LOGARÍTMOS - Questões Resolvidas

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    Questes Resolvidas

    LOGARTMOS

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    1. Calcule x = log a b nos seguintes casos:

    a) b = 625 e a = 5b) b = 81 e a = 1/3

    c) b = 0,0001 e a = 100d) b = 432 e a = 0,125

    ) log 625 = 5 = 625

    log = =

    > 0 1

    > 0

    5 = 5 = 4

    ) log

    8 1 = 1

    3

    = 81

    3

    = 3

    = 4

    ) log 0,0001 = 100 = 0,0001

    10 = 10 10 = 10

    2 = 4 = 2

    ) log, 4 32 = 0,125 = 4 32

    125

    1000

    = 2. 2

    1

    8

    18

    = 2. 2 12

    = 2

    2 = 2 3 =

    9

    2

    = 3

    2

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    2. Calcule: a) o logaritmo de 256 na base 2 2

    b) o logaritmo de 9/16 na base 2 / 3.

    ) log 256 =

    2 2

    = 256

    2

    . 2

    = 2

    2. 2

    = 2

    2

    = 2

    2 = 2

    2 = 2

    3

    2= 8 =

    16

    3

    ) log

    916

    =

    2

    3

    =9

    16

    2

    3

    =3

    2

    2

    3

    =

    3

    2

    2

    3

    =3

    2

    2

    3

    =2

    3

    = 4

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    3. Calcule o valor da expresso: log5 1 + 4log

    45 + log3 (log5 125)

    log 1 = 5 = 1

    5 = 1 = 5 = 0

    4

    =

    = = log

    log 5 = log = 5

    log log 125 =

    log log 5 =

    log 3 =

    = 1

    : + +

    0 + 5 + 1 = 6

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    4. Se log 2 = a e log 3 = b, calcule em funo de a e b o valor da expresso:

    E = 10 log (log 3 ) + log (1 + log 2 / log 3 )

    10: log log3 + log 1 +log2

    log3

    4. Se log 2 = a e log 3 = b, calcule em funo de a e b o valor da expresso:

    E = 10 log (log 3 ) + log (1 + log 2 / log 3 )

    log + log 1 +

    = log + log

    +

    = log .

    +

    = log +

    = 1 0 = 10 = + = 10

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    5. Determinar o campo de existncia da expresso log x-3 (7 x).

    log = =

    > 0 1

    > 0

    3 > 0 3 1

    7 > 0

    I

    I . > 3

    II

    II. 4

    III

    III. > 7 < 7

    3

    4

    7

    IIII

    II

    3 < < 7 4 3 , 4 4 , 7

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    6. Resolva as equaes:

    a) log 5 (x + 3) = log 5 (x + 3)

    b) log x (14 5x) = 2

    c) log 1/5 (x 4x + 4) = 0

    d) log 1/3 (x + 3x 1 ) =2

    e) log4 [log2 (log3 x)] = 1/2

    f) ( log8 x) 4 log8 x + 4 = 0

    ) . . : + 3 > 0 + 3 > 0

    + 3 > 0

    + 3 > 0 > 3 > 3

    log + 3 = log + 3

    + 3 = + 3

    = 0. 1 = 0 = 0 = 1

    0 > 3

    1 > 3

    = 0 , 1

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    b) log x (14 5x) = 2

    . . :

    > 0

    1

    1 4 5 > 0 5 > 14

    5 < 1 4

    0 < 0 2

    4 + 4 = 15

    4 + 4 = 1

    4 + 3 = 0...

    = 1 = 3

    = 1 , 3

    d) log 1/3 (x + 3x 1 ) =2

    . . :

    + 3 1 > 0

    + 3 1 =1

    3

    + 3 1 = 3

    + 3 1 = 9

    + 3 1 0 = 0..

    . = 5 = 2

    5 + 3. 5 1 > 0

    9 > 0

    = 5

    2 + 3. 2 1 > 0

    9 > 0

    = 2

    = 5 , 2

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    e) log4 [log2 (log3 x)] = 1/2

    . . :

    > 0

    log log = 4

    log log = 4

    log log = 2

    log = 2

    log = 4

    = 3

    = 8 1

    = 81

    f) ( log8 x) 4 log8 x + 4 = 0

    . . :

    > 0

    log =

    4 + 4 = 0... = 2

    log = 2

    = 8

    = 6 4

    = 64

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    7. Sabendo que a, b e c so nmeros reais e positivos e que log10 (a. b) = 12,6 e

    log10 (a . c) = 0,2 , calcule log10 b/c .

    log . = 12,6 . = 10,

    log . = 0,2 . = 10,

    Dividindo por , temos:.

    . =

    10,

    10,

    = 10

    ,,

    = 10,

    log

    = log 10

    , = 12,4

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    8. Resolva as equaes:

    a) log (2x + 4x 4) + colog (x + 1) = log 4

    b) log x + 2 log x 10 = 3

    c) log 2 (x + 1) = log 2 (x 1) + 1

    d) log x 2 . log x/16 2 = log x/64 2

    e) log 4 (x 3) log 16 (x 3) = 1

    f) log4

    (log2

    x) + log2

    (log4

    x) = 2

    ) ..: 2 + 4 4 > 0 + 1 > 0 > 1

    log 2 + 4 4 log + 1 = log 4

    log2 + 4 4

    + 1= log 4

    2 + 4 4

    + 1= 4

    2 + 4 4 = 4 + 4

    2 = 8 = 4 = 2

    = 2

    = 2

    2.2 +4.2 4 > 0

    1 2 > 0

    = 2

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    b) log x + 2 log x 10 = 3

    . . :

    > 0 1

    log + 2 .log 10

    log = 3

    log =

    + 2.1

    = 3

    + 2 = 3

    3 + 2 = 0...

