Precomposin
g Equations
Precomposin
Equations
g g
Composing
Polynomials
Precomposin
Equations
with Planar Transformations
Let’s “precompose”
the
function
with
x function
#-3x f(x)9f(x)
f(x)the=the
Let’sLet’s
“precompose”
function
= x the
the the
#-3xfunctionwith
function
“precompose”
= x #-3x9
function
9 with
=
-3
=
-3
o
g(x)
(Precompose
withjmeans
that
we’ll
would
look
at
We
o
g.
g(x)g(x) = -3
that
we’llwe’ll
looklook
at fat g.f oWe
f (Precompose
f that
f withjmeans
(Precompose
withjmeans
We would
g. would
f3 −2x+9
Let’s
compose
the
function
f
(x)
=
x
with
the
function
g(x)
=
4−x.
call g o f “postcomposing”
with
call call
g.)
with
g o fg o“postcomposing”
g.)
ff “postcomposing”
f f with g.)
—
—
—
f a g(x)
=
f ◦ g(x)
fa (g(x))
f(g(x))
= f(g(x))
f a =g(x)
g(x)
= f(g(x))
f
=
—
3
= g(x)
−=2g(x)2g(x)
+ 92g(x)
2g(x)
+
=
+ +
—
—
3
= (4 −
x)=(4—x)
− 2(4 − x) + 9
=(4—x)
—
3
2(4—x)+
—
3
=(4—x)
2(4—x)+
—
3
2(4—x)+
getTo
the
answer
in
perhaps
aperhaps
slightly
way,
first we
To get the To
same
answer
slightly
is perhaps
a answer
wedifferent
first
way,
ai4
T
-st
get
the
same
answer
is perhaps
a slightly
we write
first
way,
ai4
T
-st
Tosame
get
the
same
slightly
isdifferent
different
adifferent
we first
way,
ai4
T
-st
the‘r’twi-thformula
for
f
(x).
twi-th- the formula
f(x)
for
the
f(x)
formula
for
‘r’twi-th- the formula for f(x)
X
3
2x
-X
3
X
2x
3
2x
think
of
gthink
as
replaces
xreplaces
with
4−
x.4 x.
Second, we Second,
thinkSecond,
ofwe
thethink
function
as
replaces
x with
4replaces
x.
gSecond,
we
of the
function
asfunction
thatthat
x with
g that
we
of
the that
function
as
x with
4
gthe
—
—
x
—
x.
xx
Third, the Third,
formula
for
obtained
canf for
rewriting
beobtained
formula
the
formula
for
◦beg(x)
be
bethe
rewriting
the formula
Third,
formula
g(x)
can
rewriting
be obtained
be
the the
formula
fthea g(x)
ffora can
a g(x)
Third,
the
formula
can
be obtained
be rewriting
formula
f
for f(x) while
every
4.
with
x
x
for freplacing
(x),
except
that
we’ll
replace
every
x
with
4
−
x.
for f(x)
while
replacing
every
4.
with
x
x
for f(x) while replacing every x with x 4.
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Suppose p : R2 → R is the polynomial p(x, y) = 3x2 − xy + 2. Recall
that A(−3,4) : R2 → R2 is the planar transformation given by the formula
A(−3,4) (x, y) = (x − 3, y + 4).
2Let’s
Suppose
2
3x
2.precompose
Let’s
precompose
Suppose
p(x,
=
2
2.will
Let’s
precompose
addition
xy
withthe
addition
p pwith
++
Suppose
p(x,
2
3x
xy
the
addition
p with
y) =p(x,
+3x
The composition
p y)
◦ y)
A=(−3,4)
:2.Rxy
→
R
again
be
a polynomial
inthe
two
o y).
function
function
)
4
,
3
A(_
.
is,find
find
)
4
,
3
(x,
)
4
,
3
A(_
.That
That
is,we’ll
find
)
4
,
3
A(_
(x,
o A(_
y).(−3,4)
function
y). (x, y).
)
4
,
3
A(_
.
That
is,
we’ll
)
4
,
3
(x,
variables.
On this
page,
we’llfind
the
formula
for
p◦A
p op A(_
pwe’ll
—
—
—
First,
write
First,
write
thetheformula
forfor
down
pp
First,
write
down
thedown
formula
forformula
p p(x,
First,
write
down
the
formula
for
y).
