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    You are at:Home»Pure Math»Algebra»Logarithm Applications | Logarithm Examples and Answers | All Math Tricks
    logarithm applications | logarithm tricks | logarithm tutorial | logarithm tricks | logarithm application | logarithm questions | logarithm examples | logarithm questions
    Algebra

    Logarithm Applications | Logarithm Examples and Answers | All Math Tricks

    sivaalluriBy sivaalluriMay 27, 2019Updated:March 2, 2025No Comments9 Mins Read

    This logarithm tutorial explained logarithm applications with examples and solutions based on the logarithmic formulas as per previous exercises

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    • Logarithm questions for all class | Logarithm tutorial | Exercise – 3
      • Logarithm Examples and Answers ( Logarithm Applications )

    Logarithm questions for all class | Logarithm tutorial | Exercise – 3

    Please go through the below link for basic concepts of logarithms viz ., Meaning of Logarithm ,  Rule for writing Mantissa and Characteristic ,  Common Logarithms , Natural Logarithm , ..  etc

    Logarithm Tutorial | Exercise – 1

    Please go through the below link for the logarithm formula sheet like the logarithm addition rule and logarithm subtract rule the Base change rules, and logarithmic inequalities rule

    Logarithm Tutorial | Exercise – 2

    Logarithm Examples and Answers ( Logarithm Applications )

    Example- 1 : Find the value of logarithmic expression   \frac{1}{\log_{x} xy} + \frac{1}{\log_{y} xy}

    Solution: Here use the base change rule    \frac{1}{\log_{x} xy} + \frac{1}{\log_{y} xy}

    = log xy x +  log xy y

    = log xy xy   = 1

    Example- 2 : Find the value of logarithmic expression  log ay/by + log by/cy + log cy /ay

    Solution:  log ay/by + log by/cy + log cy /ay

    = log ( ay by cy / by  cy ay  ) = log 1 = 0

    Example – 3 : Express  log \ \frac{\sqrt[3]{a^2} }{ b^4\ \sqrt{c}}  in terms of log a, log b, & log c

    Solution:     \log \frac{\sqrt[3]{a^2}}{b^4 \sqrt{c}}        

    =  \log \sqrt[3]{a^2} - \log \left( b^4 \sqrt{c} \right)

    = log a 2/3    – [  log b4   + log c 1/3 ]

    = (2/3) log a – 4 log b – (1/3 ) log c

    Example – 4 : If the value of  4 a2 + 9 b2 = 10 – 12ab  then find the value of  log (2a + 3b)

    Solution:   log (2a + 3b)

    =  ( 2/2 ) log (2a + 3b)

    =  ( 1/2 ) log (2a + 3b)2

    = (1/2) log ( 4 a2 + 9 b2 + 12 ab )

    = (1/2) log (  10 -12 ab+ 12 ab )

    = 1/2 log 10 10 = 1/2

    Example – 5 : If log 10 3 = 0.4771, find the value of log 10 15 + log 10 2

    Solution: log 10 15 + log 10 2

    = log 10 (15 x 2) 

    = log 10 (10 x 3 ) 

    = log 10 10 + log 10 3

    = 1 + 0.4771 = 1.4771

    Example – 6 :  log x = log 9.6 – log 2.4, then find the value of  x

    Solution:  log 10  x = log10 9.6 – log10 2.4

    ⇒  log 10  x = log10 (9.6/2.4)

    ⇒ log 10  x = log10 (4) ⇒  x = 4

    Example – 7 : Find the value of  log 625  √125

    Solution: log 625  √125

    = log 54 53/2

    = 3 /( 4 x 2) log 5 5 = 3/8 ( 1) = 3/8

    Example – 8 :  log (x2 – 6x + 6) = 0 , then find the value of  x

    Solution: log 10  (x2 – 6x + 6) = 0

    ( If base is not mentioned in logarithmic expression, then it will be a common logarithm i.e base = 10)

    Now, the above logarithmic expression is converted into exponential form

    (x2 – 6x + 6) = 100   = 1  (Factoring polynomials )

    ⇒   x = 5 and 1

    Example – 9 : log x 4 + log x 16  + log x 64 + log x 256 = 10, Then find x = ?

    Solution: log x 4 + log x 16  + log x 64 + log x 256 = 10

    ⇒ log x ( 4  x 16  x  64  x 256 ) = 10

    ⇒ log x ( 4  x 16  x  64  x 256 ) = 10

    ⇒ log x ( 4 10 )= 10

    ⇒ x 10  = 4 10  ⇒ x = 10

    Example – 10 :  log x = log 1.5 + log 25, then find the value of  x

    Solution:  log 10 x = log10 1.5 – log10 25

    ⇒  log 10  x = log10 ( 1.5 x 25)

    ⇒ log 10  x = log10 (37.5) ⇒  x = 37.5

    Example – 11 : Find the number of digits in the exponential form of  4 2014  

    Solution: log 10 (4 2014) = 2014 x log 10 (4) = 2014 x 2  log 10 (22)

    =4028 log 10 2 = 4028 x 0.3010 =  1212.428

    Therefore the number of digits in 4 2014 is  1213 (1212 + 1)

    Example – 12 : If log 3 x +  log 9 x  + log 27 x  + log 729 x  =  6  then x =?

