More Related Content Similar to Ecuaciones exponenciales (segunda parte 2) (20) More from Widmar Aguilar Gonzalez (20) 1. ALGEBRA PREUNIVERSITARIA Ing. Widmar Aguilar, Msc.
EJERCICIOS RESUELTOS DE ALGEBRA
PREUNIVERSITARIA
ECUACIONES EXPONENCIALES
Ing. WIDMAR AGUILAR, Msc
Abril 2021
2. ALGEBRA PREUNIVERSITARIA Ing. Widmar Aguilar, Msc.
1)
(2√7) = 2 . 7
2 7 = 2 . 7 →
= 6 ;
2 =
X = 6 → = + 1 = 36+1
= 37 → )
2)
3 + 3 . 3 + 3 . 3 + 3 . 3 + 3 . 3 = 3. 11
3 1 + + + + ! = 3. 11
3 (121) = 3. 11 . 81.121
3 = 3.3 = 3#
= 5 → %)
3)
Elevando todo a cubo:
3. ALGEBRA PREUNIVERSITARIA Ing. Widmar Aguilar, Msc.
& ' = 3
( ) = 3
( ) = 3
Por analogía, se tiene:
= 3 → = √3 → ( ()
4)
2 − 3 . 3
*
+ = 3 . 3
*
+ − 2 . 2
2 (1+ ) = 3 ( 3
*
+ + *
+
)
. 2 = 3 (
,
*
+
)
2 = 2 .
+
2 . 3+ = 2 . 3
= ∧ =
= → (()
4. ALGEBRA PREUNIVERSITARIA Ing. Widmar Aguilar, Msc.
5)
4
*
/ +
#
= +
/
2. 2
+
/ . 2
+
/ + 5 2
+
/ = 9
2
+
/ = 0
2 0 + 50 − 18 = 0
0 =
# ± √ #,
=
# ±
0 = 2
0 = − − − − −34 56708974 % 2
De: 0 = 2
2
+
/ = 2 →
:
= 1 ; ; = 2
E =
:<,
=
<,
= 11 → (%)
6)
5. ALGEBRA PREUNIVERSITARIA Ing. Widmar Aguilar, Msc.
√ = &√ '
√ = +
Igualando los exponentes:
√2 − 1 = → 2 √2 − 1 =
Elevando toda la expresi[on al cuadrado:
&2 √2 − 1' =
4(2 − 1) =
− 8 + 4 = 0
=
± √
=
± √
= 4 + 2√3
= 4 − 2√3
. = (4 + 2√3)(4 − 2√3)
. = 16 − 4(3)
. = 4 → (=)
7)
(2() + (2() = 7
De:
= (2() − (2()
= (2() − 2(2() (2() + (2()
= (2() − 2 + (2()
+ 2 = (2() + (2()
6. ALGEBRA PREUNIVERSITARIA Ing. Widmar Aguilar, Msc.
( + 2) = (2() + 2. (2() . (2() + (2()
Como:
(2() + (2() = 7
+ 4 + 4 = 7+2
+ 4 − 5 = 0 ; = 0
0 + 40 − 5 = 0
0 =
±√ , >
=
±
0 = −5 0 = 1
= 0 → = −5 − − − −34 =6597
= 0 → = 1
E = ± 1
(2() − (2() = ± 1 ---------(1)
Como: 0 < ( <
0 < 2( < 1
Las soluión de (1) depende de; valor de X,
> 0 → (2() − (2() = -1
→ (=)
8)
(7 ) = 7
A
→ 7 .
= 7
A
Igualando exponentes: 4 . 8 = 4#
(2 ) . (2 ) = (2 )#
7. ALGEBRA PREUNIVERSITARIA Ing. Widmar Aguilar, Msc.
2#
= 2 >
→ 5 = 10
X = 2
E = − 2 + 5 = 2 − 2(2) + 5
= 5 → (=)
9)
5 . 5 + 5 = 5 . 6. 5
#
#
+
#
=
# . .
#+ ; 5 . 5 + 5 = 5 . 6
5 . 6 = 5 . 6 → 5 = 5
= 3
= √2 + 1 = B2(3) + 1
E = √7 ---------- (e)
10)