2
x
3
-x÷2
2
x
3
x÷2
3
2
x
x÷2
Second,
write
down
what
Second,
write
down
)
4
,
3
A(_
replaces
each
of of
the
of of(x,(x,
)
4
,
3
A(_
replaces
each
thecoordinates
coordinates
Second,
write
down
what
Awhat
replaces
each
of
the
coordinates
y) y)
Second,
write
down
what
)
4
,
3
A(_
replaces
each
of the
coordinates
of (x,ofy) the
(−3,4)
with.
vectorwith.
(x,
y) with.
with.
r)
r)r)
= =(i-3d
(i-3d
= (i-3d
x—3
x—3 x—3
x’ x’x’
Third,
the
formula
forp po ◦A(_
A(x,
y)(x,
found
rewriting
the formula
for
o y)
Third,
thetheformula
Third,
)
4
,
3
A(_
(x,
forp po (x,
)
4
,
3
A(_
foieplacing
isfoieplacing
with
each
x xwith
(−3,4)
each
Third,
the
y) y)isby
formula
forformula
)for
4
,
3
isisfoieplacing
each
x with
p,
except
that
we’ll
replace
each
x
with
x
−
3,
and
replace
each
y
with
y
+
4.
x xreplacing
andreplacing
replacing
3, 3,and
each
4. 4.
each
y4.withy +
y+
x 3, and
each
y with
yy+with
—
—
—
(x3)2
33(x3)2
3 (x3)2
÷2
(x-3)(r)
÷2 ÷2
(x-3)(r)
(x-3)(r)
In can
conclusion,
This
can
This
canbebesimplified.&-I4tt}
simplified.&-I4tt}
This
be
simplified.&-I4tt}
2 3)2
3)2 − 3) 3)2
◦0
A0
(x,
y) = 3(x
(x −
+
(−3,4)
)y)A(_
4
,
3
A(_
(x,
)
4
,
3
(x,
= =
3(x
(x+3)(y
4) 24)+ +
2 2
0 A(_
3(x
(x4)3)(y
3)(y
++
)p(x,
4
,
3
+
= 3(xy) y)
(x − 3)(y
24)
+
—
—
—
—
—
—
—
—
=
*
*
*
2
3x
* = 3x
*
2
—
—
—
23x
3x
2 18x+2—xy—4x+3y+
18x+2—xy—4x+3y+
2i-2
18x+2—xy—4x+3y+
2i-2 2i-2
*=xy=3x
*22xxy
*+xy3y22x
*+
*+
*
*
23x
2
22x
3y3y
+*+
+
= =
—
—
—
—
—
—
—
131
*
*
* **
* **
* **
* **
* **
126
* **
* **
126126
* **
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Let’s compose q(x, y) = x + y − 1 with the matrix
Let’s precompose q(x, y) = x + y 1 with the matrix
the matrix
Let’s precompose
q(x, y) =q(x,
x + y x1 with
Let’s precompose
y) =1 2+ y 1 with the matrix
M=
3M
1 (‘12
(‘12
31
(‘12
That is, we’ll find q ◦ M (x, y). M
M
31
31
First, write the formula for q.
First, write
the
formula
for
write the
First, writeFirst,
the formula
forformula
q. q. for q.
—
—
—
;z
;z
;z
Second,
write
howeach
M replaces
the coordinates
(x, y) (x, y)
Second,
write
what
M down
replaces
of the
the coordinates
coordinates
of (x,
theof
vector
Second,
write
down
how
M
replaces
of
y)
Second, write down how M replaces the coordinates of (x, y)
with.
11
11 ZN/ZN /x÷2
/x÷2 /x÷2
11 ZN/ZN
ZN/ZN
3 33 I)t)
I)t)I)t)
>x+2,
>x+2,
xF-xFxF- >x+2,
o M(x,
Third, the
formula
for q ◦ M
is found
rewriting
formula xfor
q, x + 2y
Third,
the formula
for(x,q y)
found
b eptheig-each
with
y) is by
o M(x, y) is found b ep ig-each x with x + 2y
Third,
the
formula
for
q
except that
replace
each
with
+ 2yy)and
each yb
with
+ y.
and we’ll
each
Third,
3x
the formula
y with
is found
ep 3xig-each
x with x + 2y
q oxM(x,
+ y.x for
and eachand
3x
y with
y.
