Teoria Determinare direttamente il valore dei seguenti logaritmi:

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Teoria Determinare direttamente il valore dei seguenti logaritmi:

1)

^log11¹= ^x

11

^x

= 1 11

^x

= 11

⁰

x = 0 .

2)

^log6⁶= ^x 6^x = 6 6^x = 6¹

x = 1 .

3)

^log7⁴⁹= ^x 7^x = 49

7

^x

= 7

²

x = 2 .

4)

^log3²⁷= ^x 3^x = 27 3^x = 3³

x = 3 .

5)

^log2¹⁶= ^x 2^x = 16

2

^x

= 2

⁴

x = 4 .

6)

^log2³²= ^x 2^x = 32

2

^x

= 2

⁵

x = 5 .

7)

^log8²= ^x 8^x = 2

2

^3x

= 2 3 x = 1

^x= ₃¹

.

8)

^log6³⁶= ^x 6^x = 36 6^x = 6²

x = 2 .

9)

^log36⁶= ^x 36^x = 6 6^2x = 6

2 x = 1

^x= ₂¹

.

10)

^log9⁸¹= ^x 9^x = 81 3^2x = 3⁴

2 x = 4

^x= ₂⁴

x = 2 .

11)

^log9²⁷= ^x 9^x = 27 3^2x = 3³

2 x = 3

2 x= 3

.

12)

^log1₃^{3= x 3}

3 1^x =



 



 ^{x 1}

3 1 3

1 ⁻



 



= 



 





x = − 1 .

13)

^log₅^{1 25= x 25} 5

1 ^x

 =



 



 2

x

5 5 1 =



 



 ^{x 2}

5 1 5

1 ⁻



 



= 



 





x = − 2 .

14)

^log₅^{1125= x 125} 5

1^x =



 



 3

x

5 5 1 =



 



 ^{x 3}

5 1 5

1 ⁻



 



= 



 





x = − 3 .

15)

^log1₃^{27= x 27} 3

1 ^x

 =



 



 3

x

3 3 1 =



 



 ^{x 3}

3 1 3

1 ⁻



 



= 



 





x = − 3 .

16)

^log1₂^{64= x 64} 2

1^x =



 



 6

x

2 2 1 =



 



 ^{x 6}

2 1 2

1 ⁻



 



= 



 





x = − 6 .

17)

^log ₂^{1 x}

2

1 =

2 1 2

1^x =



 





1 x

2 1 2

1 =



 



 ^{x 1}

2 1 2

1 



 



= 



 





x = 1 .

18)

^log ₄^{1 x}

2

1 =

4 1 2

1^x =



 





2 x

2 1 2

1 =



 



 ^{x 2}

2 1 2

1 



 



= 



 





x = 2 .

19)

^log ₄₉^{1 x}

7

1 =

49 1 7

1^x =



 





2 x

7 1 7

1 =



 



 ^{x 2}

7 1 7

1 



 



= 



 





x = 2 .

20)

^log2 ₆₄^{1 = x 2x =} ₆₄¹

6 x

2 2 1



 



= 

2

^x

= 2

⁻⁶

x = − 6

.

21)

^log2₁₂₈¹ = ^x

128 2^x = 1

7 x

2 2 1



 



= 

2

^x

= 2

⁻⁷

x = − 7

.

22)

^log ₁₂₈^{1 x}

2

1 =

128 1 2

1^x =



 



 ^{x 7}

2 1 2

1 



 



= 



 





x = 7 .

23)

^log ₉^{4 x}

3

2 =

9 4 3

2^x =



 



 ^{x 2}

3 2 3

2 



 



= 



 





x = 2 .

24)

^{log 16}₈₁ ^x

3

2 =

81 16 3

2^x =



 



 ^{x 4}

3 2 3

2 



 



= 



 





x = 4 .

25)

^{log 25}₄ ^x

5

2 =

4 25 5

2^x =



 



 ^{x 2}

2 5 5

2 



 



= 



 



 ^{x 2}

5 2 5

2 ⁻



 



= 



 





x = − 2

.

