Logarithm Applications | Logarithm Examples and Answers | All Math Tricks

In this logarithm tutorial explained about logarithm applications with examples and solutions based the logarithmic formulas as per previous exercises

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

Please go through the below link for basic concepts of logarithms viz ., Meaning of LogarithmRule for write Mantissa and CharacteristicCommon Logarithms , Natural Logarithm , ..  etc

Logarithm Tutorial | Exercise – 1

Please go through the below link for logarithm formula sheet like logarithm addition rule and logarithm subtract rule and 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 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 / b cy a) = 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 \ b^4\ \sqrt{c}

= 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 not mentioned in logarithmic expression, the it will be common logarithm i.e base = 10)

Now above logarithmic expression convert 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 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 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

Therefor 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 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 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 above logarithmic expression in terms of log 2, log 3 and then substitute these values

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

= \frac{ (3/2) \ log \ 3 \ + 3 \ \ log \ 2 \ \ - \ (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

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My self Sivaramakrishna Alluri. Thank you for watching my blog friend

2 thoughts on “Logarithm Applications | Logarithm Examples and Answers | All Math Tricks

    jyoti das

    (July 8, 2019 - 10:47 am)

    good post nice information thankful to you………

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