tema2 hoja3 logaritmos soluciones.desbloqueado

5/28/2018 Tema2 Hoja3 Logaritmos Soluciones.desbloqueado

1/11

IES Juan Garca Valdemora

Departamento de Matemticas

EJERCIC

1.-Calcula, aplicando la definicia)

327327log3 ==y

yy

Por tanto, 327log3 =

b) 642

164log

2

1 =

= y

y

Por tanto, 664log2

1 =

c) 1282128log2 == y yPor tanto, 7128log2 =

d) 32)2(32log2

==y y

Por tanto, 1032log2 =

e) 93

19log 33

3

1 =

=y

y

Por tanto,3

29log 3

3

1 =

f) ==22

)22(25,0logy

y

4322

3== yy

y

Por tanto,3

425,0log

22 =

g) =

=

2

12

1

2

1

82

1log

y

y

2

522 2

5

==

yy

Por tanto,2

5

82

1log

21

=

TEMA 2: POTENCIAS,

IOS LOGARITMOS. SOLUCION

, los siguientes logaritmos:

33

3

== y

6622 6 === yyy

727 == yy

1052

22225252

1

====

y

yy

y

3

2

3

23333 3

2

3 2 ==== yyyy

=

=

2

3

2

3

2

1

24

12

100

252225,

yy

3

4=

=

==

2

5

2

33

2

12

22

12

22

12 yyy

2

5=y

ADICALES Y LOGARITMOS

4 ESO Matemticas B

1

S

==

22

3

222

2

1y

y


2/11



h) 162

116log 33

2

1 =

=y

y

Por tanto,3

416log 3

21

=

i) ln 5 25 2 == eeeye yyPor tanto,

5

2ln

5 2 =e

j) ln 22 == ee

eey

e

e yy

Por tanto,2

3ln

2

=e

e

k) 00,0100001,0log ==y yPor tanto, 40001,0log =

l) existeno0log = exi(log xa

m) 10(10)10log( 6 == y yPor tanto, 6)10log(

6 =

n) existeno)10log( 6 = (log xa

o) 55555log5 ==y y

Por tanto,2

355log5 =

p) 1010010log 2== y yPor tanto, 1010log =

q) 216216log 55 16 == y y

Por tanto, 5

3

216log

5 1

6 =

TEMA 2: POTENCIAS,

3

42222 3

4

3 4 ==== yyyy

5

25

2

== ye

2

32

3

2

1

2

== yee

e

e y

4101014 == yy

)0te >x

6101066 == yy

)0existe >x

2

355555 2

3

2

1

=== yyy

11010 1 == yy

6666)6(66

5 35 131 ===

yyy


4 ESO Matemticas B

2

3

4

5

35

3

= y


3/11



r) =

=

5

10

5

104,0log

y

y

422 == y

y

Por tanto, 404,0log

5

1 =

s) ==334 10

14

1024

1log yy

3

5

3

102 == yy

Por tanto,3

5

1024

1log

34

=

t) 21282log 33128 == y yPor tanto,

21

12log 3128 =

u)9

3

9

1

9

3log

44

9

1 =

=y

y

Por tanto,8

7

9

3log

4

9

1 =

v)27

33

27

3log

44

3 ==yy

Por tanto,

4

5

27

3log

4

3 =

w) existeno)16(log2 = (log xa

x) 11ln33

=== ee

eye

yy

Por tanto, 31

ln3 =

e

y) basela,existeno81log 3 =

z) 1,001log >= aaa

TEMA 2: POTENCIAS,

( ) =

=

2

2

1

1 525

15

100

4504,

yy

y

4=

===

2

3

10

2

3 10

222

2

12

2

1)2(

24

yyy

2

1

3

17222)2( 3

1

73

1

7 ==== yyyy

( )4

7233

3

33 4

7

2

2

4

1

2 === y

yy

3333

3

33 4

5

2

3

4

1

2

3

4

1

====

yyyy

)0existe >x

33 = y

positrealnmerounserdebelogaritmoune


4 ESO Matemticas B

3

25

310

8

7=

1dedistintoyvo


4/11



2.-Halla el valor de las siguientesa)

4

1log243log

5

1log 163525 +

5125

51log)(

5525 == y

y

243243log)( 3 == yy

4

116

4

1log)( 16 == y

y

b) 6log7

1log5,0log 21649

62

36

186

18231 ==+=

,025,0log)( 662 == yy

7

149

7

1log)( 49 == y

y

62166log)( 216 == yy

64464log)( 4 == yy

c) = 255 5)008,025(log yy55)5(55

102310 == yy

Por tanto, )008,025(log 255

Otra forma (aplicando propie

( ) log5log5log

25log)008,025(log

5

23

5

10

5

51Prop.