26)

^{log 125}₂₇ ^x

5

3 =

27 125 5

3^x =



 



 ^{x 3}

3 5 5

3 



 



= 



 



 ^{x 3}

5 3 5

3 ⁻



 



= 



 





x = − 3

.

27)

^log ₂^{3 x}

4

9 =

2 3 4

9 ^x

 =



 





2 3 2

3 ^2x

 =



 





2 x = 1

^{x= 1}₂

.

28)

^log ₇^{6 x}

36

49 =

7 6 36

49^x =



 



 ^{2x 1}

6 7 6

7 ⁻



 



= 



 





2 x = − 1

^{x= −} ₂¹

.

29)

^log2 ²= ^x 2^x = 2 2

1

x

2

2 =

2

x= 1

.

30) log

₂³

2 = x 2

^x

=

³

2

3

1

x

2

2 =

3

x= 1

.

31) log

₂³

4 = x 2

^x

=

³

4

2^x = ³ 2²

2

^x

= 2

³² 3 x= 2

.

32) log

₃³

3 = x

3^x = ³ 3 3

1

x

3

3 =

3

x= 1

.

33) log

₃³

9 = x

^3x = ^{3 9} 3^x = ³ 3² 3

2

x

3

3 =

3

x= 2

.

34)

^{log 3 9 x}

3

1 = ^{x 3} ₉

3 1 =



 



 3

x 2

3 3 1 =



 



 ^{x 32}

3 1 3

1 ⁻



 



= 



 



 3

x= − 2

.

35)

^{log 3 5 x}

5

1 = ^{x 3} ₅

5 1 =



 



 ^x 5₃¹

5 1 =



 



 ^{x 13}

5 1 5

1 ⁻



 



= 



 





3 x = − 1

.

Dato il logaritmo e la base determinare il numero:

1)

^log3^x= ⁰ 3⁰ = x x= 3⁰

x = 1 .

2)

^log3^x= ¹ 3¹ = x x= 3¹

x = 3 .

3)

^log3^x= ² 3² = x x= 3²

x = 9 .

4)

^log3^x= ⁰ 3⁰ = x x= 3⁰

x = 1 .

5)

^log2^x= ⁴

2

⁴

= x x = 2

⁴

x = 16 .

6)

^log8^x= ² 8² = x x= 8²

x = 64 .

7)

^log10^x= ² 10² = x x= 10²

x = 100 .

8)

^log10^x= ⁴ 10⁴ = x x= 10⁴

x = 10000 .

9)

2

x 1

log₂ =

2

²

x

1

=

²

1

2

x =

^{x= 2}

.

10)

2

x 1

log₄ =

4

²

x

1

= x = 4

^{21 x= 4}

x = 2 .

11)

3

x 1 log₂₇ =

x 27

³

1

=

³

1

27

x = ^{x =}

³

^{27 x = 3} .

12)

^log₄^{1x= −1 x} 4

1 ¹ =



 



 ^{− 1}

4 x 1

−



 



= 

x = 4 .

13)

^log4^x= −²

4

⁻²

= x x = 4

⁻²

2

4 x 1



 



= 

16 x= 1

.

14)

^log11^x= −²

11

⁻²

= x x = 11

⁻²

2

11 x 1 



 



= 

121 x= 1

.

15)

^log₅^{1 x= 3 x} 5

1³ =



 



 ³

5 x 1



 



= 

125 x= 1

.

16)

^log₄^{1 x= −3 x} 4

1 ³ =



 



 ^{− 3}

4 x 1

−



 



= 

x = 4

³

x = 64 .

17)

^log₂^{1 x= −6 x} 2

1 ⁶ =



 



 ^{− 6}

2 x 1

−



 



= 

x = 2

⁶

x = 64 .

18)

^{log0,1x= 2 log}₁₀^{1 x= 2 x} 10

1 ² =



 



 ²

10 x 1 



 



= 

100 x= 1

.

19)

^log0,2^x= ² log x 2

5

1 = _x

5 1² =



 



 ²

5 x 1



 



= 

25 x= 1

.