+
each y with 3x + y.
(x+2(3x+)+ (3x+)- I
(x+2
+
I
(x+2 + (3x+)-
I
In conclusion
It can be simplified as
It can beIt simplified
q ◦ simplified
Mas
(x, y) =as(x + 2y) + (3x + y) − 1
can be
qoM(x,y) = 4x+3y— 1
qoM(x,y)
= 4x+3y— 1
qoM(x,y)
= 4x+3y— 1
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
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*
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*
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127
127
127
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As we
we saw
saw in
in the
the previous
previous chapter,
chapter, ifif TT :: R2 →
—* R
R2 is
is aa planar transformatransforma
As
tion, and
and ifif S is
is is
thethe
setset
of
tion,
is the
the set
set ofofsolutions
solutionsofofp(x,
p(x,y)y)==q(x,
q(x,y),y),then
thenT(S)
T (S)
−1
−1
solutions
of of
p(x,
= q(x,
WeWe
cancalled
say this
economically
of
solutions
p ◦y)ToT
(x, y)
= q ◦y)ToT.
(x, y).
thismore
(POTS),
and we
as follows:
can
describe it more economically as follows:
The
withT −1
T
Theequation
equationfor
forSSprecomposed
composed with
Third, rewrite the equation
for 5, replacing each x with x y and each y
an equation
equation for
for TT(S).
isis an
(S).
with —x + 2y.
—
To compose
precompose
an equation
there
three
steps
to be
followed.
To
an equation
withwith
T −1T’,
, there
are are
three
steps
to be
followed.
1: Write
Step 1:
Write the
the original
original equation.
equation.
Step
−1
2: Find
Step 2:
WriteT how
T’write
replaces
of the
vector
(x, y).
Step
and
whatthe
T −1coordinates
replaces each
of the
coordinates
of
the vector (x, y) with.
Step 3: Rewrite the equation from Step 1, except replace every x and
every y the
withequation
the formulas
in Step
2.
Step 3: Rewrite
from identified
Step 1, except
replace
every x and
every y with the formulas identified in Step 2.
We’ll practice these three steps—practice precomposing equations—with the
next
problem.
We’ll
practice (POTS) using these three(-i+2-(x-)÷iQ
steps in the next problem.
(x-(-x2)÷3
Problem: Suppose
Suppose that
that SS isis the
the subset
subset of
of the
the plane
plane that
that is the set of
Problem:
=
2
solutions of
of the
the equation
equation xy
xy +
2 − xx +
4y
10.
solutions
+ 10.
+ 33 = 4y
—
Now simplify the equation so that the answer, an equation that has N(S)
S
as its set of solutions, is
+10
( ).
Let
Let N
N be
be the
the invertible
invertible matrix
matrix N
N=
=
133
128
2 1
. Give
Give an
an equation
equation that
that
1 1
N(S)
is theof.set of so1utions.
N(S)
the
of
solutions
N (S) is
is N(S)
the set
set
of
solutions
of.
is the set of so1utions.
N(5)
find anfor
equation
for
N(S),
have to the
precompose
Solution:
To
equation
N(S),
precompose
the
equa the equa
Solution:Solution:
To find
find an
anTo
equation
for
N (S), we
we have
have to
towe
compose
equation
find
an equation
for N(S), we have to precompose the equa
for
S with
.
1
N
tion
S tion
with
for Solution:
Sfor
with
N −1To
.N’
tion for S with N
.
1
down thefor
equation
for S.
First,
down
the
S.
First, write
writeFirst,
downwrite
the equation
equation
for
S.
First, write down the equation for S.
+3
+3
÷ 10
÷ 10
−1 write down how N
−1 replaces the coordinates
Second,
1
of (x, y)
Second, write
find
Ndown
and
N
replaces
each ofofthe
coordinates
of
Second,
howwrite
N’what
replaces
the
coordinates
(x, y).
