of of
the
6.
That’s
because
0.
—
—
—
—
/
//
4
/
//
4
—
—
6
=
0.
//
4
/
To state a more general version of the previous example, a is a zero
• To
state
a more
of
previous
aexample,
zero
• general
Togeneral
stateversion
aversion
moreofgeneral
versionexample,
ofexample,
the previous
a is a zer
of x a.
• To
state
a more
thethe
previous
a isa aiszero
of x a.
of of
x x−
a. a.
—
—
—
x-a
278
/
278
312
/
278
x-a
/
x-a
2
Thefunction
functionx
x
. .•
The
2x
The
function
2
−
9has
has
zeros
3and
and
−3.
9 9has
zeros
3 3and
zeros
—3.
—3.
——
Thezeros
zerosofof
ofa apolynomial
apolynomial
polynomial
are
called
roots.
• ••The
are
called
roots.
The
zeros
are
called
roots.
x
ehas
hasnono
nozeros.
zeros.
• ••
e’e’
has
zeros.
xx
Theset
setofof
ofzeros
zerosofofof
sin(x)
{. . .—27r,
,—27r,
−2π,
−π,
0,r,π,
2π,
3π, . . .}.
• ••The
sin(x)
isisis
—7r,
2r,
The
set
zeros
sin(x)
—7r,
2r,
0,0,r,
3ir,.
3ir,.
{...
{...
..
That
is,
the
set
of
zeros
of
sin(x)
is
the
set
That
Thatis,is,the
theset
setofofzeros
zerosofofsin(x)
sin(x)isisthe
theset
set
{ nπTi|Tin ∈
Z}
{n7r
Z}
{n7r
Z}
,
Sin()
Sin()
313
279
279
,
Finding
zeros
of
functions
zeros
functions
Finding
ofof
zeros
Finding
of functions
zeros
functions
Finding
Iff(x)
f (x)
isfunction,
function,
then
the
zeros
off(x)
f (x)
are
the
solutions
ofthe
the
equation
equation
then
thethe
zeros
of of
f(x)
are
thethe
solutions
of of
the
If If
f(x)
isIfis
af(x)
then
the
zeros
ofare
f(x)
are
the solutions
of
the equation
is a function,
equation
then
zeros
solutions
aafunction,
f (x) f(x)=O.
= 0.
f(x)=O.
f(x)=O.
Examples.
Examples.
Examples.
Examples.
The
zeros
ofthe
the
linear
polynomial
2x+5 +are
are
the
solutions
ofthe
the
of of
the
zeros
the
linear
2x2x+
solutions
. The
of zeros
of
thepolynomial
linear
polynomial
are
the solutions
. zeros
The
oflinear
polynomial
solutions
.•The
of
52x5are
+the5the
the
5
.
Thus,
equation
2x
+
5
=
0.
Subtract
5:
2x
=
−5.
Then
divide
by
2:
x
=
−
2: 2:x =
equation
2x2x
5: 5:2x2x
= 5:
—5.—5.
bydivide
equation
2xThen
=Then
—5.divide
Then
=
5 Subtract
0. Subtract
+ 5+ 5 2x0. +0.Subtract
equation
=
divide
by
xby=2: xThus,
2 Thus, Thus,
5
isthe
the
only
zero
of2x
2x+
5. + 5.
−is2isthe
only
zero
of of
2x
5.
isonly
the
only
zero
of+5.2x
+
zero
—.
—.
—
—
—.
—
2.5
2.52.5
2 2
2
The
zeros
oflog(x
logofe (x
2)are
are
the
solutions
oflog(x
logofe (x
0.This
ThisThis
of of
2)−2)=2)=0.=0.2)This
• The
zeros
log(x
2)−2)are
the
solutions
log(x
• zeros
Theof zeros
log(x
2)the
aresolutions
the solutions
log(x
= 0.
••The
of
0
equation
isequivalent
equivalent
−=
1,and
and
thus
isequivalent
equivalent
to toxtox x 3.
equation
is isequivalent
to toxtoxx2to
1,=1,and
is isequivalent
to=3.x3. 3.
equation
is equivalent
2e =
=
= thus
1,thus
and
thus
is equivalent
equation
2 2x== =
That
is
the
only
zero
ofthe
the
function
loge (x
−2).2). 2).
function
loge(x
2).
That
is, is,
3is,is33is
the
zero
of of
the
loge(x
That
is,the
3only
isonly
the
only
zero
offunction
the function
loge(x
That
zero
—
—
—
—
—
—
—
—
—
—
—
—
1o
(x-2)
1o (x-2)
1o(x-2)
z(t3z(t3 z(t3
2
2
The
zeros
ofx
x 4x+3
−4x+3
are
the
solutions
ofthe
the
equation
x 4x+3
−4x+3
zeros
of of
thethe
solutions
of of
the
equation
0.=0.0. = 0.
