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Logarithms

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  1. Asked: 3 years agoIn: Logarithms

    Simplify:[1/logxy(xyz)+1/logyz(xyz)+1/logzx(xyz)]

    a3e26
    Added an answer about 3 years ago

    Given expression: logxyz xy+ logxyz yz+ logxyz zx = logxyz (xy*yz*zx)=logxyz (xyz)2 2logxyz(xyz)=2*1=2

    Given expression: logxyz xy+ logxyz yz+ logxyz zx 

    = logxyz (xy*yz*zx)=logxyz (xyz)2 

    2logxyz(xyz)=2*1=2

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  2. Asked: 3 years agoIn: Logarithms

    If log10 2=0.30103,find the value of log10 50.

    f1279
    Added an answer about 3 years ago

    log10 50=log10 (100/2)=log10 100-log10 2=2-0.30103=1.69897.

    log10 50=log10 (100/2)=log10 100-log10 2=2-0.30103=1.69897.

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  3. Asked: 3 years agoIn: Logarithms

    Evaluate: log3 27

    e621e
    Added an answer about 3 years ago

    Let log3 27=33 or n=3. ie, log3 27 = 3.

    Let log3 27=33 or n=3. 

    ie, log3 27 = 3.

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  4. Asked: 3 years agoIn: Logarithms

    If log 2=.30103,find the number of digits in 256 .

    f4e43
    Added an answer about 3 years ago

    log 256 =56log2=(56*0.30103)=16.85768. Its characteristics is 16. Hence,the number of digits in 256 is 17.

    log 256 =56log2=(56*0.30103)=16.85768. 

    Its characteristics is 16. 

    Hence,the number of digits in 256 is 17.

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  5. Asked: 3 years agoIn: Logarithms

    Find the value of x which satisfies the relation 

    Log10 3+log10 (4x+1)=log10 (x+1)+1

    fd51d
    Added an answer about 3 years ago

    log10 3+log10 (4x+1)=log10 (x+1)+1 Log10 3+log10 (4x+1)=log10 (x+1)+log10 (x+1)+log10 10 Log10 (3(4x+1))=log10 (10(x+1)) =3(4x+1)=10(x+1)=12x+3 =10x+10 =2x=7x=7/2

    log10 3+log10 (4x+1)=log10 (x+1)+1 

    Log10 3+log10 (4x+1)=log10 (x+1)+log10 (x+1)+log10 10 

    Log10 (3(4x+1))=log10 (10(x+1)) 

    =3(4x+1)=10(x+1)=12x+3 

    =10x+10 

    =2x=7

    x=7/2

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  6. Asked: 3 years agoIn: Logarithms

    3log8x=​​​​​​​​log4(x+6)

    29082
    Added an answer about 3 years ago

    3 log8x = log4 (x+6)⇒ 3 log23x = log22 (x+6) (∵ logamb = 1/m logab)⇒ 3/3 log2x = 1/2 log2 (x+6)⇒ 2 log2x = log2 (x+6)⇒ log2x2 = log2 (x+6)  (∵ n log a = log an)⇒ x2 = x+6  (By taking anti log)⇒ x2 - x - 6 = 0⇒ x2 - 3x + 2x - 6 = 0⇒ x(x-3) + 2(x-3) = 0⇒ (x+2) (x-3) = 0⇒ x+2 = 0 or x-3 = 0⇒ x = -2(NotRead more

    3 log8x = log4 (x+6)

    ⇒ 3 log23x = log22 (x+6) (∵ logamb = 1/m logab)

    ⇒ 3/3 log2x = 1/2 log2 (x+6)

    ⇒ 2 log2x = log2 (x+6)

    ⇒ log2x2 = log2 (x+6)  (∵ n log a = log an)

    ⇒ x2 = x+6  (By taking anti log)

    ⇒ x2 – x – 6 = 0

    ⇒ x2 – 3x + 2x – 6 = 0

    ⇒ x(x-3) + 2(x-3) = 0

    ⇒ (x+2) (x-3) = 0

    ⇒ x+2 = 0 or x-3 = 0

    ⇒ x = -2(Not possible) or x = 3)

    (∵ Domain of log x is x > 0)

    Hence, solution is x = 3

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  7. Asked: 3 years agoIn: Logarithms

    Evaluate log34 34

    79de0
    Added an answer about 3 years ago

    We know that loga a=1,so log34 34=0. 

    We know that loga a=1,so log34 34=0. 

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