Ecuaciones logaritmicas parte2

ALGEBRA PREUNIVERSITARIA Ing. Widmar Aguilar, Msc.
EJERCICIOS RESUELTOS DE ALGEBRA
PREUNIVERSITARIA
PARTE 2 DE ALGEBRA
ECUACIONES LOGARITMICAS
Ing. WIDMAR AGUILAR, Msc
Abril 2021

111)
2. √ . 8 = ; > 0, X ≠ 1
2.
√
( ) . 2 =
= 2. . 3
= 3.6. 2.
= 3.6. 2. = 18 ; x> 0, X ≠ 1
= 18 → ( )
112)
! + ( ) = 1 ; > 0, ≠ 1
3 − + ( ) = 1
%
− + ( ) = 1
% & %
−
& %
+ ( ) = 1
% & %
− '
()*%
& %
+ ( ) = 1
1 − + (1 + ( ) ) = 1 +
x+ x − 2 =0
( x+ − 2) = 0

= 0 ; x+ − 2 = 0
= 0 → = 3,
= 1
x+ − 2 = 0
=
- ±√ &/
=
- ±
= 1 → = 3 = 3
= −2 → = 3-
=
0
+ + = 1 + 3 +
0
; > 0, ≠ 1
+ + =
1
0
→ (2)
113)
| + 1| + 3 = 1 ; ≠ 1, > 0
+ 2 + 1 + 3 = 1
+ 5 = 0 ; ≠ 1, > 0
( + 5) = 0
5
= 0
+ 5 = 0
= 0 → = 3,
= 1
= −5 → = 3-6
. = (1). 3-6
. = 3-6
→ (7)

114)
− 16 = ; > 0
3 − 16 = − 2
3 − 16 = 2 − 2
= 16 − 2
= log ! = 8 ; x> 0
= 8 → ( )
115)
6 . 6 − 6 − 12 = 0 ; > 0
6 − 6 − 12 = 0
6 =
±√ &;/
=
&1
6 = 4 ; 6 = −3
6 = 4 → = 5;
6 = −3 → = 5-
. = 5;
. 5-
; > 0
. = 5 → (=)

116)
>?@ 6- %A
>?(6- )
= 3 ∶ 35 − > 0 y x<5
− 35 < 0 D y x< 5
ln(35 − ) = 3 ln(5 − )
ln(35 − ) =ln(5 − )
(35 − ) = ln(5 − )
35 − = 125 − 75 + 15 −
125 − 75 + 15 − − 35 + = 0
15 − 75 + 90 = 0
− 5 + 6 = 0
( − 2)( − 3) = 0
5
= 2
= 3
Un valor de x es:
= 2 → ( )
117)
4 + 4 + − − − − − + 4H
= 4
2 .
+ 2 .
+ − − − − − + 2 H
= 2

2 + 4 + 6 + − − − + 2I = 12
∑ 2K
H
LM = 12 ; 2 ∑ K
H
LM = 12
2.
H(H& )
= 12
I(I + 1) = 12 ; I + I − 12 = 0 ; n>0
(I + 4)(I − 3) = 0
I − 3 = 0 ; I > 0
I = 3 → (N)
118)
O
;D = 2
;
P%
! = 5
; y >0
;D = 2 → 4 = D ; D = 16
;
%
! = 5
; 16 = 5 ; (16 ) = 5
(2/2) 16 = 5 ;
2;
+ = 5
4+ = 5
= 1
| | = 2 = 2 → ( )

119)
;@ ( )A! = 1
;@ ( )A! = 2
;@ ( )A = 2
;@ ( )A = ;4
( ) = 4
= 16
= 16
= ± Q(4 ) = ±4 → ( )
120)
''
'R
(
.
+
.
+
.;
+ − − − − − +
H(H& )
)-H
= I
De:
.
+
.
+
.;
+ − − − − − +
H(H& )
= ∑
L(L& )
H
LM
∑
L(L& )
H
LM = ?
L(L& )
=
S
L
+
T
L&
→ 1 = U(K + 1) + VK