. − . − + 1 = 0
−
+
− + 1 = 0
− − . + = 0
( − ) − ( . − ) = 0
( − 1) − ( − 1) = 0
8. ALGEBRA PREUNIVERSITARIA Ing. Widmar Aguilar, Msc.
( − )( − 1) = 0
De: − 1 = 0 → = 1 → = 1
= 1
Y de : − = 0 → =
=
+ = 1 + = → (=)
11)
3 ,
+ 3 = 27 (3 + 3 ,
)
3 . 3 + 3 . 3 = (3 ) (3 + 3 , )
3 . 3 + 3 . 3>
= 3 . 3#
+ 3 . 3
3 (3 + 1) = 3 (3#
+ 3 )
+
=
>
3 = 243 → 3 = 3#
= 5 → (()
12)
8 − . 2 + 2 . 2 − = 0
9. ALGEBRA PREUNIVERSITARIA Ing. Widmar Aguilar, Msc.
8+8.2 = + . 2
8 1 + ! = ( 1 + 2 )
8
,
! = ( 1 + 2 ) → . 2 = 8
. 2 = 8 → . 2 = 2. 2
Por analogía: x = 2 → (()
13)
3 ,
+ 3 −
.#
C* = C+
3 ,
+ 3 − 5. 3 ( )
= 247. 3 (
3 ,
+ 3 ,
− 5. 3( )
= 247. 3( )
3 ,
+ 3 = 252. 3( )
3 . 3 + 3 . 3 = 252. 3 . 3
3 . 3 + 3 . 3 = 252. 3 . 3>
3 3 + ! = 252. 3 (1)
3 ! = 252. 3 → 3 = 729
3 = 3
2x = 6 → = 3 → (=)
14)
10. ALGEBRA PREUNIVERSITARIA Ing. Widmar Aguilar, Msc.
2 ,
+ (2) ,
= 2 . 5
2 . 2 + (2 ) . 2 = 2 . 5
2. 2 + (2 ) = 2 . 5
(2 ) + 2. 2 − 80 = 0 ; 2 = 0
0 + 20 − 80 = 0
(t + 10 ) ( t - 8 ) =0
E
0 = −10
0 = 8
0 = −10 − − − − − 34 =6597 , 34 G 940 3=8; % 2
2 = 8 → 2 = 2
= 3
= √ + 7 = √9 + 7
= 4 → (%)
15)
*H,
, + = 7
7 + 7 = 7 (7 + 7 )
7 + 7 = 7 . 7 + 7 . 7
7 (7 − 1) = 7 (7 − 1)
7 = 7
= 8
11. ALGEBRA PREUNIVERSITARIA Ing. Widmar Aguilar, Msc.
= √4 − 7
= B4(8) − 7 = √25
= 5 → (=)
16)
Descomponiendo cada radical:
B2 + √3 = I + I
B2 − √3 = I − I
(I + I + I − I ) = 6
*
(2I ) = 6
*
→ (√6) = 6
*
√6 = (6
*
)
*
H ; elevando todo al cuadrado:
6 = (6
*
)
+
H
6 = (6
*
)
+
++
→ 6 = (6
*
)
*C+
6 = 6
* .
2 ; 6 = 6
* .
2 ( )
12. ALGEBRA PREUNIVERSITARIA Ing. Widmar Aguilar, Msc.
6 = 6
J*
++ C* ….. elevando todo a: 2( )
6
(+ C*)
= 6
*
2( )
= 8 → 2( )
= (2 )
2 − 1 = 39
X = 20 → (()
17)
5 . 5 -5 . 5 + 5 . 5 − 5 . 5 = 5 . 744
5 5 −
#
! + 5 25 −
#
! = 5 . 744
5
#
! + 5
#
! = 5 . 744
5
#
+
#
! = 5 . 744
5
#
! = 5 . 744 → 5 = 5 . 5
5 = 5 ; = 4
= √ + 9 = √16 + 9
= 5 → (=)
18)
13. ALGEBRA PREUNIVERSITARIA Ing. Widmar Aguilar, Msc.
2 . 2 − 3 . 3 = 3 . 3
2 . 2 = 3 . 3 + 3 . 3
2 . 2 = 3 (3 + 9)
2 . 2 = 3 . 3. 2
2 . 2 = 3 . 3
( ) = → ( ) = !
2 = −1
= − → (;)
19)
4 + 4 . 4 + 4 . 4 +4 . 4 = 4 . 85
4 (1 + 4 + 16 + 64) = 4 . 85
4 (85) = 4 . 85
4 = 4
= 2 → (()
20)
2
C
= (2 )
K*
2
C
= 2 & K*'
Igualando los exponentes:
14. ALGEBRA PREUNIVERSITARIA Ing. Widmar Aguilar, Msc.
8 = 2(4 , )
(2 ) = 2(2 ,
)
2 = 2 ,
3 − 9 = 2 + 3
= 12 → (%)
21)
=
√
( ) =
√
√ √
= (
√
) .
