LawsOf Logarithms

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    Logarithms

    Laws of logarithms

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    log a xy = loglog aa x + log x + log aa yyloglog aa x + log x + log aa yy = log a xy

    Examples:log 5 (3v 5) = log 5 3 + log 5 5

    log 3 5 + log 3 4 = log 3 (5 v 4)= log 3 20

    P roduct Rule

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    Quotient Rule

    = log 3 16 log 3 5

    )y x

    ( loga

    3

    165

    log ( )

    )( log4

    205

    )y x ( log a == loglog aa x x loglog aa yy

    loglogaa x x loglog

    aayy ==

    loglog 5 5 2020 loglog 5 5 44 === log= log

    5 5 5 5

    =1=1

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    log x m = m log x m log x = log x m

    log x 5 3 =

    4 log 9 3 = log 9 34

    = log 9 81= log 9 92

    = 2 log 9 9

    = 2 (log 9 9 = 1)

    P ower Rule

    3 log 3 log x x 5 5

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    E xpress the following s single log rithms

    loga 3 + log a 4 log a5

    log 3 + log 4 log 5 = log (3 v 4) log 5 (Rule 1)= log 12 log 5

    = (Rule 2)

    = log 2.4

    125al

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    Question:Given that log2 3 = 1.58and log2 5 = 2.32,Find value of each of the following.(a) log2 75(b) log2 0.3

    (c) log2 5

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    SOLUTION

    G iv en that log 2 3 = 1.58 and log 2 5 = 2 .32

    (a) log2 75 = log2 [3v25]= log2 3 + log2 25 Rule 1= log2 3 + log2 52= log2 3 + 2 log2 5= 1.58 + 2(2.32)= 6.22

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    (b) log2 0.3 = log2 (3 10)= log2 3 log2 10 Rule

    2 = log2 3 [log2 (5v2)]= log2 3 [log2 5 + log2 2]

    = 1.58 (2.32 + 1)= 1.74(c) log2 5 = (1/2) log2 5

    = (1/2)(2.32)

    S olutionS olution

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    EX AM PLEE valuat e log5 12 .

    log 5 12 =10

    10

    12

    5

    log log

    1 07920 6990 . .

    !

    544 1 .!

    Use calculato r

    U se at l east 4 significant figu res

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    CHAN GE OF B AS EC hange of base-a to base b is as follows:

    loga b =

    For example, to change log32 2 to base-2

    log32 2 =

    b

    b

    l og bl og a

    2

    132log

    1

    5!

    1

    bl og a

    !

    52

    1

    2log !

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    EXERCIS E

    G iv en that log 2 5 = 2.32 find th e valu e

    for each of th e following without u s ingcalculato r . ( without changing to ba se- 10)(a ) log5 4

    (b ) log5 2(c) log4 50

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    G iv en that log2

    5 = 2 .32

    (a ) log5 4

    22 32

    !.

    5

    4

    2

    2

    loglog

    !

    5 2

    2

    22

    loglog!

    =0. 862 1

    Change to base-2

    2

    25

    !log

    Rul e 3

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    (b ) log5 2 =

    (c) log4 50

    2

    15log

    12 32

    !

    .

    =0.431

    2

    2

    50

    4!

    l l

    2

    22

    2 25

    2

    v!

    log ( )log

    22 2

    2 5

    2!

    l l

    5 2

    12l !

    21 2 5

    2!

    log

    3 225 0 ..!

    =2.82