1 = (A+B)i +A
U = 1 ; U + V = 0
A = 1 ; U = −V
L(L& )
=
L
−
L&
= -(
L&
−
L
)
Sea: f(i) =
L&
→ W(K − 1) =
L- &
=
L
Se tiene que la sumatoria es una serie telescópica:
∑
L(L& )
H
LM = ∑ W(K) − W(K − 1) = W(I) − W(0)
H
LM
W(I) =
H&
; W(0) = 1
∑ −
L(L& )
H
LM = −(
H&
− 1) =
H
H&
''
'R
(∑
L(L& )
H
LM )-H
= I
''
'R
(
H
H&
)-H
= I
''
'R
(
H&
H
)H
= ''
'R
(
,
)H
(
H&
H
)H
= (
,
)H
, por analogía:
H&
H
=
,
10I + 10 = 11I
I = 10
Como: E = log(I + 10I)
X = log(100 + 100)
X = log(200) = log(100.2)
X = log 10 + 2
X = 2 + log 2 → ( )

121)
( + 7) − ;(3 + 1) = /( − 9) − ;( − 1)
( + 7) − /( − 9) = ;(3 + 1) − ;( − 1)
( + 7) − %( − 9) = ;[
( & )
-
]
( + 7) − ( − 9) = ;[
( & )
-
]
( + 7) − ( − 9) = ;[
( & )
-
]
[
@ &1A
-0
! = ;[
( & )
-
]
[
&1
-0
! = ;[
( & )
-
]
; [
&1
-0
! = ;[
( & )
-
]
&1
-0
! =
&
-
( + 7) ( − 1) = ( − 9) ( − 1)
2 6
+ 2 ;
− 68 − 4 + 194 + 130 = 0
Por rufini:

2 6
+ 2 ;
− 68 − 4 + 194 + 130 =
= ( − 5)(2 ;
+ 12 − 8 − 44 − 26)
= 2( − 5)( ;
+ 6 − 4 − 24 − 13)
El polinomio divisible es:
( ) = ;
+ 6 − 4 − 24 − 13 → (=)
122)
] √125
^
= ; = > 0 ; = ≠ 1
=
%
= 125
'
^
=
%
= 5
%
^ → = = (5
%
^)%
= = 5
'
= = √5 → (N)

123)
_ ` a
'
b
c'
= ; 7 = (
'
)
> 0 ; ≠ 1
d (
'
)
@ √ A
c'
=
_
√
(
'
)
( )
c'
=
(
'
c . 1)
'
c' =
(
'
&
)
'
c' =
e'
( c') = ; igualando exponentes,
&
( - )
=
+ = 2 − 2
= 3
( − 3) = 0
= 0 − − − −I Nfgh 2
= 3 → = ± √3
= − √3 → I Nfgh 2

= √3 → (2)
124)
log (2 − 1)H
+ log ( − 1) , H
= I
2 − 1 > 0 D − 1 > 0 → > 1
log (2 − 1)H
+ log ( − 1) , H
= 10H
log (2 − 1)H
= 10H
− log ( − 1) , H
log (2 − 1)H
= log
,i
( - )'R()*i!
(2 − 1)H
=
,i
( - )'R()*i
( − 1) , H
= (
,
-
)H
-------------(1)
Para que se cumpla (1), se tiene:
j
10log (I) = I
− 1 =
,
-
log(I ,) = I
10H
= I ,
→ I = 10 D I = 10
I = 10 : n E Z
− 1 =
,
-
→ ( − 1)(2 − 1) = 10
2 − 3 − 9 = 0

=
±√0&1
;
=
±0
;
j
= 3
= −
= − → I Nfgh 2 D7 kf2 l > 1
l = 3 → (=)
125)
+ ; + ; + ;,0 + R + %R =
66
+ + m + ' + R + %R =
66
1 + + + +
,
+
,
! =
66
,
,
=
66
= 5 ; x >0
= 26
= 32 → ( )
126)
log = ; > 0
− 2 log = 0

( − 2) = 0
5
= 0
= 2
log x =0 → = 10,
= 1
l = 2 → = 10
= 100 → (N)
127)
]( − 2√7 + 27) n]
= 1
`=. ]@ − 2√7 + 27A = 1
o`
. ]@ − 2√7 + 27A = 1
]@ − 2√7 + 27A = ]7
− 2√7 + 27 = 7
− 2√7 + 7 = 0
=
√` ±√;`-;`
= √7 ; a> 0
= √7 → (7)
128)
(9 -
+ 7) = 2 + (3 -
+ 1)