+
+
( ) = ( )
*
+
.
+
+
( ) = ( )
*
H = ( )
*
H
.
+
+
( ) = ( )
*
J
.
+
+
( ) = (( )
*
*<)
Por analogía, se tiene:
= → (%)
15. ALGEBRA PREUNIVERSITARIA Ing. Widmar Aguilar, Msc.
22)
+
= + 1
Sumando y restando al exponente izquierdo 2,
+ ,
= + 1
& +, ' ( , )
= + 1
+& +K*'
+( K*) = + 1
+,
= ( + 1). ( , )
.
+
= ( + 1). ( , )
Por analogía: = + 1
= +
=
,
= + 1+
,
=
+, ,
,
=
, , ,
,
=
( , )
( , )
= 3 → (=)
23)
(B4 − √15) >
+ (B4 + √15) >
= (2√2)
16. ALGEBRA PREUNIVERSITARIA Ing. Widmar Aguilar, Msc.
(I4 − √15)
>
+ (I4 + √15)
>
= 8
Haciendo : 0 = (4 + √15)L >
0 = (
& ,√ #'& √ #'
√ #
)L >
0 = (
√ #
)L >
→ (4 + √15) (L >)
= M+
√0 +
√M+
= 8
0 +
M
= 8 → 0 − 80 + 1 = 0
0 =
±√
=
± √ #
= 4 ± √15
0 = 4 + √15 ; 0 = 4 − √15
Si: 0 = 4 + √15
(4 + √15) = (4 + √15) * >
− 10 = 2
= 12
Si: 0 = 4 − √15
&4 − √15' = (4 − √15) + >
− 10 = −2
= 8
+ = 12 + 8
+ = 20 → (;)
17. ALGEBRA PREUNIVERSITARIA Ing. Widmar Aguilar, Msc.
24)
0 =
#
2(3 M, ) − 13(6M) + 6(2 M) = 0
2(3 )M
. 3) − 13(2.3)M
+ 6(2 )M
= 0
6. 9M
− 13. 2M
. 3 M
+ 6. 4M
= 0
6(4M
+ 9M
) = 13. 2M
. 3 M
3.2.2 M
+2.3.3 M
− 13. 2M
. 3 M
= 0
3. 2 M,
− 13. 2M
. 3 M
+ 2. 3 M,
= 0
3. 2 M,
− 4. 2M
. 3 M
+ 2. 3 M,
− 3 . 2M
. 3 M
= 0
(3. 2 M,
-3 . 2M
. 3 M
) + ( 2. 3 M,
− 2 . 2M
. 3 M) = 0
3. 2M
(2M,
− 3. 3M
) + 2. 3M
( 3M,
− 2. 2M
) = 0
3. 2M
(2M,
− 3M,
) − 2. 3M
( 2M,
− 3. 3M
) = 0
(2M,
− 3M,
) (3. 2M
− 2. 3M
) = 0
De: 2M,
− 3M,
→ 2.2M
= 3.3M
( )M
= ( )
0 = −1
= 5(−1) = −5
3. 2M
− 2. 3M
→ ( )M
= ( )
0 = 1
= 5(1) = 5
18. ALGEBRA PREUNIVERSITARIA Ing. Widmar Aguilar, Msc.
. = (5)(−5) = −25
. = −25 → (()
25)
( )+
= 2 + 1
+ ,
= 2 + 1
+ , ,
= 2 + 1 →
+, ( , )
= 2 + 1
+K+
(+ K*) = 2 + 1
.
+
= (2x+1). ( , )
:or analogias, se tiene:
= 2 + 1 ---------(1)
= − 5 = ( − 5)
Restando -5 a toda la ecuación (1),
− 5 = 2 + 1 − 5
− 5 = 2 − 4
= (2 − 4) = 2 − 4
= 2(2 + 1) − 4 = 4 + 2 − 4
= 2 → (%)
26)
19. ALGEBRA PREUNIVERSITARIA Ing. Widmar Aguilar, Msc.
3.3 + 3 . 3 + 3 . 3 = 3 .117
3 (3 + 9 + 27) = 3 .117
3 . 3.13 = 3 . 117
3 . 3 = 3 . 3
3 = 3
= 4 → (;)
27)
4 . 4 − 6 = 2. 9 . 9
4 . 4 − 2 . 3 = 2. 9 . 9
4. 4 − 2 . 3 = 18. 9
Si: 2 = ; → 4 = ;
3 = ( → 9 = (
4; − ;( − 18( = 0
; =
N±√N+, N+
=
N± N
; = (
; = −2(
; − (! (; + 2() = 0
O2 − . (3 )P (2 + 2. 3 ) = 0
De: 2 + 2. 3 = 0 → 2 = − 2. 3
! = −2 − − − −34 =6597 .