    = 1 = 2

    = 1 log = 1

    = 1 0 = 1 0

    = 2 log

    = 2

    = 1 0 = 100

    = 10 , 100

    log = 1

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    c) log 2 (x + 1) = log 2 (x 1) + 1

    . . :

    + 1 > 0 > 1

    1 > 0

    log + 1

    log 2= log

    1 + log 2

    log + 1

    log 2

    = log[

    1 . 2]

    log + 112 .log 2

    = log 2 2

    log

    = . log

    log + 11

    2= log 2

    2

    2.log + 1 = log 2 2

    log + 1 = log 2 2

    + 1 = 2 2

    + 2 + 1 = 2 2

    2 3 = 0... = 1 = 3

    = 1

    = 3 3 1 > 0 8 > 0

    = 3

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    d) log x 2 . log x/16 2 = log x/64 2

    . . :

    > 0 1, 16 64

    log 2

    log .

    log 2

    log

    16

    =log 2

    log

    64

    1

    log .

    1

    log l o g 16 =

    1

    log l o g 64

    log =

    1

    . 4=

    1

    6 . 4 = 6

    4 + 6 = 0

    5 + 6 = 0... = 2 = 3

    = 2 log = 2 = 2 = 4

    = 4

    = 3 log = 3 = 2 = 8

    = 8

    = 4 , 8

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    e) log 4 (x 3) log 16 (x 3) = 1

    . . :

    3 > 0 > 3

    log 3 log 3

    log 16= 1

    log 3 =

    2= 1

    2= 1 = 2

    log 3 = 2

    3 = 4

    3 = 1 6 = 1 9 = 19

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    f) log 4 (log 2 x) + log 2 (log 4 x) = 2

    . . :

    > 0

    log log

    log 4+ log log = 2

    log log 2

    + log log = 2

    log log + 2. log log = 4

    log log + log log = 4

    log log . log = 4

    log .log

    log 4

    = 2

    log .log

    4= 16

    log

    =

    .

    4= 16

    = 64

    = 64

    = 2

    = 2

    = 4

    log = 4

    = 2 = 1 6

    = 16

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    9. Resolva as inequaes :

    a) log 5 (4x 1) < log 5 3

    b) log (x + 4) log (2x 2)

    c) log (x 2x) 3

    d) log 2 (x 3) + log 2 (x 2) < 1

    ) . . :4 1 > 0 >1

    4

    5 > 1

    4 1 < 3

    4 < 4

    < 1

    1

    4

    1

    III

    I

    II

    = 1

    4< < 1

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    b) log (x + 4) log (2x 2)

    . . :

    + 4 > 0 > 4 < 4

    2 2 > 0 2 > 2 > 1

    1 2 0 < < 1

    + 4 2 2 2 2 4

    3 6

    3 6

    2

    I

    II

    III

    1

    2

    4

    IIIIII

    = 1 < 2

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    c) log (x2x) 3

    . . :

    2 > 0

    Para resolver uma inequao logartmica, fica mais fcil quando temos, em ambos oslados da desigualdade, logaritmos na mesma base.

    Do lado esquerdo temos um logaritmo na base .

    Vamos escrever o 3 do lado direito como um logaritmo na base tambm:

    log

    = 3 = 1 2

    = 2 = 8 3 = log

    8

    log

    2 log

    8

    1 2 0 < < 1

    2 8 2 8 0

    I

    II

    c) log (x 2x) 3

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    2 > 0

    2 8 0

    I

    II

    : = 2

    : 0 2

    0

    2

    : < 0 > 2

    > 0

    : = 2 8

    : 2 4

    2

    4

    : 2 4

    0

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    : < 0 > 2 : 2 4

    0

    2

    4

    2

    = 2 < 0 2 < 4

    2 , 0 2 , 4

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    d) log 2 (x 3) + log 2 (x 2) < 1

    . . :

    3 > 0 > 3

    2 > 0 > 2 > 3 I

    Vamos escrever o 1 do lado direito como um logaritmo na base 2 tambm:

    log = 1 = 2 = 2 = 2 1 = log 2

    d) log 2 (x 3) + log 2 (x 2) < 1

    log 3 + log 2 < log 2

    log 3 . 2

    log 5 + 6 < log 2

    2 > 1

    5 + 6 < 2 5 + 4 < 0

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    5 + 4 < 0

    =

    5 + 4

    : 1 4

    1

    4

    : 1 < < 4

    < 0

    : > 3

    1

    3

    4

    = 3 < < 4

    3 , 4

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    10. Resolva as equaes:

    a) 2 x = 7

    b) 9 x 7.3 x + 10 = 0

    ) 2 = 7 log 7 =

    =log7

    log2

    0,845

    0,301 2,8

    ) 3 7 . 3 + 1 0 = 0

    3 =

    7 + 1 0 = 0... = 2 = 5

    = 2 3 = 2

    = log 2= log2log3

    0,3010.477 0,63

    = 5 3 = 5

    = log 5=

    log5

    log3

    0,699

    0.477 1,5

    = 2,8

    = 0,63 ; 1,5

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    ISERJ 2012Professora Telma Castro Silva