20)

^{log0,5x= 3 log1}₂^{x= 3 x} 2

1³ =



 



 ³

2 x 1



 



= 

8 x= 1

.

Determinare la base dei seguenti logaritmi:

1)

^logx⁴= ²

x

²

= 4 x

²

= 2

²

x = 2 .

2)

^logx⁹= ² x² = 9 x² = 3²

x = 3 .

3)

^logx¹⁴⁴= ²

x

²

= 144 x

²

= 12

²

x = 12 .

4)

^logx²⁷= ³

x

³

= 27

^x3 = ³³

x = 3 .

5)

^logx⁸= ³ x³ = 8

x

³

= 2

³

x = 2 .

6)

^logx¹⁰⁰⁰= ³ x³ = 1000 x³ = 10³

x = 10 .

7)

^logx¹⁰⁰⁰⁰= ⁴ x⁴ = 10000 x⁴ = 10⁴

x = 10 .

8)

^logx³²= ⁵ x⁵ = 32

x

⁵

= 2

⁵

x = 2 .

9)

^logx²⁴³= ⁵ x⁵ = 243 x⁵ = 3⁵

x = 3 .

10)

^logx⁶⁴= ³ x³ = 64

x

³

= 4

³

x = 4 .

11)

^logx¹²⁵= ³ x³ = 125 x³ = 5³

x = 5 .

12)

^logx⁶⁴= ⁶ x⁶ = 64

x

⁶

= 2

⁶

x = 2 .

13)

^logx¹⁶⁹= ² x² = 169 x² = 13²

x = 13 .

14)

^logx²= −¹

x

⁻¹

= 2 x

⁻¹

= 2

¹

1 1

2 x 1

−

− 



 



= 

2 x= 1

.

15)

^logx¹⁶= −⁴ x⁻⁴ = 16

x

⁻⁴

= 2

⁴

4 4

2 x 1

−

− 



 



= 

2 x= 1

.

16)

^logx⁸= −³ x⁻³ = 8

x

⁻³

= 2

³

3 3

2 x 1

−

− 



 



= 

2 x= 1

.

17)

^logx²⁵= −² x⁻² = 25 x⁻² = 5²

2 2

5 x 1

−

− 



 



= 

5 x= 1

.

18)

2

2 1

log_x =

x

²

2

1

= x

²¹

= 2

¹²

x = 2 .

19)

2

3 1

log_x =

x

²

3

1

= x

²¹

= 3

²¹

x = 3 .

20)

3

2 1

log_x³ = 3 3 1

2

x =

³

1 3 1

2

x = x = 2 .

21)

5

2 2

log_x^{5 2} = 5 5 2 2

2

x = x

⁵²

= 2

⁵²

x = 2 .

22)

3

2 9

log_x³ 1 = ^{3 3}

2

9

x = 1 ³

1 3

2

9

x 1



 



=  ³

2 3

2

3

x 1



 



=  3

x= 1

.

23)

3

2 25

log_x³ 1 = ^{3 3}

2

25

x = 1 ³

1 3

2

25

x 1 



 



=  ³

2 3

2

5

x 1



 



=  5

x= 1

.

Applicando le proprietà sui logaritmi trasformare i seguenti logaritmi neperiani in somme algebriche di logaritmi:

1)

^{log2a2b3 =}

b log 3 a log 2 2 log b log a log 2

log + ² + ³ = + +

=

.

2)

log3a⁴ b =

log b

2 a 1 log 4 3 log b log a log 4 3 log b log a log 3

log

²

1

4

+ = + + = + +

+

= .

3)

=

+ 3 x

x log 5

2

(

^{x 3}

)

^{log5 2logx log}

(

^{x 3}

)

log x 5

log ² − + = + − +

=

.

4)

₃ ⁼

2 3

y y x log27

=

−

=

−

= ³

1 2

3 3 3

2

3y log y log3 x y logy

x 27 log

(

^{log3 logx}

)

₃^5logy

3 y 3log y 1 log 2 x log 3 3 log

3 + + − = + +

=

.