0
0
+ 7! = 2 + + 1!
0 &
0
! −
&
! = 2
p
q em%
q
% e%
%
r = 2
q em%
q
% e%
%
= 4
(0 & )
( & )
= 4
9 + 63 = 12. (3 + 3)
(3 ) − 12. 3 + 27 = 0
3 =
±√ ;;- ,/
=
±
s
3 ' = 9
3 = 3
3 ' = 9 → 3 ' = 3 → = 2
3 = 3 → 3 = 3 → = 1
+ = 2 + 1
+ = 3 → (N)
129)
( − 3 + 6) − ( − 1) = 2
- &
-
! = 2
- &
-
= 4

− 3 + 6 = 4 − 4
− 7 + 10 = 0
( − 5)( − 2) = 0
5
= 5
= 2
+ = 5 + 2
+ = 7 → ( )
130)
log √7 + 4 + √2 + 3 = 1 + log (1,5)
7 + 4 > 0 D 2 + 3 > 0 → > −
;
1
log Q(7 + 4)(2 + 3) = 10 + log (
6
,
)
log Q(7 + 4)(2 + 3) = og (
6
,
. 10)
Q(7 + 4)(2 + 3) = 15
(7 + 4)(2 + 3) = 15
14 + 29 − 213 = 0
=
- 0±√ 0 & 0 /
/
=
- 0±
/
j
= 3
= −
0
/
= −4.28
→ I Nfgh 2
= 3 → (N)

131)
(− − 8 − 14). @ &; &;A9 = 1
(− − 8 − 14).
q( &; &;)
= 1
(− − 8 − 14) = ( + 4 + 4)
(− − 8 − 14) = ( + 4 + 4)
2 (− − 8 − 14) = ( + 4 + 4)
(− − 8 − 14) = ( + 4 + 4)
(− − 8 − 14) = ( + 4 + 4)
;
+ 16 + 92 + 224 + 196 = + 4 + 4
;
+ 16 + 91 + 220 + 192 = 0
Factorizando se tiene:
( + 3)( + 4)( + 9 + 16) = 0
De: + 9 + 16 = 0
=
-0±√/ - ;
=
-0±√ 1
⎩
⎪
⎨
⎪
⎧
= −3
= −4
= −
0
+
√ 1
= −2,44 − − − I Nfgh 2
; = −
0
−
√ 1
= −6.56 − − − I Nfgh 2
= −4 → (7)

132)
log d
1
+ 5 + log + 7! = 1 + log
0
!
log d
1
+ 5 + log d + 7 = 1 + log
0
!
Df = 7x+15 >0 y 2x+21 >0
CS → > −
6
1
log (d
1 & 6
!
&
! = 10 + log (
0
)
log (d
1 & 6
!
&
! = log (
0
.10)
d
1 & 6
!
&
! = 45
1 & 6
!
&
! = 45
(7x+15)(2x+21) = 45 . 9
14 + 177 + 315 = 18225
14 + 177 − 17910 = 0
=
- 11±√ 11 & ,, 0 ,
/
=
- 11± , 1
/
j
= 30
= −
601
;
= −42.64 − − − −I Nfgh 2
X = 30 → ( )

133)
xyza
%
{
&/b
,
−
>yz %
& !
,
=
1
0
>yza
%e 'm
{
b
,
−
>yz
em
%
!
,
=
1
0
9. log
%&
1
! − 9. log
&
! = 7. 10
9[log( + 216) − 27] − 9[log( + 6) − 3]
= 7
log9. log( + 216) − 9. 27 − 9. log( + 6) +
9. 3 = 7
log9. log( + 216) -log9(log9+log3)- 9. log( + 6) +
9. 3 = 7
log( + 216) − log9-log3- log( + 6) + 3 =
1
0
log( + 216) − log9- log( + 6) =
1
0
log
%&
&
! − 9 =
1
0
log( − 6 + 36) − log 9 =
1
0
9. ( log( − 6 + 36) − 9) = 7
9. (log
- &
0
! = 7
log
- &
0
! =
1
0

