Logarithm Applications | Logarithm Examples and Answers

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

Table of Contents

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 a^y/b^y + log b^y/c^y + log c^y /a^y

Solution: log a^y/b^y + log b^y/c^y + log c^y /a^y

= log ( a^y b^y c^y / b^y c^y a^y) = 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 5^3/2

= 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 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 x^1/2= 1/2 log ₁₀x

Therefore the above equation becomes

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

⇒ 1/2 log ₁₀x = 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 ₁₀x = 2 / log ₁₀ x

⇒ log ₁₀x = 4 / log ₁₀ x

⇒ ( log ₁₀x )² = 4

⇒ ( log ₁₀x ) = 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

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