• The
—x
2
x
4x+3
—
2
x
4x+3
zeros
of x
are
the solutions
of
the equation
•zeros
The
—=4x+3
2
x
—are
2
4x+3
are
solutions
equation
=
••The
—
2
—
2
Using
the
quadratic
formula,
we
can
find
that
the
solutions
are
1are
and
3. 3.
that
thethe
solutions
areare
1 1and
Using
the
quadratic
formula,
wewe
can
find
3.1 3.and
that
the solutions
Using
the quadratic
formula,
we
can
find
that
solutions
and
Using
the
quadratic
formula,
can
find
2
Therefore,
and
are
zeros
ofx
−4x
4x+
3. + 3.
thethe
zeros
of of
2
x
Therefore,
1 and
3 1are
3.+3.4x
arezeros
the
zeros
of
2
x
+
Therefore,
3the
are
2x 4x
Therefore,
11and
33and
—
—
—
II I
33
314
280280
280
3
√
√
• The
zeros
of the2+
function
x − 3 are of
solutions
. The zeros
of the
function
are solutions
= 0. x − 3 = 0.
/‘x √23 +
2+ \/x of 32 +
This equation
is equivalent
x − and
3 = thus
−2, and
has no solutions
This equation
is equivalent
to /x 3to= —2,
it hasthus
solutions
√no it
3 has no zeros.
since square-roots
never negative.
since square-roots
are never are
negative.
Therefore,Therefore,
2 + /x 23 +
has xno−zeros.
—
—
—
—
3
An illustration
when
not to divide
An illustration
of whenofnot
to divide
asked
find the solutions
of the equation
If asked toIf find
thetosolutions
of the equation
(x − 1)(x − 2) = 3(x − 1)
(x—1)(x—2)=3(x—1)
be to
tempted
to divide
sides
of the by
equation
(x − 1) leaving
you mightyou
be might
tempted
divide both
sidesboth
of the
equation
(x 1) by
leaving
with
2) = 3,simpler
a muchequation.
simpler equation.
while
(xis− 2) = 3 is
However, However,
while (x —2)
you with you
(x —2)
= (x
3, −
a much
= 3
a muchequation
simpler (the
equation
(thetosolution
is 5)equivalent
it is not equivalent
certainly certainly
a much simpler
solution
it is 5) to
it isit not
to theequation
original (x—1)(x—2)
equation (x−1)(x−2)
= That’s
3(x−1).
That’sthe
because
the function
to the original
= 3(x—1).
because
function
(x a−zero
1) has
1), and multiplying
dividing
an by
equation by
(x 1) has
(thea zero
zero (the
is 1), zero
and ismultiplying
or dividingoran
equation
function
zero
in theofdomain
of the isequation
is not
a valid way
has athat
zerohas
in athe
domain
the equation
not a valid
way
a functiona that
of obtaining
an equivalent
of obtaining
an equivalent
equation. equation.
steps
below the
describe
theprocess
generalforprocess
solving equations
of the
below
solvingforequations
of the
The stepsThe
describe
general
form h(x)f
(x) = h(x)g(x).
After
thesteps,
general
we’ll
return to the
form h(x)f(x)
= k(x)g(x).
After listing
thelisting
general
we’llsteps,
return
to the
example
of (x
− 1)(x
− 2)1).
= 3(x − 1).
example of
(x 1)(x
—2)
= 3(x
—
—
—
—
Steps
for solving
(x) = h(x)g(x)
solving
Steps for
h(x)f(x)h(x)f
h(x)g(x)
=
Stepthe
1: Find
theofdomain
of the equation.
CallD.this set D.
Step 1: Find
domain
the equation.
Call this set
Stepthe
2: Find
zerosthat
of h(x)
D. set
CallZ.this
Z. The numbers
Step 2: Find
zeros the
of h(x)
are inthat
D. are
Callinthis
Thesetnumbers
will be solutions
of h(x)f
(x) = because
h(x)g(x)ifbecause
in Z will in
be Zsolutions
of h(x)f(x)
= h(x)g(x)
h(z) = 0if h(z) = 0
then =
h(z)f
(z)==00f
= 0==h(z)g(z).
0g(z) = h(z)g(z).
then h(z)f(z)
Of(z)
= (z)
Og(z)
Stepthe
3: Find
the solutions
of the h(x)f(x)
equation =
h(x)f
(x) = with
h(x)g(x)
Step 3: Find
solutions
of the equation
h(x)g(x)
the with the
D −function
Z. The h(x)
function
h(x)
hasinnoDzeros
restrictedrestricted
domain Ddomain
Z. The
has no
zeros
Z, in D − Z,
you can
with by
dividing
h(x) to
the equivalent
theobtain
equivalent
so you cansostart
withstart
dividing
h(x) tobyobtain
f (x) = g(x).
equation equation
f(x) = g(x).