- &
0
= 10 1/ 0
− 6 + 36 = 9. 10 1/ 0
− 6 + 9@1 − 10 1/ 0
A = 0
=
±d - ( - ,()*{/()*q)
=
±}
= 3 + ~
= 3 − ~
+ = 6 + ~ − ~
+ = 6 → (=)
134)
2 log pd5 +
6
;
+ d
6
;
r = 30 − 2
2 log pd5 +
6
;
+ d
6
;
r = log
,
!
2 log pd5 +
6
;
+ d
6
;
r = log 15
2 log pd
6
;
+ d
6
;
r = log 15
2 log p5d
6
;
+ d
6
;
r = log 15
2 log p6. d
6
;
r = log 15

Log (p6. d
6
;
r = 15
36.
6
;
= 15
5 = 10
X = 2 → (=)
135)
8 + (4 % ) = (36 % )
8 + (4 % & % ) = (36 % & % )
8 + (4 & % ) = (36 & % )
1 + = €
8 + (4•) = (36•)
8 + € (4) = € (36)
8 + € (2 ) = € (3 . 2 )
8 + 2€ (2) = 2€ 3 + 2€ 2
8 = 2 t(1) → € = 4
1 + = € → 1 + = 4
= 3 ; x> 0
= 3 = 27 → ( )
136)

1( − 2) + 1( − 5) = 2 12
x-2 > 0 y X- 5 >0 → ‚ƒ = > 5
1{( − 2)( − 5)… = 12
( − 2) ( − 5) = 4
− 7 + 10 = 4
− 7 + 6 = 0
( − 6)( − 1) = 0
5
= 1 − − − I Nfgh 2
= 6
= 6 → (N)
137)
&
√
&
= 4 ; > 0 , ≠ 1
+ 6 = 4. @ √ 2 + 2A
− 4
√
2 + 6 − 8 = 0
− 4 2 − 2 = 0
− 8 2 − 2 = 0
−
/
− 2 = 0
− 2 − 8 = 0
=
±√;&
=
±
5
= 4
= −2

= 4 → = 2;
= −2 → = 2-
. = 2;
. 2-
. = 4 → ( )
138)
Q = √ : x>0
= √
= (√ )
= ( )
= !
4 =
( − 4) = 0
De:
= 0 → = 2,
= 1
− 4 = 0 → = 4 → = 2;
+ = 1 + 2;
+ = 17 → (2)
139)

>yz@ 6- %A
>yz(6- )
= 3
CS= 35 − > 0 ⋂ 5 − > 0
CS = x < √35
%
→ < 3,27
log(35 − ) = 3log(5 − )
log(35 − ) = log(5 − )
35 − = (5 − )
35 − = − + 15 − 75 + 125
15 − 75 + 90 = 0
=
16±√16 -6;,,
,
=
16± 6
,
5
= 3
= 2
+ = 2 + 3
+ = 5 → (=)
140)
& ( + − 6) = 4
+ 1 > 0 ; x≠ 0 → > −1 D ≠ 0
& ( + − 6) = & ( + 1);
( + − 6) = ( + 1);
;
+ 2 − 11 − 12 + 36 = ;
+ 4 + 6 + 4 + 1
2 + 17 + 16 − 35 = 0
Factorizando:

( − 1)( + 7)(2 + 5) = 0 ; > −1 D ≠ 0
= 1
= −7 − − − − − I Nfgh 2
= −
6
--------- no cumple
= 1 → ( )
141)
Q1 + 6 + Q4 6 − 2 = 4 ; 1+ 6 >0 y
4 6 − 2 > 0
X > 5;
Q1 + 6 + Q4 6 − 2 = 4
Q1 + 6 + Q2 6 − 2 = 4
( Q1 + 6 + Q2 6 − 2) = 16
1 + 6 + 2Q(1 + 6 )(2 6 − 2) + 2 6 − 2 = 16
3 6 − 1 + 2Q(1 + 6 )(2 6 − 2) = 16
3 6 + 2Q(1 + 6 )(2 6 − 2) = 17
2Q( 6 + 1)(2 6 − 2) = 17-3 6
2Q2 6 − 2 = 17 − 3 6
2√2 (Q 6 − 1 = 17 − 3 6

