In the last article, we have studied to solve problems on the use of laws of logarithms to prove given logarithmic expression. In this article, we shall study to solve more problems on these laws to prove given relation using given logarithmic expression.

Example 01:

Solution:

Given

Dividing both sides by xy

Proved as required.

Example 02:

Solution:

Given

Dividing both sides by xy

Proved as required.

Example 03:

Solution:

Given

Dividing both sides by xy

Proved as required.

Example 04:

Solution:

Given

Proved as required.

Example 05:

Solution:

Given

Proved as required.

Example 06:

Solution:

Given

Proved as required.

Example 07:

Solution:

Given

Proved as required.

Example 08:

Given 2log₂(x + y) = 3 + log₂x + log ₂y, Show that x² + y² = 6xy

Solution:

Given 2log₂(x + y) = 3 + log₂x + log ₂y

log₂(x + y)² = 3 log₂2 + log₂x + log ₂y

log₂(x + y)² =  log₂2³ + log₂x + log ₂y

log₂(x + y)² =  log₂8 + log₂x + log ₂y

log₂(x + y)² =  log₂8xy

 (x + y)² =  8xy

x² + 2xy + y² = 8xy

x² + y² = 6xy

Proved as required.

Example 09:

If a² + b² = 3ab, show that

Solution:

a² + b² = 3ab

Adding 2ab on both sides

a² + 2ab + b² = 3ab + 2ab

(a + b)² = 5ab

Proved as required.

Example 10:

If a² -12ab+ 4b² = 0, show that

Solution:

a² -12ab+ 4b² = 0

a² + 4b² = 12ab

Adding 4ab on both sides

a² + 4ab + 4b² = 12ab + 4ab

(a + 2b)² = 16ab

Proved as required

Example 11:

If a² + b² = 7ab, show that

Solution:

a² + b² = 7ab

Adding 2ab on both sides

a² + 2ab + b² = 7ab + 2ab

(a + b)² = 9ab

Proved as required.

Example 12:

If b² = ac, prove that log a + log c = 2 log b

Solution:

b² = ac

Taking log on both sides

Log b² = log (ac)

2 log b = log a +log c

Log a + log c = 2 log b

Proved as required.

In the next article, we shall study to solve more problems on these laws to prove given logarithmic relation.

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The post Use of the Laws of Logarithms: Set – III appeared first on The Fact Factor.

In this article, we shall study the laws of logarithm and their proofs. Laws of logarithm are very important in mathematics and every student should have confidence in using them.

Laws of Product:

log_a(mn) = log_am + log_an, where a, m, n are positive real numbers with a ≠ 1

Proof:

Let log_am = x and log_an = y   ……. (1)

By the definition of logarithm, we have

a^x = m and a^y = n

Now mn = a^x . a^y

mn = a^{x + y}

By definition of logarithm

log_a (mn) = x + y

Substituting values of x and y from relation (1) we get

log_a (mn) = log_am + log_an (Proved)

Corollary:

log_a(mnp) = log_am + log_an + log_ap

Proof:

L.H.S. =  log_a (mnp)

∴ L.H.S. = log_a (mn) + log_a (p)  (Law of product)

∴ L.H.S. = log_a m + log_a n + log_a p  (Law of product)

log_a(mnp) = log_am + log_an + log_ap (Proved)

Law of Quotient:

log_a(m/n) = log_am – log_an, where a, m, n are positive real numbers with a ≠ 1, n ≠ 0

Proof:

Let log_am = x and log_an = y   ……. (1)

By the definition of logarithm, we have

a^x = m and a^y = n

Now m/n = a^x / a^y

m/n = a^{x – y}

By definition of logarithm

log_a (m/n) = x – y

Substituting values of x and y from relation (1) we get

log_a (m/n) = log_am – log_an (Proved)

Corollary:

log_a(mn/pq) = log_am + log_an – log_ap – log_aq

Proof:

L.H.S. = log_a(mn/pq)

∴ L.H.S. = log_a(mn) – log_a(pq)   (Law of quotient)

∴ L.H.S. = (log_am + log_an) – (log_ap + log_aq)  (Law of product)

∴ L.H.S. = log_am + log_an – log_ap – log_aq

log_a(mn/pq) = log_am + log_an – log_ap – log_aq (Proved)

Law of Exponent:

log_amⁿ = n. log_am where a, m, n are positive real numbers with a ≠ 1

Proof:

Let log_am = x     ……. (1)

By the definition of logarithm, we have

a^x = m

mⁿ = (a^x)ⁿ = a^nx

By definition of logarithm

log_amⁿ  = n x

Substituting value of x from relation (1) we get

log_amⁿ  = n log_am  (Proved)

Corollary:

log_a(x^py^q/z^rw^s) = (p log_ax + q log_ay) – (r log_az + s log_aw)

Proof:

L.H.S. = log_a(x^py^q/z^rw^s)

∴ L.H.S. = log_a(x^py^q) – log_a(z^rw^s)  (Law of quotient)

∴ L.H.S. = (log_ax^p + log_ay^q) – (log_az^r + log_aw^s)  (Law of product)

∴ L.H.S. = (p log_ax + q log_ay) – (r log_az + s log_aw)  (Law of exponent)

∴ L.H.S. = p log_ax + q log_ay – r log_az – s log_aw 

log_a(x^py^q/z^rw^s) = (p log_ax + q log_ay) – (r log_az + s log_aw) (Proved)

Logarithm of 1 to Any Base

Log_a1 = 0 i.e. Logarithm to any base is always zero.

Proof:

Let log_a1 = x ………. (1)

By definition of logarithm

a^x = 1

a^x = a⁰

x = 0

Substituting this value in equation (1) we get

log_a1 = 0 (Proved)

Logarithm of Any Number to the Same Base:

log_aa = 1 i.e. Logarithm of any number to the same base is 1

Proof:

Let log_aa = x ………. (1)

By definition of logarithm

a^x = a

a^x = a¹

x = 1

Substituting this value in equation (1) we get

log_aa = 1 (Proved)

Important Property:

where a and m are positive real numbers with a ≠ 1

Proof:

Let log _am = x

By definition of logarithm

a^x = m

Substituting in equation (1)

(Proved)

Change of Base Law:

where a, b, and m are positive real numbers with a ≠ 1 and b ≠ 1

Proof:

Let log_am = x, log_bm = y and log_ba = z    ……. (1)

By the definition of logarithm, we have

a^x = m  ……….. (2)

b^y = m  ……….. (3)

b^z = a  ……….. (4)

From equations (2) and (3)

a^x = b^y    ……….. (5)

Substituting value of a from equation (4) in (5) we get

(b^z)^x = b^y

b^zx = b^y

zx = y

x = y/z

Substituting the values of x, y, and z from equation (1) we get

(Proved)

Corollary:

Proof:

We have

Substitute m = b

(Proved)

In the next few articles, we shall study how to use the laws of logarithm to evaluate or simplify given logarithmic expression.

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The post Laws of Logarithms appeared first on The Fact Factor.

In this article, we shall study the concept of logarithms and interconversion between exponential form and logarithmic form.

Laws of Indices:

• a^m × aⁿ = a^{m + n}
• a^m ÷ aⁿ = a^{m – n}           (a ≠ 0)
• (a^m)ⁿ = a^mn
• (ab)^m = a^m × b^m
• (a ÷ b)^m = a^m ÷ b^m    (b ≠ 0)
• a^-m  = 1/a^m                   (a ≠ 0)
• a⁰ = 1

Definition of Logarithm:

If m = a^x, where a > 0 and m > 0 then x is called the logarithm of m to the base a and is written as x = log_am and read as a log of m to the base a where m, a, x are real numbers.

Notes:

• The logarithm of a negative number and zero are not defined.
• Logarithm to the base 10 are called common or Briggsian logarithms.
• Logarithms to the base e, where e is an irrational number whose value is e = 2.7182… are called natural or Naperian logarithms.

Conversion from the Exponential Form into Logarithmic Form:

Conversion from the Logarithmic Form into the Exponential Form:

In the next article, we shall study to solve the problems based on the definition of a logarithm.

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The post Concept of Logarithm appeared first on The Fact Factor.

In the last article, we have studied the laws of logarithms and their proofs. In this article, we shall study to solve problems on laws of logarithms (law of product, Law of quotient, Law of exponent, etc.) to evaluate or simplify given logarithmic expression.