Second,
write
down
how
1
N
replaces
the
coordinates
of
(x,
the vector (x, y) with.
y)
N
-I
-I
-I_/i -I
2
(I
-I_/i
(I
z(i)—i(i)—I 2) -I
z(i)—i(i)—I 2)
2
N
_
_
-_
-_
N
1
‘V
(
\+2)
‘V
N
1
(
/j)
2
\+2)
(i
(i
—1
—1
/j)
2
129
134
129
129
First, write down the equation for S.
1 write down how N
1
Second,
1
replaces
the
ond,
find
N
replaces
eac
N(5) write
ond,
write
down
N’
replaces
the
coordinat
ose
an equation
with N
T’,
thereand
are how
three
steps
to be what
followed.
econd,
write
down how N
1 replaces the coordina
ector (x,
y) with.
÷
Solution:STooluffiitniodna:nTeoqufiantdioannfoerquNa(tSio)n, wfoer hNav(Se)t,owpcerohmcaopvmeospteostpehrethceoqmueaqptouiosaenthe equa
ite the original
StfiornSfoowlruitStihotwiinNo:thT1f.oN’rfiSndwiatnheN1.quation for N(S), we have to precompose the equequation.
a
-I_/i
-I
Solution: To
+3
+3
10
÷(I
10 Solution:ToffiindanequfiantdioannfoerquNa(tSio)n,wfoerhNav(Se)t,owpcerohmcaopvmespteostphreethcoqmueaqptouiosaenthe qua
tion for S with N1.
-I_/i
-I
ti nSfoolruStiowni:thT1foN’rfiSndwaitnheN1.quation for N(S), we have to precompose the qua each x with x − y -I
rewrite
the equation
for
S, except
replace
S. Third, the
(I
z(i)—i(i)—I
First, writeFidrostw,nwtrhitee edqouwantiothnefoerquSa.tion forreplaces
2) and
ite how T’
coordinates
of the
vector
(x, y). fotrioSnwfoirthSNwith. N1.
N
_
_
N
y with
−x +12y.
z(i)—i(i)—I
2)
First, writeFidrostw,nwtrhitee edqouwantiothneforquSa.tion for S.
2
1 replaces
Second,
write
down
how
1
N
the coordinates
of (x, y)
Second,
find
N
and
write
what
N
replaces
each
ofofthe
coordinates
of
Second,
write
down
how
N’
replaces
the
coordinates
(x, y).
S
o
l
u
t
i
o
n
:
T
o
f
i
n
d
a
n
e
q
u
a
t
i
o
n
f
o
r
N
(
S
)
,
w
e
h
a
v
e
t
o
p
r
e
c
o
m
p
o
s
e
t
h
e
q
u
a
Solution: To ffiind an equationfor N(S), we have to pcroreplace
mcopmspesthethqueaqtuioan
write the equation from Step 1, except
every x andFirst, writedownthe quationfor S.
ose the qua
÷ 10 Second,
+3 the
write
downintfiorhow
1
replaces the coordinates of (x, y)
nSSfoowlruitStihownNi:thT1f.oN’rfiSndwaitnheN1.quation for N(S), we have to precompN
vector (x,
y) with.
ery y with +3 the
identified
÷ 10 formulas
tion for S with N1.Step 2.
+3 ÷ 10
(i
First, writeFidrostw,nwtrhitee edqouwantiothneforquSa.tion for S.
÷ 10
+
3
-I_/i
-I
-I
(-i+2-(x-)÷iQ
First, write down the quation for S.
(i
1
1
e these
three
steps—practice
precomposing
equations—with
the
(I
Second, wfinrSidetecNodnodw, nawhrdiotewdrN’riotwenwheopawltacNN1es threepcloacersdienthaecthcsof rtdh(xien,aycto)e.srdofin(axt,ey)s of
\+2)
First, write down the equation for S. each
-_
-_
‘V
N
1
(
N(
1
‘V
(x-(-x2)÷3
-I_/i -I
-I—1
(-i+2-(x-)÷iQ
theSvecotndr,(wx,riyt)ewdoitwhn. how N1 replaces the co rdinates of (x, y)
(-x2)÷3
(I
z(i)—i(i)—I
/
2
j)
N
2)
\+2)
—1
2
+3 ÷ 10
_
1 1
Second, wfinrSidetecNodnodw,nawhrdiotewdrN’riotwenwheopawlt cNN1es threepcloacersdienthaecthsofrtdh(xien,aycto)e.srdofin(axt,eys )of
Sthe vecotndr,(wx,riyt)ewdoitwhn. how N1 replaces the co rdinates of (x,y)
_
-I_/i
/j)
+3
N2
2)
÷ 10
z(i)—i(i)—I
(I-I_/i -I -I
-I_/i -I
I
Suppose that
(Iz(i)—i( )—I 2) 2 S is the subset of the plane that is the set-I -I (Iof
I
_
/
i
1 1 tes
=
()—I 2) -I 2
summarize
written above(Iz(i)—iwith
the following diagram.