25

5

=+=

=

d) ==

2

3

2 22

125,04log yy

2

2

)2(22

2

1

2

332

=

yy

TEMA 2: POTENCIAS,

expresiones:

10

501

2

15

10

1

2

15

10

1

)(

==

+=

51255

5

1)5( 51

2

5

1

2=== y

yy

533 5 == yy

24222)2( 2424 ==== yyyy

+=

=

3

1

2

1

6

13

3

1

2

1

6

164log

)(4

22222

125 6

1

6 16 ===

yyy

12777)7( 1212 ==== yyyy

3

113666)6( 133 ==== yyyy

3443 == yy

=

2

5225 51000

8)5(5)008,025( yy

455546 == yy

4

dades)

log45log65log105log

l1000

8

log)5(log008,0log

553Prop.

6

5

10

2

5

52

5

2

5

5

==+

=

+=+

=

=

2

1

2

3

2

2

3

2

2

1000

1252

22

125,04 yy

22

2

22

2

22 3

2

1

2

5

2

1

2

92

==

=

yyy


4 ESO Matemticas B

4

5

28

10

565==

101=

2

1

4

2= y

=3

6

1=

2

1

2

10

125

15

4145

125

1

log5g

1log5

2

5

10

5

==

=

+

=aa

2

1

2

3

2

2

8

12

3


5/11



Por tanto,2

125,04log

2

3

2 =


313

l2log22log

4log2

125,04log

2

1

22

9

2

2

22Prop.

2

3

2

==

=

=

=

e) ==

55

2

2 225,0

16log yy

2222 1017

52

17

== yy

Por tanto,1

1

25,0

16log 5

2

2 =


10171

10172log

1017

22

1

)2(log

25,0

16log

1log2

2

1

24

25

2

2

===

=

=aa

TEMA 2: POTENCIAS,

3

dades)

2log2

12log

2

52log2

og

(2log2log125,0

222

1

22

5

2

2

2

22

22

1

8

1

1000

125125,0

242

2

3

2

1

3

3

2

==

=

=

====

=

=

=

5

2

1

1

8

52

1

242

22

22

22

1

)2(2

25,0

16 yy

10

17=y

0

dades)

2log

2

2log

22

2log 2

17

2

5

1

2

1

8

2

5

1

2

1

1

8

2

5

1

=

=

=


4 ESO Matemticas B

5

2log32log2

6

2log)

1log22

2

1

22

3

3

==

=

=

aa

=

5

2

1

8

2

22y

2log3Prop.

10

17

2

5

1

==


6/11



3.-Halla el valor de en cada casEn todos los apartados aplicamo

a) 1727log 22 == xxx

b) 72

17log 2

1

== xxx

c) 72log 4247 === xxx

d) 41

491

41

491log ==

xx

e) 22

1log 2

1

2 ==

xxx

f)8

1

3

1log

3

1

8

1 =

= xxx

g) 772)7(log 27 == xx

h)3

1

2

1

3

1log 2

1

==

x

xx

i) 0,03001,0log 3 == xx

j) == 31 273

127log xx

13

13

9

9 === xxx

k)3

3 13log

x

exex ==

TEMA 2: POTENCIAS,

:

s la definicin de logaritmo y luego desarroll

71

71717 22 ===== xxxx

497)(7 22 == xx

774 2 = xrsimplifica

8

4

2

444

71

71)(

491 =

== xxxx

2

2

2

1

2

1

2

1 === xx

arracionaliz

2

1

8

13 =x

77

72== x

9333

112

1

2

1

==== xxx

10001000

1000

111 33

3 === xx

x

==== 333331

3

1

)(327

127

1xxx

x

9683

1

333

111

e

x

e

x

e

xe ===


4 ESO Matemticas B

6

amos

77

71 = xrracionalia

87

10=x

33)3(


7/11



l) 3015625,0log 3 == xx

4.-Sabiendo que 301,02log = ya) 2log)32log(12log 2

Prop.1

2 ==

b)Prop.10000

2log0002,0log =

=

4301,014301,0 ===

c) 6log516log6log Prop.351

5===

d) == l)100027log(27000log,43431,113477,03 =+=+

e)Prop.2

6log32log6

32log =

3log2log2log25

12

5===

=

f)simpli10000

125log0125,0log

=

)10log2(log

Prop.3parntesiQuitar

3 =+=

g) 55 log100

48log48,0log

==

04,00954,02408,0

10log3log2logProp.

5

2

5

1

5

4

=+=

=+=

h) 42Prop.4

6,0log1log6,0

1log =

TEMA 2: POTENCIAS,

64

11

1000000

156251015625, 3

33 == x

xx

477,03log = calcula:

477,0)301,0(23log2log23logProp.3

+=+=+

Prop.3

42log10log2log10000log2log ==

699,3

,0301,0(51)3log2(log

51)32log(

51

Prop.1 +=+=

+=+= l33log310log3log)103og(Prop.3

33

Prop.1

33

431

2

5

Prop.1

5log2(log2log)32log(2log +==

0255,0477,0301,0233log2log

23

2

3

==

Prop.2rfica)108log(080log1log

80

1log ==

903,11301,0310log2log3ys

===

5

1

5

4

2Prop.5

2

51

54

5

1

2

4

1log32log

10

32log

10

32

=

=

0638,

5

1301,0

5

410log

5

23log

5

12log

5

4

3.+=+=

444 l10

32log

10

6log6,0log0 =

===


4 ESO Matemticas B

7

464 == x

079,1477,0602,0 =+=

1log10log

==

aa

1556,0)477 =

=10og

parntesisQuitarProp.3

)3 =

Prop.1

3 )102log( =

1Prop.

5

2

0 =

15

2477,0 =

3Prop.

4

1

10

32og =


8/11



(lo4

1

10

32log

4

1

1Prop.2Prop.

=

=

i)

=

=

10

2log

10

36log6,3log

2

5,01477,02301,02 =+=

j)1Prop.log)1036log(360log ==

,0213log22log2 ==++=

k) 31Prop.3 9log5log)95log( +=13log

3

22log10log =+=

l) += lo2,3log)7,22,3log(1Prop.

3

3log910log2log5

lo10

3log3

10

2log

parntesisquitar

2Prop.

35

+=

=

+

=

5.-Pasa a forma algebraica:a) AC l22loglog3log

2

1+=

32

1

log2logloglog AC +=

2

2loglog

log2

loglog

23

23

23

BAC

BA

C

BA

C

=

=

+

=

b) yxz lo3loglog3

2log

3

1+=

3

2

3

1

loglogloglog syxz +=

TEMA 2: POTENCIAS,

) ( )1477,0301,04

110log3log2g +=+

+=+=

22log210log3log2log

3Prop.

22

Prop.12Prop.

2

56

2

1Prop.

22 3log2log1)32log(10log36 +=+=+

556,21477,0201 =++

2Prop.

3 2 3log2log10log3log2

10log +=+

=

017,1477,03

2301,0 =+

+

=+= log3

10

32log7,2log32,3log7,2

g3Prop.

3

3,0510log43log92log510log3

l2log5)10log3(log310log2g3Prop.

35

=+=

=+

Bg

2

sg

3


4 ESO Matemticas B

8

( ) 0555,0222,04

1=

=13log

3Prop.

2 1 =+

3Prop.