Laws of Logarithms

• Log a + Log b = Log (ab) (Law of Product)
• Log a – log b = log (a/b)    b ≠ 0 (law of Quotient)
• Log a^m = m Log a (Law of exponent)
• Log (1) = 0
• Log_aa = 1
• 
• 
• 

Example 01:

• Simplify log₁₀5 + 2 log₁₀4

Solution:

log₁₀5 + 2 log₁₀4

= log₁₀5 + log₁₀4² (Law of exponent)

= log₁₀5 + log₁₀16

= log₁₀(5 x 16) (Law of product)

= log₁₀80

Ans: log₁₀5 + 2 log₁₀4 = log₁₀80

Example 02:

• Simplify 2 log 7 – log 14

Solution:

2 log 7 – log 14

= log 7² – log 14 (Law of exponent)

= log 49 – log 14

= log (49/14) (Law of quotient)

= log (7/2)

Ans: 2 log 7 – log 14 = log (7/2)

Example 03:

• Simplify log₁₀3 + log ₁₀2 – 2log₁₀5

Solution:

log₁₀3 + log ₁₀2 – 2log₁₀5

= log₁₀3 + log ₁₀2 –log₁₀5² (Law of exponent)

= log₁₀3 + log ₁₀2 –log₁₀25

= log₁₀((3 x 2)/25) (Law of product and quotient)

= log₁₀(6/25)

Ans: log₁₀3 + log ₁₀2 – 2log₁₀5 = log₁₀(6/25)

Example 04:

• Simplify log 5 + log 3 – Log2

Solution:

log 5 + Log 3 – Log2

= log ((5 x 3)/2) (Law of product and quotient)

= log (15/2)

Ans: log 5 + Log 3 – Log2 = log (15/2)

Example 05:

• Simplify ½ log 9 + 1/3 log 27

Solution:

½ log 9 + 1/3 log 27

=  log 9^½ + log 27^1/3(Law of exponent)

=  log 3 + log 3

= 2 log 3

= log 3² (Law of exponent)

= log 9

Ans: ½ log 9 + 1/3 log 27 = log 9

Example 06:

• Simplify 2 log₁₀4 – ½ log₁₀16 + 1

Solution:

2 log₁₀4 – ½ log₁₀16 + 1

= log₁₀4² – log₁₀16^½ + log₁₀10 (Law of exponent)

= log₁₀16 – log₁₀4 + log₁₀10

= log₁₀((16 x 10)/4) (Law of product and quotient)

= log₁₀ (40)

Ans: 2 log₁₀4 – ½ log₁₀16 + 1 = log₁₀ (40)

Example 07:

Simplify 2log3 – ½ log 16 + log 12

Solution:

2log3 –  ½ log 16 + log 12

= log3² – log 16^½ + log 12 (Law of exponent)

= log9 – log 4 + log 12

= log ((9 x 12)/4) (Law of product and quotient)

= log (9 x 3)

= log 27

Ans: 2log3 –  ½ log 16 + log 12 = log 27

Example 08:

Simplify ½ log₅36 + 2log₅7 – ½ log ₅12

Solution:

½ log₅36 + 2log₅7 – ½ log ₅12

= log₅36^½ + log₅7² –log ₅12^½ (Law of exponent)

= log₅6 + log₅49 –log ₅2√3

= log₅((6 x 49)/ 2√3) (Law of product and quotient)

= log₅((3 x 49)/ √3)

= log₅((√3 x √3 x 49)/ √3)

= log₅(49√3)

Ans: ½ log₅36 + 2log₅7 – ½ log ₅12 = log₅(49√3)

Example 09:

• Simplify 2 log 3 + 3 log 2

Solution:

2 log 3 + 3 log 2

= log 3² + log 2³ (Law of exponent)

= log 9 + log 8

= log (9 x 8) (Law of product)

= log 72

Ans: 2 log 3 + 3 log 2 = log 72

Example 10:

• Simplify 2 log 5 + 3 log 4 – 4 log 2

Solution:

2 log 5 + 3 log 4 – 4 log 2

= log 5² + log 4³ – log 2⁴ (Law of exponent)

= log 25 + log 64 – log 16

= log ((25 x 64)/16) (Law of product and quotient)

= log (25 x 4)

= log 100

Ans: 2 log 5 + 3 log 4 – 4 log 2 = log 100

Example 11:

• Simplify log₁₀2 + 3

Solution:

log₁₀2 + 3

= log₁₀2 + 3 x 1

= log₁₀2 + 3 x log₁₀10 (Number and base same rule)

= log₁₀2 + log₁₀10³ (Law of exponent)

= log₁₀2 + log₁₀1000

= log₁₀(2 x 1000) (Law of product)