    Solution: log 3 x +  log 9 x  + log 27 x  + log 729 x

    = log 3 x +  log 32 x  + log 33 x  + log 36 x

    = log 3 x +   (1/2) log 3 x  + (1/3) log 3 x  + (1/6) log 3 x

    = [1 + (1/2) +( 1/3) + (1/6) ] log 3 x

    ⇒ 2 log 3 x  = 6

    ⇒  log 3 x  = 3

    ⇒   x  = 33    ⇒ x = 27

    Example – 13 : If log x ( 1/ 1024) = – 5, then find the value of x

    Solution:   log x ( 1/ 1024) = – 5

    Now convert into exponential form

    i.e  x -5 = 1/1024 = 1 / 45

    ⇒  x -5 = 1/1024 = 1 / 45     = 4 -5

    ⇒ x = 4

    Example – 14 : If 2 log 10 ( x +3) =log 10 169, then find the value of  x

    Solution:  2 log 10 ( x + 3 ) = log 10 169

    ⇒  log 10 ( x + 3 )2 = log 10 169

    ⇒  ( x + 3 )2 = 132

    ⇒  ( x + 3 ) = 13 ⇒   x  = 10

    Example – 15 :  If  log (2x +13)  – log (1.5 – x ) = 1 + log 3 , then find the value of  x

    Solution:   log (2x +13)  – log (1.5 – x ) = 1 + log 3

    ⇒  log (2x +13) / (1.5 – x ) = log 10 + log 3

    ⇒  log (2x+13) / (1.5 – x ) = log 30

    ⇒  (2x +13) / (1.5 – x ) = 30

    ⇒   2x +13  = 45 – 30x  

    ⇒ 32 x = 47 -13  ⇒x = 1

    Example – 16 :   log 10  x –   log 10  √x  = 2 log x  10, then find the value of  x

    Solution:  log 10  x –   log 10  √x  = 2 log x  10,

    Here log 10  √x  = log 10  x1/2   = 1/2 log 10 x

    Therefore the above equation becomes

    ⇒ log 10  x –   1/2 log 10 x = 2 log x  10,

    ⇒  1/2 log 10 x =  2 log x  10,

    ( Now use the base change rule for log x  10 ;    i. e log x  10 = 1/ log 10  x )

    ⇒  1/2 log 10 x =  2 / log 10  x

    ⇒  log 10 x =  4  / log 10  x

    ⇒ ( log 10 x )2  =  4

    ⇒ ( log 10 x ) =  4 / 2 = 2  ( Now convert into exponential forms)

    ⇒  x  = 10 2   = 100

    Example – 17 : Find the value of the expression   1/ log 3 2   +  2/ log 9  4 – 3/ log 27  8

    Solution: 1/ log 3 2   +  2/ log 9  4 – 3/ log 27  8

    Here use the base change rule for all terms in the above expression ( i.e  1/  log 3 2  = log 2 3 )

    ⇒  log 2 3   +  2 log 4  9  – 3  log 8  27

    ⇒  log 2 3   +  2 log 22 32 – 3  log 23  33

    ⇒  log 2 3   +  ( 2 x 2 /2 )  log 2 3   – (3 x 3 /3)  log 2  3

    ⇒  log 2 3   +  2  log 2 3   – 3  log 2  3

    ⇒  3  log 23   – 3  log 23  = 0

    Example – 18 :  If log 2 = 0.301, log 3 = 0.477 , find the number of digits in 10810

    Solution: Let x =   10810

    ⇒ log x = log (108)10

    ⇒ log x = 10 log (108)

    ⇒ log x = 10 log (27 x 4 )

    ⇒ log x = 10 log (33 x 22 )

    ⇒ log x = 10  ( 3 log 3 + 2 log  2)

    ⇒ log x = 10 ( 3 x 0.477 + 2 x 0.301)

    ⇒ log x =  20.33

    Thus x has 21 digits

    Example – 19  :  If log 2 = 0.301, log 3 = 0.477 , find the value of logarithmic equation   \frac{\log \sqrt{27} + \log 8 - \log \sqrt{1000}}{\log 1.2}

    Solution:    \frac{\log \sqrt{27} + \log 8 - \log \sqrt{1000}}{\log 1.2}

    Here rewrite the above logarithmic expression in terms of log 2, and log 3 and then substitute these values

    =  \frac{\frac{3}{2} \log 3 + 3 \log 2 - \frac{3}{2} \log 10}{\log \left( \frac{3 \times 4}{10} \right)}

    =  \frac{\frac{3}{2} \log 3 + 3 \log 2 - \frac{3}{2} \log 10} {\log 3 + 2 \log 2 - \log 10}

    =[ ( 1.5 x 0.477 ) + ( 3 x 0.301 ) – ( 1.5)  ]  / [ 0.477  + (2 x 0.301) – 1 ] = 1.5

    Example – 20  :  Find x  if log ( x – 13) + 4 log 2 = log ( 3x + 2)

    Solution:  log ( x – 13) + 4 log 2 = log ( 3x + 2)

    ⇒ log ( x – 13) + log 16 = log ( 3x +2)

    ⇒ log [ ( x – 13)  x  16 ] = log ( 3x +2)

    ⇒ log ( 16x – 208) = log ( 3x + 2)

    ⇒ 16x – 208 =  3x + 2

    ⇒ 16x – 3x  =  210   ⇒  x  = 14

     

    Thanks for reading this article. I hope you liked this article of “ Logarithm Applications |Logarithm tutorial | Exercise – 3 ”.  Give feedback and comments, please.

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