20. ALGEBRA PREUNIVERSITARIA Ing. Widmar Aguilar, Msc.
2 − . (3 ) = 0 → 2 =
+.
+
! = ( ) → ! = ( )
= −2
= √2 − 2
= B2(−2) − 2
= √6 → (%)
28)
√ , = 9
( ,
)
*
= (−3)
( ,
)
*
= (−3)( )( )
= (−3 )
( ,
)
*
= (− ) .
( ,
)
*
= (− )
+
!.( )
( ,
)
*
= (− )
*
, !.( )
( ,
)
*
= (− )
*
, !(
*
C
*)
Por analogía:
= -
E = 9 − 6 + 4
21. ALGEBRA PREUNIVERSITARIA Ing. Widmar Aguilar, Msc.
= 9(−1/3) − 6 − ! + 4
= 1 + 2 + 4
= 7 → (%)
29)
32# K+
= (2#*+C
)
*
A
( 32# K+
)# R
2#*+C
32# .# K+
= 2#*+.#C
(2#
)#+ K+
= 2#*+.#C
, igualando exponentes:
5. 5 ,
= 5 . 5
5 ,
= 5 , igualando nuevamente
exponentes:
2 + 3 = 12-x
3 = 9 → = 3 → (()
30)
C+S
=
>
− = ( )( − )
>
− = − ,
22. ALGEBRA PREUNIVERSITARIA Ing. Widmar Aguilar, Msc.
>
=
3 − 20 = 2 − 1
= 19 → (;)
31)
(128
+ CT
)#
= 2#
[(2 )
+ CT
]#
= 2#
2# . . + CT
= 2#
, igualando exponentes,
5 . 7. 7 = 5
5 . 7 = 5
7 =
# CJ
#
; 7 = 5
7 . 7 = 5 . 5
#
! =
#
!
#
! =
#
!
2 = 8
= 4 → (%)
32)
2 + 128 = 3. 3 ,
23. ALGEBRA PREUNIVERSITARIA Ing. Widmar Aguilar, Msc.
(2 ) − 3. 2 . 8 + 128 = 0
(2 ) − 24. 2 + 128 = 0
2 =
±√ + #
=
±
2 = 16 2 = 8
2 = 2 → = 4
2 = 2 → = 3
+ = 4 + 3
+ = 7 → (%)
33)
C+
= 2
(
C+
) = 2
C+
= 2 → ( )
C+
= 2
Por analogía,
= 2 → = 2
= ( ) /
= [( )
*
+]
= (
√
)
Por analogía: x =
√
=
√
→ (()
24. ALGEBRA PREUNIVERSITARIA Ing. Widmar Aguilar, Msc.
34)
A, W
+ 5.7 = 7
A
+ 5
W
A
.
W
+ 5.7 = 7
A
+ 5
W
A
.
W
- 7
A
= 5
W
− 5.7
A
(
W
− 7) = 5 (
W
− 7)
A
= 5
A
= √5
A #.
A
A
A
= (√5
A
)( B#)
A A
Por analogía: x = √5
A
= √5
A
→ (%)
35)
,
= 9 − 1 = 8
. = 2.2
Por analogía,
= 2
= √ + B +
H
= √2
+
+ B2
+
H
= 2 + 2
25. ALGEBRA PREUNIVERSITARIA Ing. Widmar Aguilar, Msc.
= 6 → (=)
36)
2 . 2 − 2 . 2 + 2 . 2 − 2 = 50
2 (8 − 4 + 2 − 1) = (2.25)
2 (5) = 2 . 5
5 = 5
2 = 1
=
= = ( )
= → (%)
37)
27
K
= √3
(27
K
) = 3
(3 ) ( K )