—
—
Step 4:the
Collect
and the solutions
Step 3. These
Step 4: Collect
zeros the
fromzeros
Stepfrom
2 andStep
the 2solutions
from Stepfrom
3. These
are the solutions
of theequation.
original equation.
are the solutions
of the original
281
315
Examples.
• Let’s return to the equation (x−1)(x−2) = 3(x−1) and proceed through
the four steps outlined above.
. Let’s
of the
2 16)equation,
(x
1og(x) =so(2
16).
(Step
1): find
(x −the
1)(xsolutions
− 2) = 3(x
− 1)equation
is a polynomial
the domain
is R.(Step
Written
in the
notation
of the
general
D=
1): The
domain
of the
equation
is steps
because
weR.can only take
(0, oc),above,
the
logarithm
a positive
(Step
2): The of
function
thatnumber.
we’d like to divide both sides of the equation by
—
—
is (x(Step
− 1).2):
Its The
zero is
1, which
solution
of (x − 1)(x
− 2) =
zeros
of (x
2will be
16) aare
the solutions
of 2
16 3(x
= −
0, 1)or
since
([1] − 1)([1]
2) = 0(−1)
0 =
3(0)are
= 43([1]
1). However,
Written of
inthese
the
equivalently,
the −
solutions
of x
2 == 16,
which
and−—4.
= (2
notation
of the
above,
= {1}.of (x
two zeros,
onlygeneral
4 is insteps
the Z
domain
(0, oc),
2 16)
16) log(x),
so
it’s the
zero to
of find
2 the
(x
16) solutions
that concerns
in this problem.
(Step
3): only
We need
of theusequation
(x − 1)(xWith
− 2) the
=
notation
of
the
general
steps
listed
Z
above,
= {4}.
3(x − 1) with the restricted domain of R − {1}. On this domain, the function
(x −(Step
1) has
so find
we can
by of
(x(x
− 1).16)The
resulting
3): no
Wezeros,
need to
the divide
solutions
2
loge(x)
16) with
= (x
2equivalent
equation
is (x −domain
2) = 3.(0,We
add The
2 to function
see that (x
the solution
is xzeros
= 5.in the
the restricted
oo)can{4}.
2
16) has no
set
(0, 4):
oc) The
{4}.
we cannumber
divide the
equation
2 16)ofloge(x) =
(x
(Step
zeroTherefore,
of (x − 1)—the
1—and
the solution
2
16)−by2) (x
2
the equivalent
equation
loge(x)5—are
= 1.
The
(x(x
− 1)(x
= 3(x16)− to
1) obtain
with domain
R − {1}—the
number
the
only solution
of 1)(x
this −
equation
is −
x =
1To
= e.
solutions
of (x −
2) = 3(x
1).e
repeat, the set of solutions of the
equation
1)(xsolutions
− 2) = 3(x
5}. = (x
4):−The
(Step (x
of −
21) is4){1,
(x
loge(x)
2 4) are 2 (from Step 2)
and e (from Step 3).
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
(x -;)Cx-z
5
316
283
Examples.
• Let’s find the solutions of the equation (x2 − 16) loge (x) = (x2 − 16).
—2) because
= 3(x —1) and proceed through
• Let’s
to the
equation
(x is1)(x
(Step 1):
Thereturn
domain
of the
equation
(0, ∞),
we can only take
the
four
steps
outlined
above.
the logarithm of a positive number.
—
polynomialof equation,
domain
1): (x
3(x are1) the
is a solutions
(Step (Step
2): The
zeros1)(x
of —2)
(x2 −= 16)
x2 − 16 so
= the
0, or
of which
the general
steps−4.
above,
D = JR.
is R. Written
in the notation
equivalently,
the solutions
of x2 = 16,
are 4 and
However,
of these
2
2
both
thelog
equation
by
(Step
2):4 The
we’d likeofto(x
divide
two zeros,
only
is infunction
(0, ∞), that
the domain
− 16)
= sides
(x −of16)
e (x),
2
of (x
1)(x With
2) = the
3(x 1)
(x only
1).zero
Its zero
which
be a solution
so it’s isthe
of (xis −1, 16)
thatwill
concerns
us in this
problem.
([1] general
2) listed
= O(—1)
1). Written in the
since
1)([1] steps
= 0
notation
of the
above,
Z == 3(0)
{4}. = 3([1]
thetogeneral
steps
above,ofZ(x=2 −
{1}.
(Stepnotation
3): We of
need
find the
solutions
16) log (x) = (x2 − 16) with
—
—
—
—
—
—
—
—
—
e
3): We (0,
equation
(Step domain
need
to find the
the restricted
∞)−{4}.