5)

₄ ⁼

4

z y x log81

=

−

=

−

= log81x⁴y log⁴ z log3⁴x⁴y logz⁴¹

( )

^logz

4 x 1 log 3 log 4 z 4log y 1 log x log 4 3 log

4 + + − = + −

=

.

6)

₂⁴₃ = d c

b a log5

=

−

− +

+

=

−

= log5a⁴b logc²d³ log5 loga⁴ logb logc² logd³ d

log 3 c log 2 b log a log 4 5

log + + − −

=

.

7)

⁵₄ ³ =

cd b a log49

=

−

− +

+

=

−

= log49a⁵b³ logcd⁴ log7² loga⁵ logb³ logc logd⁴ d

log 4 c log b log 3 a log 5 7 log

2 + + − −

=

.

Tenendo presente le proprietà sui logaritmi ridurre ad un unico logaritmo neperiano ciascuna delle seguenti espressioni:

1)

^{+ − log25+ log27=}

2 9 1 log 5 log

= +

− +

= +

− +

= log5 3log3

2 3 2 log 2 5 log 3 log 5 2log 3 1 log 5

log ^{2 2 3}

3 log 5 3 log 3 5 log 3 log 2 5

log + − + =

=

.

2)

^{log12− log6+ log125− 1}₅^{log32− log25=}

(

^⋅

)

⁻

(

^⋅

)

^{+ − − =}

= ^{2 3} log2⁵ log5²

5 5 1 log 3 2 log 2

3 log

(

⁺

)

^{+ − − =}

− +

= ^{2 3} log2⁵ log5²

5 5 1 log 3 log 2 log 2

log 3 log

=

−

− +

−

− +

= log2 2log5

5 5 5 log 3 3 log 2 log 2 log 2 3 log

5 log 5 log 2 2 log 5 log 3 3 log 2 log 2 log 2 3

log + − − + − − =

=

.

3)

^{2log2− 1}₃^{log8+ 2log3− log}₅^{3− 3log2− 1}₂^log25=

(

⁻

)

^{− − =}

− +

−

= ³ log5²

2 2 1 log 3 5 log 3 log 3 log 2 2 3log 2 1 log 2

=

−

− +

− +

−

= 2log2 log2 2log3 log3 log5 3log2 log5

=

−

= log3 2log2

=

−

= log3 log2²

4 log3 4 log 3

log − =

.

4)

^{log9− 1}₆^{log64− log12+ log}₃^{8− 1}₄^{log81+ log18=}

(

^⋅

)

^{+ − +}

(

^⋅

)

⁼

−

−

= ^{2 6 2 3} log3⁴ log2 3²

4 1 3 log2 3 2 log 2 6log 3 1 log

(

⁺

)

^{+ − − + + =}

−

−

= 2log3 log2 log2² log3 log2³ log3 log3 log2 log3²

(

+

)

+ − − + + =

−

−

= 2log3 log2 2log2 log3 3log2 log3 log3 log2 2log3

= +

+

−

− +

−

−

−

= 2log3 log2 2log2 log3 3log2 log3 log3 log2 2log3

= +

= log2 log3

=

log

(

2⋅3

)

= log6

.

Applicando le proprietà calcolare i seguenti logaritmi :

1)

3 ₃ ⁼

3 3 log 27

=

− +

= log

3

27 log

3

3 log

3³

3

=

− +

=

³

1 2 3

1 3 3

3

3 log 3 log 3

log

=

− +

= log 3

3 3 1 2log 3 1 log

3 _{3 3 3}

osservando che

^log3³= ¹

si ha:

=

− +

= 3

1 2 3 1

− =

= +

6 2 3 18

6

= 19

.

2)

_{2 4} =

2 2 log 16

=

− +

= log

2

16 log

2

2 log

2⁴

2

=

− +

= log

₂

2

⁴

log

₂

2

²¹

log

₂

2

⁴¹

=

− +

= log 2

4 2 1 2log 2 1 log

4 _{2 2 2}

osservando che

^log2²= ¹

si ha:

=

− +

= 4

1 2

4 1 + − =

4 1 2 16

4 17

.

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