) 2 We
o). rdofin(axt,eys )of
Second, wfinrSidetecNodnodw,nawhrdiotewdrN’riotwenwheopawlt cNN1es threepcloacerdsienthaecthsofrtdh(xien,aycwe’ve
e equationz(i)—i( )—I 2xy
2
4y
x + 10. what
+ 3 can
z(i)—i()—I 2) 2
Sthe vecotndr,(wx,riyt)ewdoitwhn. how N1 replaces the co rdinates of (x,y)
N
-I
_
_
N
-_
-_
2
-_
_
N
N
of solutions
_
—
-_
-_
‘V
N
1
(
‘V1N((
N
1
‘V
(
N
1
‘V
/j)
2
N(S) is the set
of.
-I -I_/i -I
(i
(I
-I -I_/i -I so that the answer, an
the equation
(i equation that has N(S)
(Iz(i)—i()—I 2) 2
(i Now simplify
N
(i
(i
\
+
2)
—1
N z(i)—i()—I 2) 2
as its set of solutions,
is equation
\+2)
e equation
that has
N(S) —1 \+2)
—1an
—1 \+2)so that the answer,
S
\+2) —1 2/j)2/j\)+2) 129
ons, is
—1
134
-_
1N‘V2/‘V1Nj)((
_
_
-_
2/j)
NN
(i
-_
/j)‘V1N‘V1N((
2
(i
(i
129
129
N(5)
—1 \+2)
—1 \+2)
/2j)2/j)
+10
129 129
134
129
129 129
134
129
129 129
e invertible matrix N
=
( ).
134
129
129
Solution:
To129find an equation
for N(S), we have to precompose
134
129
tion for S with N’
Give an equation that
First, write down the equation for S.
128
S is the set of solutions of xy + 3 = 4y 2 − x + 10, so an equation for the set
N (S) is found by composing xy + 3 = 4y 2 − x + 10 with N −1 . That is, N (S)
131 down how N’ replaces
write
is the set of solutions ofSecond,
(x − y)(−x
+ 2y) + 3 = 4(−x + 2y)2 −the
(x coordinates
− y) + 10. of (x, y).
131
135
Exercises
For #1-5, match the numbered planar transformation on the left with its
lettered inverse on the right.
1.) A(−6,1)
A.) A(−4,5)
2.)
3 0
0 21
B.)
1 −9
0 1
8 0
C.)
0 17
3.) A(4,−5)
1
0
8
0 7
4.)
5.)
1 9
0 1
D.) A(6,−1)
1
E.)
0
3
0 2
For #6-10, use your answers from #1-5 to find the given vectors.
6.) A−1
(−6,1) (x, y)
7.)
3 0
0 12
−1 x
y
−1
8.) A(4,−5)
(x, y)
−1 x
0
8
y
0 7
1
9.)
−1 1 9
x
10.)
0 1
y
136
For #11-15, S is the subset of the plane that is the set of solutions of the
equation
3xy = x2 − y + 2
Match the numbered subset of the plane on the left with its lettered equation
on the right. You’ll be using your answers from #6-10 to answer these questions. You’ll also be using (POTS), which says that an equation for T (S) can
be found by composing the equation for S with T −1 .
11.) A(−6,1) (S)
A.) 3(8x)( y7 ) = (8x)2 − ( y7 ) + 2
Exercises
3
0
x the formx 2
For12.)
all polynomials
#1-8, write
B.) 3(in
1 (S)
3 )(2y) = ( 3 ) − (2y) + 2
0 2
+Dx+Ey+F
Ax
+
2
Bxy+Cy
1.)13.)