= log₁₀(2000)

Ans: log₁₀2 + 3 = log₁₀(2000)

Example 12:

• Simplify log (x² – 3x + 2) – log (x – 1) + log (x – 2)

Solution:

log (x² – 3x + 2) – log (x – 1) + log (x – 2)

= log (x – 2)(x – 1) – log (x – 1) + log (x – 2)

= log ((x – 2)(x – 1)(x – 2))/(x – 1)) (Law of product and quotient)

= log (x – 2)(x – 2)

= log (x – 2)²

= 2 log (x – 2) (Law of exponent)

Ans: log (x² – 3x + 2) – log (x – 1) + log (x – 2) = log (x – 2)²

Example 13:

Simplify

Solution:

Example 14:

Simplify

Solution:

Example 15:

Simplify

Solution:

Example 16:

Simplify

Solution:

Example 17:

Simplify

Solution:

Example 18:

Simplify

Solution:

Example 19:

Simplify

Solution:

Example 20:

Simplify

Solution:

Example 21:

Simplify

Solution:

Example 22:

Simplify

Solution:

Example 23:

Simplify

Solution:

In the next article, we shall study to solve more problems on the laws of logarithms to prove given logarithmic expression.

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The post Use of the Laws of Logarithms: Set – I appeared first on The Fact Factor.

In the last article, we have studied the concept of logarithm and interconversion between exponential form and logarithmic form. In this article, we shall study to solve problems based on the definition of logarithm.

Definition of Logarithm:

If m = a^x, where a > 0 and m > 0 then x is called the logarithm of m to the base a and is written as x = log_am and read as a log of m to the base a where m, a, x are real numbers.

Note the change in position of the terms during conversion from one form to another.

Example 01:

• Find the value of log₂32

Solution:

Let log₂32 = x

By definition of logarithm

∴ 2^x = 32

∴ 2^x = 2⁵

∴ x = 5

Ans: log₂32 = 5

Example 02:

• Find the value of log₁₀100000

Solution:

Let log₁₀100000 = x

By definition of logarithm

∴ 10^x = 100000

∴ 10^x = 10⁵

∴ x = 5

Ans: log₁₀100000 = 5

Example 03:

• Find the value of log₅3125

Solution:

Let log₅3125 = x

By definition of logarithm

∴ 5^x= 3125

∴ 5^x = 5⁵

∴ x = 5

Ans: log₅3125 = 5

Example 04:

• Find the value of log₅125

Solution:

Let log₅125 = x

By definition of logarithm

∴ 5^x = 125

∴ 5^x = 5³

∴ x = 3

Ans: log₅125 = 3

Example 05:

Find the value of log _1/2  8

Solution:

Let  log _1/2  8  = x

By definition of logarithm

∴ (1/2)^x = 8

∴ (2^-1)^x = 2³

∴ 2^-x = 2³

∴ -x = 3

∴ x = -3

Ans: log _1/2  8 = -3

Example 06:

• Find the value of log₈16

Solution:

Let log₈16 = x

By definition of logarithm

∴ (8)^x = 16

∴(2³)^x = 2⁴

∴ 2^3x = 2⁴

∴ 3x = 4

∴ x = 4/3

Ans: log₈16 = 4/3

Example 07:

• Find the value of log₆₄4

Solution:

Let log₆₄4 = x

By definition of logarithm

∴ 64^x = 4

∴ (2⁶)^x = 2²

∴ 2^6x = 2²

∴ 6x = 2

∴ x = 1/3

Ans: log₆₄4 = 1/3

Example 08:

• Find the value of log₁₆32

Solution:

Let log₁₆32 = x

By definition of logarithm

∴ 16^x = 32

∴ (2⁴)^x = 2⁵

∴ 2^4x = 2⁵

∴ 4x = 5

∴ x = 5/4

Ans: log₁₆32 = 5/4

Example 09:

• Find the value of log₉(1/81)

Solution:

Let Log₉(1/81) = x

By definition of logarithm

∴ 9^x = 1/81

∴ (3²)^x = 1/3⁴

∴ 3^2x = 3^-4

∴ 2x = -4

 ∴ x = -2

Ans: log₉(1/81) = -2

Example 10:

• Find the value of log₂(1/256)

Solution:

Let Log₂(1/256) = x

By definition of logarithm

∴ 1/256 = 2^x

∴ 1/2⁸ = 2^x

∴ 2^-8 = 2^x

∴ x = -8

Ans: log₂(1/256) = -8

Example 11:

• Find the value of log₈₁ 27

Solution:

Let log₈₁ 27 = x

By definition of logarithm

∴ 81^x = 27

∴ (3⁴)^x = 3³

∴ 3^4x = 3³

∴ 4x = 3

∴ x = ¾

Ans: log₈₁ 27 = 3/4

Example 12:

• Find the value of log₁₀0.001

Solution:

Let log₁₀0.001 = x

By definition of logarithm

∴ 10^x = 0.001

∴ 10 ^x = 1/1000

∴ 10 ^x = 1/10³

∴ 10 ^x = 10^-3

∴ x = -3

Ans: log₁₀0.001 = – 3

Example 13:

• Find the value of log₅0.008

Solution:

Let log₅0.008 = x

By definition of logarithm

∴ 5^x = 0.008

∴ 5^x = 8/1000

∴ 5^x = 1/125

∴ 5^x = 1/5³

∴ 5^x = 5^-3

∴ x = -3

Ans: log₅0.008 = -3

Example 14:

• Find the value of

Solution:

Example 15:

• Find the value of

Solution:

Example 16:

• Find the value of

Solution:

Example 17:

• Find the value of

Solution:

Example 18:

• Find the value of

Solution:

Example 19:

• Find the value of

Solution:

Example 20:

• Find the value of log₂(log_xx²)

Solution:

Log₂(log_xx²) = Log₂(2log_xx) = Log₂(2(1))

= Log₂2 = 1

Ans: Log₂(log_xx²)  = 1

Example 21:

• Find the value of log₅(log_xx²)

Solution:

Log₂(log₃3) = Log₅(1) = 0

Ans: Log₅(log_xx²) = 0

Example 22:

Find the value of

Solution:

Example 23:

Find the value of

Solution:

In the next article, we shall see laws of the logarithm, and problems based on them.

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The post Problems Based on Definition of Logarithm appeared first on The Fact Factor.

In the last article, we shall study problems based on the change of base rule. In this article, we shall study the use of definition and the laws of logarithms to find the value of ‘x’ (solve for x).

Example 01:

Solve for x, if log₅x = 2

Solution:

Given: log₅x = 2

∴ x = 5² = 25

∴ x = 25

Example 02:

Solve for x, if log₂(1/2) = x

Solution:

Given: log₂(1/2) = x

∴ log₂2^-1 = x

∴ x = -1 Log22

∴ x = – 1(1)

∴ x = -1

Example 03:

Find x, if log₄(x – 2) = 2

Solution:

Given: log₄(x – 2) = 2

∴ x – 2 = 4²

∴ x – 2 = 16

∴ x = 18

Example 04:

Find x, if log₅(x² – 5x + 11) = 1

Solution:

Given: log₅(x² – 5x + 11) = 1

∴ x² – 5x + 11 = 5¹

∴ x² – 5x + 6 = 0

∴ (x – 3)(x – 2) = 0

∴ x – 3 = 0 or x – 2 = 0

∴ x = 3 or x = 2

Both x = 3 and x = 2 satisfy given relation

x = 3 or x = 2

 Example 05:

Find x, if log₂x – log₂(x – 1) = 5

Solution:

Given: Log₂x – log₂(x – 1) = 5

∴ x = 32x – 32

∴31x = 32

∴ x = 32/31

Example 06:

Find x, if log x + log(x + 1) = log 6

Solution:

Given: log x + log(x + 1) = log 6

∴ log x(x + 1) = log 6

∴x(x + 1) = 6

∴ x² + x – 6 = 0

∴ (x + 3)(x – 2) = 0

∴ x = -3 or x = 2

x = -3 is not possible because log of negative number not defined.

∴ x = 2

Example 07:

Find x, if log (1 + x) – log (1 – x) = 1

Solution:

Given: log (1 + x) – log (1 – x) = 1

Example 08:

Find x, if log₃(x + 6) = 2

Solution:

Given: log₃(x + 6) = 2

∴ x + 6 = 3² = 9

∴ x = 3

Example 09:

Find x, if log (x + 3) + log (x – 3) = log 16, find x

Solution:

Given: log (x + 3) + log (x – 3) = log 16

∴ log [(x + 3)(x – 3)] = log 16

∴ log (x² – 3²) = log 16

∴ x² – 9 = 16

∴ x²  = 25

∴ x = ± 5

∴ x = -5 is not possible because the log of negative numbers not defined.