8( 6 − 1) = (17 − 3 6 )
8 6 − 8 = 289 − 102 6 + 9 6
6 − 102 6 + 297 = 0
6 =
, ±√ , - //
=
, ±0
6 = 99 → = 500
-----no cumple la igualdad
6 = 3 → = 5
= 125 → (7)
142)
+ (3 − 2) = 2 + 63; 3 − 2 > 0
> → > 0.63
6 + (3 − 2) = 2 + 63
(6 . (3 − 2)) = 2 + 63
(6 . (3 − 2)) = (2 . 63)
(2.3) (3 − 2) = 2 . 3 . 7
2 . 3 . 3 − 2 . 3 . 2 = 2 . 3 . 7
3 − 2. 3 = 3 . 7
(3 ) − 2(3 ) − 63 = 0
3 =
±√;& 6
=
±
s
3 ' = 9
3 = −7

3 ' = 9 → 3 ' = 3 → = 2
3 = −7 − − − I 2 K‡€2 2I ‡ ˆ27 2‡
= 2 → (=)
143)
;@ ( ‰)A = 0
;@ ( ‰)A = ;1
( ‰) = 1
( ‰) = 3
‰ = 3
‰ = 2 = 8
X = ‰ + 2‰ + 1
X = 8 + 2(8) + 1
X = 81 → ( )
144)
log(25 − ) − log ( + 1) = 0 ; 25 − > 0 D ( + 1) > 0
−5 < < −1 Š − 1 < < 5
log
6-
( & )
! = log 1
6-
( & )
= 1

25 − = ( + 1)
25 − = + 2 + 1
2 + 2 − 24 = 0
=
- ±√;& 0
;
=
- ± ;
;
5
= 3
= −4
+ = 3 − 4
+ = −1 → (7)
145)
+ 4 + 3 = 0 ; > 0
= €
€ + 4€ + 3€ = 0
€(€ + 4€ + 3) = 0 ; € = 0
€ + 4€ + 3 = 0
€ =
-;±√ -
=
-;±
→ 5
€ = −3
€ = −1
= 0 → = 2,
= 1
= −3 → = 2-
=
/
= −1 → = 2-
=
+ + = 1 +
/
+

+ + =
/
→ ( )
146)
3log(5 − ) = log (35 − )
5-x > 0 y 35- > 0 → < √35
%
→ < 3.27
log(5 − ) = log (35 − )
(5 − ) = 35 −
125-75x+15 − = 35 −
15 − 75 + 90 = 0
− 5 + 6 = 0
( − 3)( − 2) = 0
5
= 2
= 3
. = (2)(3)
. = 6 → (N)
147)
@ ( , )A = 1
@ ( , )A = 2
( , ) = 2
( , ) = 3

, = 3 = 9
= 100
E = log
X = , 100
X = 9 → ( )
148)
+ ; + / = 6 ; > 0
+ + % = 6
+ + = 6
3 = 6
= 2
= 2 = 4 → (7)
149)
E = √
%
2 ( & )
+ 5 ‹( &1)
= 7 {( & /)
2 + 3 > 0 D + 7 > 0 D 2 + 18 > 0 → > −
De; 7 n =

2 + 3 + + 7 = 2 + 18
= 8
E = √
%
X = '
%
2
X = '
%
2 = 9(1)
X = 9 → (N)
150)
E = √
− 8 2 = 3 ; > 0 , ≠ 1
−
/
= 3
−
/
= 3
− 3 − 4 = 0
=
±√0&
=
±6
= 4 → = 2;
= −1 → = 2-
=
E = √ = √16 = 4
E = √ = Q1/2 =
√
→ (=)

151)
√ + 7= − √ − 7= = 7=
log D = √ + 7= + √ − 7=
7=. D = (√ + 7= − √ − 7=)(√ + 7= + √ − 7=)
7=. D = ( + 7= − ( − 7=))
7=. D = 27=
D = 2
D = 10 = 100 → (=)
152)
+ log(1 + 2 ) = 5 + 72
+ log(1 + 2 ) = 5 + 72
+ log(1 + 2 ) = log(5 . 72)
= log(5 . 72) − log(1 + 2 )
= log
((6 .1 )
&
,10 = log
((6 .1 )
&
10 =
0./.6
&
5 . 2 =
0 ./.6
&
→ 2 =
0./
&