Thesolutions
function of
(x2the
−16)
has no (x
zeros 1)(x
in the 2) =
2
domain,
withTherefore,
the restricted
domain
of the
JR equation
{1}. On this
set (0,3(x
∞) −1){4}.
we can
divide
(x −
16) logthe
=
e (x)function
2
2
divide by (x
1). The
(x by
1) (x
has−no16)
zeros,
so we the
can equivalent
(x − 16)
to obtain
equation
logeresulting
(x) = 1. equivalent
The
1
equationofisthis
2) = 3. isWex =
cane add
(x equation
only solution
= e.2 to see that the solution is x = 5.
—
—
—
—
—
—
—
2
1and
solution
of 2)
(Step
4):solutions
The zeroofof(x(x
1)—the
number
(Step 4):
The
− 16)
loge (x)
= (x2 −
16) arethe4 (from
Step
1) with domain JR {1}—the number 5—are the
(x 1)(x
2) = 3(x
and e (from
Step 3).
solutions of (x 1)(x 2) = 3(x 1). To repeat, the set of solutions of the
equation (x 1)(x —2) = 3(x 1) is {1, 5}.
—
—
—
—
—
—
—
—
—
—
(a
282
317
x
• Let’s •find
thefind
solutions
of xex =
.
Let’s
the solutions
of exeX
= e’.
(Step 1):
The1):implied
domaindomain
is R. is R.
(Step
The implied
eX has
(Step 2):
ex has
zeros.
(Step
2): no
no zeros.
x
= e
cx
(Step 3):
Dividing
xex = xex
ex by
yields
equivalent
equation
x = 1.x
(Step
3): Dividing
yields
the equivalent
equation
by eX the
Thus, 1Thus,
is the1 only
is thesolution.
only solution.
=
1.
(Step 4):
There
no are
zeros
so 1 2,is SO
the1 only
4): are
(Step
There
nofrom
zerosStep
from2,Step
is thesolution.
only solution.
i
x
2
• Let’s •find
thefind
solutions
of (x +of2)e(x2x+=2)e
(x +=2)e
Let’s
the solutions
(xx .+ 2)ex.
(Step 1):
The1):implied
domaindomain
is R. is R.
(Step
The implied
(Step 2):
The2):zero
of zero
(x +of2)(xis +−2.
It’s—2.
in the
of the equation,
so
(Step
The
2) is
It’s domain
in the domain
of the equation,
so
it’s a solution.
it’s a solution.
(Step 3):
Dividing
(x + 2)e(x2x+=2)e
(x + 2)e(xx +
by2)ex
(x +by2) (x
gives
the us
equivalent
(Step
3): Dividing
’
2
the equivalent
+2)usgives
2x
e
2x
x
x
x
2
= cx.
cx to
= 1 which
the same
equation
e = ee
. We
canWe
then
e tobyget
equation
candivide
then by
divide
= 1 is
which
is the same
ex get
2x−x
x
equation
as e as=e
1 or =
more
e = 1.cxApplying
the logarithm
yields yields
equation
2
1 orsimply,
more simply,
= 1. Applying
the logarithm
that x =
logxe (1)
= 0. = 0.
that
= log(1)
x
2
(x +=2)e
(Step 4):
−2 4):
and—2
0 are
solutions
of (x +of2)e
(Step
andthe
the solutions
(x2x+=2)e
(xx .+ 2)ex.
0 are
1’
318
284
*
*
*
*
*
*
*
*
*
*
*
Chapter Summary
The solutions of h(x)f (x) = h(x)g(x)
are the solutions of f (x) = g(x)
together with the solutions of h(x) = 0.
319
*
*
Exercises
For #1-6, find the zeros, if there are any, of the given functions.
1.) x − 2
2.) 3x + 4
3.) x2 − 81
4.) x2 + 7
√
5.) 2 x − 7
6.) e3x−2
Find the solutions of the following equations in one variable.
√
7.) x3 x = 9x3
8.) (x − 2)(x3 − 2x)2 = −5(x − 2)
9.) (x − 4)2 loge (x + 2) = 2(x − 4)2
10.) loge (x − 1)ex = −2 loge (x − 1)
11.) −8(x2 − 25) = (x2 − 25)[loge (x)2 − 6 loge (x)]
12.) (2x + 1)5 = 9(2x + 1)3
√
13.) (x2 − 4) x = −(x2 − 4)
14.) ex x1 + x = 2ex
√
√
15.) e3x x = ex−2 x
16.) (x + 3) loge (2x) = (x + 3) loge (x + 1)
17.) (x − 4)5 = (x − 4)6
18.) (x − 2)(x − 3) = (x − 2)(x + 1)
19.) x = x2
320