2

=

=

10

27

798,114477,091

)10log3log3(310g

=+

=+


9/11



33 2

3 logloglog sy

xz +

=

=3

3 2

3 loglog sy

xz

333 2

33 23

=

=

y

sxz

y

sxz

3

92

y

sxz

=

c) BAD log3log2log2 =2 lologlog100log AD =

=

=

=

43

2

4

3

2

3

2

log100log

:log100

log

loglog100

log

CBA

D

CB

A

D

CB

A

D

43

2

43

2

100

100

CB

DA

CB

A

D

=

=

d) CBA loglog3

1

2

1log +=

3

31

21

llog10loglog

loglog10loglog

BA

BA

+=

+=

5

3

3

log10

loglog

llog10

loglog

CB

A

CB

A

=

+

=

=

5 2

3 :

10

loglog DB

C

A

TEMA 2: POTENCIAS,

Clog4

43 logCB

Dlog5

2

5 2

52

logg

log

DC

DC

2

5 2og

D

D


4 ESO Matemticas B

9


10/11



5 23

10loglog A

DB

CA

=

6.-Toma logaritmos en las siguiea)

5

3

5

3

loglog

=

=

z

yxA

z

yxA

yxA l5loglog3log +=

b) 253 loglog == BzyxByxB log

2

5log

2

3log ++=

c) 2 loglog

=

=

D

XC

AD

XC

DXC loglog2log

+=

d) 54

5

loglog

=

=

AD

C

BAD

BAD log2

1log5log +=

e) loglog =

= ECB

AE

2

1

2

1

logloglog

= BAE

BAE log2

1log

2

1log =

f) 33 2 loglog =

= FCB

AF

3

1

3

2

logloglog

= BAF

BAF log3

1log

3

2log =

TEMA 2: POTENCIAS,

5 23

10

DB

C

=

tes expresiones y desarrolla:

Prop.1

53

Prop.2logloglog)log(log ==

xAzyxA

zog

Prop.1

2

5

2

3

253 logloglog =

= BzyxBzyx

zlog

Prop.1

2

Prop.2

2

log)log(loglog ==

CADXC

A

DXCA lo2

1loglog2loglog

2

1

parntessquitar

=

Prop.1

45

Prop.24loglog)log(log =

CBAD

C

B

Clog4

2

1

2

1loglogloglog

=

=

B

E

CB

AE

C

A

4

1

2

1

2

1

Prop.1

4

1

loglogloglog

+=

CBAE

Clog4

1

3

13

2

1

22

loglogloglog

=

=

B

F

CB

AF

C

A

6

1

3

1

3

2

Prop.1

6

1

loglogloglog

+=

CBAF

Clog6

1


4 ESO Matemticas B

10

Prop.3

5loglog + zy

Prop.3

2

5

2

3

logloglog ++ zyx

Prop.3

2

1

2 logloglog

+ ADX

A

Prop.3

42

1

5 logloglog += CBA

Prop.24

1

2

1

C

parntesisuitarProp.3

Prop.26

1

3

2

C

parntesisuitarProp.3


11/11



7.-Sabiendo que 301,02log = , loxalogBASEDECAMBIO =

a)l

5

3log

2log

3log

32log32log

5

3 ===

b)2log

10

3log

4log

3,0log3,0log

24

==

c)2log

3log

2log

27log27log

2

1

3

2 ===

d)lo3

log

2log

2log

8log

2log2log

38 ===

e)1

3log

2log

3log

8log8log

2

1

3

3 ===

f)2

1log

3log

5,0log

3log3log

5

1

55

5,0 ===

g) == l

21log

03,0log03,0log

33

2

1

( ) (

,02

1

47,03

1

2log2

1

10log23log3

1

=

=

TEMA 2: POTENCIAS,

477,03g = y utilizando el cambio de base

a

x

b

b

log

log

155,3477,0

301,05

3og

2log=

=

8,0602,0

523,0

301,02

1477,0

2log2

10log3log=

=

=

=

508,91505,0

431,1

301,02

1

477,03

2log2

1

3log3

==

=

3

1

2

2=

786,32385,0

903,0

477,0

2

1

301,03

3log

2log==

=

317,0301,0

0954,0

301,00

477,05

1

2log1log

3log5

1

=

=

=

(

=

=

=

log3

1

2log21

10

3log

3

1

2log21

100

3log

2og

100

3g 2

3

1

2

1

3

) ( )373,3

1505,0

523,13

1

301

2

=

=


4 ESO Matemticas B

11

calcula:

9

)=

2log21

10log32

tema2 hoja3 logaritmos soluciones.desbloqueado

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