2 (1 + 2 ) = 72
(2 ) + 2 − 72 = 0
2 =
- ±√ & //
=
- ± 1
s
2 ' = 8
2 = −9
2 ' = 8 → 2 ' = 2 → = 3
2 = −9 → 2 = −3 − − − −I Nfgh 2
= 3 → (N)
153)
2 + = 1024
2 .
+ = 1024
+ = 1024
2 = 1024
= 512
( ) = 20)
. = 9
= 9
= ±3
= 3 → = 2 = 8
= −3 → = 2-
=
/
+ = 8 +
/

+ =
6
/
→ (2)
154)
5 − 3 ; = 56 ; x> 0
5 −
;
= 56
5logx.log4-3logxlog2 =56.log2.log4
. (5 4 − 3 2) = 56. 2. 4
. ( 46
− 2 ) = 56. 2. 4
.
;‹
%! = 56. 2. 4
. ( 21) = 56. 2. 4
7 . ( 2) = 56. 2. 4
7 = 7.8 4 ; x >0
= log 4/
= 4/
De: X = √
Œ
X =
'
Œ = (4/
)
'
Œ
X = 4 → ( )
155)

+ = 6 ; > 0
+ log = 6
+ log − 6 = 0
=
- ±√ & ;
=
- ±6
= 2 → = 10 = 100
= −3 → = 10-
=
,,,
Una raíz → = 100 → (=)
156)
6
‹ − 6 − 12 = 0 ; x > 0
6 . 6 − 6 − 12 = 0
6 − 6 − 12 = 0
6 =
±√ &;/
=
±1
6 = 4 → = 5;
= 625
6 = −3 → = 5-
=
6
. = 5;
. 5-
= 5
. = 5 → (N)

157)
− 8 2 − 5 ‹ = 0 ; > 0, x ≠ 1
− 8 2 − 3 ‹6
= 0
− 8 2 − 3 = 0
−
/
− 3 = 0
−
/
− 3 = 0
−
;
− 3 = 0
− 3 − 4 = 0
=
±√0&
=
±6
= 4 → = 2;
= 16
= −1 → = 2-
=
. = 16. = 8 → (N)
158)
+ ! = ' + ! ; + > 0 ; ≠ 1
+ > 0 ; ≠ 1 → > 0 ⋂ ≠ 1

xyz &
'
!
>yz
=
>yz &
'
!
>yz
'
log 2 = log
2 =
=
= ±
√
→ =
√
=
√
=
√
→ (7)
159)
log (2 − 1) ,,
+ log ( − 1) ,()* RR%M ,,
2 − 1 > 0 D − 1 > 0 → > 1
2003log(2 − 1) + log( − 1) ,,
= 2003
2003log(2 − 1) + 2003log( − 1) = 2003
log(2 − 1) + log( − 1) = 1
log[(2 − 1)( − 1)] = 10
(2 − 1)( − 1) = 10
2 − 3 + 1 = 10
2 − 3 − 9 = 0
=
±√0&1
;
=
±0
;
j
= 3
= −

= − − − − −I Nfgh 2
= 3 → (=)
160
( 7 + )( `7 + ` ) = `7 ,
> 0 , ≠ 1
( 7 + 2) ( 2 + ` ) = 10
n
+ 2! ( ` + 2) = 10
n &
n
)! ( ` + 2) = 10
(2 ` + 1)( ` + 2) = 10 `
2 ` + 5 ` + 2 = 10 `
2 ` − 5 ` + 2 = 0
` =
6±√ 6-
;
=
6±
;
` = 2 → = 7
` = → = 7
'
= √7
Un valor de x → = √7 → (N )

161)
,,()* &
,()*'RR
=
;
√
; x> 0
, ()* &
,
()*
'R
=
;
√
,()* &
,()*'R
=
;
√
&
=
;
√
−
;
√
+ 1=0
=
^
√%
±d
'm
%
-;
=
^
√%
±
√%
=
√
±
√
=
√
+
√
=
√
=
√
−
√
=
√
. =
√
.
√
=
. = 1 ( )