= 3 → 3 . K
= 3
3
KH
= 3, 8X6;7;3%4 943 30 G,
9 ,
= 1 → 3 ,
= 3>
2 + 8 = 0
26. ALGEBRA PREUNIVERSITARIA Ing. Widmar Aguilar, Msc.
= −4
E = x+1
= −4 + 1
= −3 → (;)
38)
(√ + 1)(√ , )&√ K*'
+
= 2
Haciendo: (√ + 1) = 0
0MY+
= 2
0MY+
= (√2) = (√2)(√ )+
0MY+
= (√2)(√ )(√+)+
Por analogía,
0 = √2
√ + 1 = √2
√ = √2 − 1
= (√2 − 1) → (()
39)
5
C+
+ . (125)H = (125)
K+
J . (625)
KH
*<
27. ALGEBRA PREUNIVERSITARIA Ing. Widmar Aguilar, Msc.
5
C+
+ . (5 )H = (5 )
K+
J . (5 )
KH
*<
5+. 5 H . 5 = 5 J . 5H. 5H. 5
5+
, H . 5 = 5 J
,H. 5 ,H
5
A
H . 5 = 5
A
J . 5
W
H
5
A
H = 5
A
J
,
W
H → 5
A CH
H = 5
A K*H
J
#
=
# ,
10 − 8 = 5 + 14
5 = 22
=
#
→ (=)
40
+K+
= 4
+K+
= √2 = √2
√
H
+K+
= √2
√
√+
+
K+
Por analogía, se tiene:
= √2 → (;)
28. ALGEBRA PREUNIVERSITARIA Ing. Widmar Aguilar, Msc.
41)
( )(
*
+
)H
=
√
(4 )( C*)H
=
√
4
CH
=
√
4 = (
√
)
*
+CH = (
√
)
H
4 = (√2)
H
(2
*
+)
H
= 2
2 = 2 → 4 = 2
2 = 2 → 2 = 1
= → (()
42)
√
H
.
*
H
= ( I
+√+
)√
√
H
.
*
H
= ( I
+√+
)√
29. ALGEBRA PREUNIVERSITARIA Ing. Widmar Aguilar, Msc.
(
*
H
)
*
H
= (( )
*
+√+
)√
(
*
H
)
*
H
= (2
*
+√+)√ = (2
C√+
)√
(
*
H
)
*
H
= (2 √ )
C√+
, luego:
*
H
= 2 √
*
H
= 2 √ .
+
+ = (2 )
√+
+ = (2 )
*
√+
*
H
= (2 )(
*
+
)
*
+
= ( )(
*
+
)
*
+
.
+
+
*
H
= ( )(
*
H
)
*
H
=
= √
K*
= Z !
*
H
*
H!K*
= I( )
W
H
W
H
= !
= 4 → (%)
43)
,
= [( + 1)( , )
]
,
= [( + 1) ( , )
2 = ( + 1)
+,
. ( + 1)
30. ALGEBRA PREUNIVERSITARIA Ing. Widmar Aguilar, Msc.
( + 1)
+, ,
= 2
( + 1)( , )+
= 2
De: 2 = √2
( + 1)( , )+
= √2
+.+
+
= √2
√
+
( + 1)( , )+
= √2
√
+
Por analogía,
+ 1 = √2
= √2 − 1 → (=)
44)
Z I B √ ]C+^
^C^
^C^
^+^
= √
Si : 3 ]
= 0
3 ]
= 0 →
]+^ = 0
Z I B √ M+
Y
Y
*
Y+
= √
Z I ( √ M+
)
*
Y
Y
*
Y+
= √
31. ALGEBRA PREUNIVERSITARIA Ing. Widmar Aguilar, Msc.
Z I ( √ .MC* M)
Y
*
Y+
= √
I B √ .MC* M,
Y
*
Y+
= √
I . ( √ .MC* M, )
*
Y
*
Y+
= √
I . ( √ .MC+ ,MC*
)
*
Y+
= √
B √ .MC+ ,MC*,
*
Y+
= √
B √ .MC+,MC*
*
Y+
= √
( √ .MC+,MC*
)M+
= √
√ ,M
= √
→ √2 + 0 = 2√2
0 = √2
3 ]
= √2
3]
= √2 → 3]
=
√
3]
= *
+
= ( )
*
+
Por analogía: n = = 0.5 → (;)
32. ALGEBRA PREUNIVERSITARIA Ing. Widmar Aguilar, Msc.
45)
[( )
C
] =
*
+√+
( . C
) =
*
+√+
. . C
=
*
+√+
*K*C
=
*
+√+
+C
=
*
+√+ , luego:
=
√
→ =
.
*
+
=
+
= ( )+
Descomponiendo el exponente (3/2),
= !
*
+
Por analogía, se tiene:
= → (()
46)
33. ALGEBRA PREUNIVERSITARIA Ing. Widmar Aguilar, Msc.
: K*, + K+
H KH,: K* = !