Precompose
3 = +y 5) 4=with
).
12
A(4,−5) (S)the equation
C.)x3(x y−+4)(y
(x −A(
4)2 − (y + 5) + 2
—
—
1 the equation 2x
2.) Precompose
2 3xy + y 1 = 2xy + y
).
3
,
2
2 with A(_
0
2
8
14.)
(S)
D.) 3(x − 9y)y = (x − 9y) − y + 2
0 7
3.) Precompose
the equatioii x + 2 = xy y 1 with
1 9
15.)
(S)
E.) 3(x + 6)(y − 1) = (x + 6)2 − (y − 1) + 2
0 1
13).
4.) Precompose the equation x
2+x
= 2y
2 5y + 2 with (21
—
—
( ).
—
—
—
For #16-19,
suppose that S is the set of solutions of the equation
For
#5-8, suppose that S is the set of solutions of the equation
3x22 +xy+y=x—y+5
+ xy + y92 = x − y + 5
3x
For #5-8, use that the equation for S precoinposed with T’ is an equation
that an equation for T (S) can be found by composing the equation for
forUse
T(S).
S with T −1 to find answers for #16-19. Write all polynomials in the form
)? What
2
is the equation for
5.) What is A
Ax2 + Bxy + Cy 2 + Dx + Ey + F
137
M(s)
7
--
3.) Precompose the equation x + 2
xy
=
x+x
4.) Precompose the equation 2
—
y
—
2
2y
1 with
21
5y + 2 with (21
Exercises
=
—
—
1
—3
What
is the
equation
for A(3,2) (S)?
16.)
What
For #5-8,
suppose
that
isA−1
the ?set
ofinsolutions
of the equation
(3,2)
For #1-8,
write
all Sis
polynomials
the
form
9
2
31
+xy+y=x—y+5
+
Ax
+
2
Dx+Ey+F
Bxy+Cy
1.) Precompose the equation x
—
2.) Precompose the equation 2x
2
y+3
y
—
4 with A(
).
12
Z
3
3xyA(
1 = 2xy + y
).
3
,
2
+ y(5)
2 with A(_
—
3.) Precompose the equatioii x + 2
=
—
=
4.) Precompose the equation 2
x+x
xy
y
—
=
—
2
2y
1 with
—
( ).
5y + 2 with (21
13).
For #5-8, use that the equation for S precoinposed with T’ is an equation
For #5-8,
suppose
S is the set of solutions of the equation
17.)
What that
is A−1
or T(S).
(2,−4) ? What is the equation for A(2,−4) (S)?
9
3x 2 +xy+y=x—y+5
)? What is the equation for A(
9
5.) ‘Vhat is A’
)(S)?
32
6
For #5-8, use that the equation
for
S precoinposed with T’ is an equation
for T(S).
1 4
18.) Suppose M =
. What is M −1 ? What is the equation for M (S)?
0 1
)? What is the equation for
2
5.) What is A
M(s)
7
--
131
138
( ).
4.) Precompose the equation x
2+x
—
=
2
2y
—
5y + 2 with (21
1
—3
For #5-8, suppose that S isthe set
solutions of the equation
3 5 of
19.) Suppose N32= y
. What
is N −1 ? What is the equation for N (S)?
=
x
2
y+5
+ 1 2+ y
—
B
N(S)
For #5-8, use that the equation for S precomposed with T’ is an equation
for T(S).
For #20-28, let
5.) What is
What is the 
equation for

x − 1 if x ∈ (−∞, 0);
f (x) = x2
if x ∈ [0, 4]; and

57
if x ∈ (4, ∞).
Find the following values.
20.) f (−2)
23.) f (1)
26.) f (4)
21.) f (−1)
24.)
f (2)
131
27.) f (5)
22.) f (0)
25.) f (3)
28.) f (6)
Multiply the following matrices.
3 4
−2 4
29.)
0 −2
1 1
2 −2
0 1
30.)
1 −3
2 −3
Find the solutions of the following equations in one variable.
31.) loge (x)2 − 5 loge (x) + 6 = 0
32.) (x3 − 3x2 + 2x − 3)2 = −1
139