162)
+ + ; + = 7.5 ; > 0
+ ^ + + 2 = 7.5
+
;
+ + 2 = 7.5
3 +
;
+ = 7.5
6
;
=
16
,
= 2
= 2 = 4 → (=)
163)
√ + 14 + √ + 7 − log(1.2) = 1
+ 14 > 0 D + 7 > 0 → > −7
log•√ + 14. √ + 7 Ž − log(1.2) = 1
log•√ + 14. √ + 7 Ž = 10 + log (
,
)
log (Q + 21 + 98) = log (
,.
,
)
√ + 21 + 98 = 12

+ 21 + 98 = 12
+ 21 − 46 = 0
=
- ±√ & /;
=
- ± 6
5
= 2
= −23
= −23 → I Nfgh 2
= 2 → (=)
164)
>yz@√ %& 0A
>yz@√ & A
= 3
√ + 19 > 0 D √ + 1 > 0 → > 0 → √ > 0
log@√ + 19A = 3 log@√ + 1A
log@√ + 19A = log@√ + 1A
√ + 19 = @√ + 1A
√ + 19 = √ + 3 + 3√ + 1
3 + 3√ − 18 = 0
3(√ ) + 3√ − 18 = 0
√ =
- ±√0&
=
- ± 6

5 √ = 2
√ = −3
√ = 2 → = 2
√ = −3 → I Nfgh 2
= 4 → (7)
165)
log ( ) + − 4 4 = 0 : x>0
. + 2 − 4 2 = 0
+ 2 − 4(2) = 0
+ 2 − 8 = 0
=
- ±√;&
=
- ±
= 2 → = 10 = 100
= −4 → = 10-;
=
,^------solución extraña
El menor valor → = 100 → (2)
166)

15 + 2 50 − 6 = 3 ; > 0, ≠ 1
6
+
6,
− = 3
6
+
6,
− = 3 →
6
+
6,
− = 3
6& 6,-
= 3
6,&>yz (
'‹
m
)
= 3
xyz
‹R.'‹
m
!
= 3 →
>yz( 6)
= 3
xyz@6%A
= 3 →
6
= 3
log = log 5
= 5 → (N)
167)
&>yz ( - )
>yz(1 & )&>yz( - )&
=
7 + 1 > 0 D − 3 > 0 D − 6 > 0 → > 6
&>yz ( - )
>yz(1 & )&>yz( - )&
=
>yz (;.( - ))
>yz[(1 & ).( - )]&
=
>yz (;.( - ))
>yz[ (1 & ).( - )]
=

log@4. ( − 3)A = . log[3(7 + 1). ( − 6)]
log@4. ( − 3)A = log Q3(7 + 1). ( − 6)
4. ( − 3) = Q3(7 + 1). ( − 6)
[4. ( − 3)] = 3(7 + 1). ( − 6)
16( − 3) = (7 + 1). (3 − 18)
5 − 27 − 162 = 0
=
1±√ 1 & ;,
,
=
1±
,
j
= 9
= −
/
6
= −
/
6
→ I Nfgh 2
= 9 → ( )
168)
log ( + 8);
− log ( − 1);
= 4
( + 8);
> 0 D ( − 1);
> 0 → < −8 Š
−8 < < 1 Š > 1
4 log( + 8) − 4 log( − 1) = 4
log( + 8) − log( − 1) = 1
log ( + 8) − log( − 1) = 1

log
( &/
-
! = log 10
&/
-
= 10
+ 8 = 10 − 10
9 = 18
= 2 → (7)
169)
8 − 2 = ; > 0, ≠ 1
. 8 − . 2 =
. ( 8 − 2) =
. log (
/
)! =
. log(4) = 1
=
;
= ;10
^
^ ,
= ;10
; = ( ;10)
= 4( ^ ,).( ^ ,
= 10 ^ ,
-------(c)

170)
M = 5 ‹/
+ 9 %6
+ √ √23
^
• = 681. 5+ 5 %0
+ √ √23
^
• = 5. 681+ 5 % +
√
(√23
^
)
Por regla de la cadena:
• = 81 +5 + 23
• = 3;
+25+
• = 4 +25+ =
/&6,&
• =
60
→ (N)

Ecuaciones logaritmicas parte2

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