,
: K*, + K+
H KH,: K* = (3) ( , )
: K*, + K+
H KH,: K* = 3
; ,
+ 3 ,
= 3 . (3 ,
+ ; ,
)
; ,
+ 3 ,
= 3 ,
+ ; ,
. 3
; ,
+ 3 ,
= 3 ,
+ ; ,
. 3
; . ; +9.3 = 27. 3 + ; . ;. 3 . 3
; . ; 1 −
.
! = 9(3.3 − 3 )
; . ;
.
.
! = 3 . (3.3 − 3 )
; . ;(3.3 − 1) = 3 . 3 . 3 (3.3 − 1)
; . ; = 3 . 3
; ,
= 3 ,
= 3 ( , )
; ,
= 27 ,
; = 27 → (=)
47)
2 −
#
(2 ) + 1 = 0
2 = 6
6 −
#
6 + 1 = 0
34. ALGEBRA PREUNIVERSITARIA Ing. Widmar Aguilar, Msc.
86 − 656 + 8 = 0
6 =
#±√ #+ .
=
#±
6 = 8
6 =
6 = 8 → 2 = 6 → 2 = 2
= 3
6 = → 2 = 6 → 2 = 2
= −3
*
+
= = −1 → (()
48)
3 . 3 + 3 − 2. 3 . 3 − 3
H
+ = 3
3 . 3 + 3 − 2. 3 . 3 − 3 = 3
3 . 3 − 2. 3 . 3 = 3
3. 3 = 3
3 ,
= 3
+ 1 = 4
= 3 → (;)
35. ALGEBRA PREUNIVERSITARIA Ing. Widmar Aguilar, Msc.
49)
[
(_K*)__+C
__ ]
*
_
`_ =
√
(_K*)
_ .`
_+C
_
`_ = *
(_K*)
_ .
_+C
_
`
= 3
*
_K C _
_ .
_+C C_+
_ = 3
*
_.
C
_ = 3
*
_
`_
= 3
*
→ (
`
)_ = *
(
`
)_ = ( )
*
`
= → = 3 → (=)
50)
= √81
J*CJ*
= (81)
*
J*CJ*
= 81
J*
36. ALGEBRA PREUNIVERSITARIA Ing. Widmar Aguilar, Msc.
= 81
J*
= (81)( )(J*)
El lado izquierda quedaría:
(√ ) = (81)( )(J*)
(√ )( √ )( √ )
= (81)( )(J*)
√ = 81
*
= 81
E = √
H
= ( )
*
H =
*
!
*
H
= (81)
*
H = √3
H
= 3 → (()
51)
= √81
CJ*J*C
= (81)
*
CJ*J*C
= 81
CJ*C
= 81 ( C*)(J*C*)
= 81 (
*
J*
)
(
*
J*
)
= (81 )(
*
J*
)
(
*
J*
)
= ( )(
*
J*
)
(
*
J*)
El lado izquierda quedaría:
37. ALGEBRA PREUNIVERSITARIA Ing. Widmar Aguilar, Msc.
(√ ) = ( )(
*
J*
)
(
*
J*)
(√ )( √ )( √ )
= ( )(
*
J*
)
(
*
J*
)
√ =
*
=
E = √
H
= ( )
*
H =
*
!
*
H
= ( )
*
H = √3
H
= 3 → (()
52)
(2 )
C*
= (2
*
)
KA
(2 )
( C*)
= (2
*
)
+( KA)
, igualando exponentes,
3. 3 ( )
= (1/3). 3 ( ,#)
3 ( ),
= 3 . 3 ( ,#)
3 = . 3 ,
3 − 2 = 2 + 9
= 11 → ( )
38. ALGEBRA PREUNIVERSITARIA Ing. Widmar Aguilar, Msc.
53)
3 ,
+ 3 −
.#
C* = C+
3 ,
+ 3 − 5. 3 ( )
= 247. 3 (
3 ,
+ 3 ,
− 5. 3( )
= 247. 3( )
3 ,
+ 3 = 252. 3( )
3 . 3 + 3 . 3 = 252. 3 . 3
3 . 3 + 3 . 3 = 252. 3 . 3>
3 3 + ! = 252. 3 (1)
3 ! = 252. 3 → 3 = 729
3 = 3
2x = 6 → = 3 → (()
54)
= 2
= √4 = 4
*
+
= 4 (
*
+
)
= ( )
*
+
= ( )
*
+
