Logarithm Applications | Logarithm Examples and Answers

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 Logarithm , Rule for write Mantissa and Characteristic , Common 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 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: $log \ \frac{\sqrt[3]{a^2} }{ b^4\ \sqrt{c}}$ = $log \ \sqrt[3]{a^2} \ - \ log \ b^4\ \sqrt{c}$

= 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 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 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 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 x^1/2= 1/2 log ₁₀x

Therefor 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 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 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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