∴ x = 5

Example 10:

Find x, if log (3x + 2) + log (3x – 2) = log 5

Solution:

Given: log (3x + 2) + log (3x – 2) = log 5

∴ log [(3x + 2)(3x – 2)] = log 5

∴ log ((3x)² – 2²) = log 16

∴ 9x² – 4 = 5

∴ 9x²  = 9

∴ x²  = 1

∴ x = ± 1

x = -1 is not possible because log of negative number not defined.

∴  x = 1

Example 11:

Find x, if log (3x + 2) – log (3x – 2) = log 5

Solution:

Given: log (3x + 2) – log (3x – 2) = log 5

∴ 3x + 2 = 15x – 10

∴ 12 = 12x

∴x = 12/12 = 1

Example 12:

Find x, if

Solution:

Given:

∴ log(x + 6) = 2 log x

∴ log(x + 6) = log x²

∴ x + 6 = x²

∴ x² – x – 6 = 0

∴ (x – 3)(x + 2) = 0

∴ x – 3 = 0 and x + 2 = 0

∴ x = 3 or x = -2

x = -2 is not possible because log of negative number not defined.

∴ x = 3

Example 13:

Find x, if log₃x + log₃4 = 2

Solution:

Given: log₃x + log₃4 = 2

∴ log₃4x = 2

∴ 4x = 3² = 9

∴ x = 9/4

Example 14:

Find x, if log₂(x² + 7) = 4

Solution:

Given: log₂(x² + 7) = 4

∴ x² + 7 = 2⁴

∴ x² + 7 = 16

∴ x² = 9

∴ x = ±3

both x = 3 and x = -3 satisfy the given relation

∴ x = 3 and x = -3

Example 15:

Find x, if log₂x + log₂(x + 2) = 3

Solution:

Given: log₂x + log₂(x + 2) = 3

∴ log₂x(x + 2) = 3

∴ x(x + 2) = 2³

∴ x² + 2x = 8

∴ x² + 2x – 8 = 0

∴ (x + 4)(x – 2) = 0

∴  x + 4 = 0 or x – 2 = 0

∴ x = -4 and x = 2

x = -4 is not possible because the log of negative numbers not defined.

∴ x = 2

In the next article, we shall study to solve more problems on the laws of the logarithm to find the value of x.

Click Here for More Subtopics in Logarithms

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The post Use of the Laws of Logarithms: Set – VI appeared first on The Fact Factor.

Laws of Logarithms

• Log a + Log b = Log (ab) (Law of Product)
• Log a – log b = log (a/b)    b ≠ 0 (law of Quotient)
• Log a^m = m Log a (Law of exponent)
• Log (1) = 0
• Log_aa = 1
• 
• 
• 

Example 01:

If , then Show that abc = 1

Solution:

Given

∴Log a = k(y – z) ……………. (1)

∴ Log b = k(z – x) ……………. (2)

∴ Log c = k(x – y) ……………. (3)

Adding equations (1), (2) and (3)

Log a+ Log b + Log c  = k(y – z) + k(z – x) + k(x – y)

∴ Log (abc) = k(y – z + z – x + x – y)

∴ Log (abc) = 0

∴ Log (abc) = log (1)

∴ abc = 1

(Proved as required)

Example 02:

If , then Show that a^xb^yc^z = 1

Solution:

Given

∴ Log a^x = k(xy – xz) ……………. (1)

∴ Log b^y = k(yz – xy) ……………. (2)

∴ Log c^z = k(xz – yz) ……………. (3)

Adding equations (1), (2) and (3)

Log a^x + Log b^y + Log c^z  = k(xy – xz) + k(yz – xy) + k(xz – yz)

∴ Log (a^x b^y c^z ) = k(xy – xz + yz – xy + xz – yz)

∴Log (a^x b^y c^z ) = 0

∴ Log (a^x b^y c^z ) = log (1)

∴ a^x b^y c^z = 1

(Proved as required)

Example 03:

If , then Show that a^{y + z}b^{z + x}c^{x + y} = 1

Solution:

Given

∴ Log a^y+z = k(y² – z²) ……………. (1)

∴ Log b^z+x = k(z² – x²) ……………. (2)

∴Log c^x+y = k(x² – y²) ……………. (3)

Adding equations (1), (2) and (3)

Log a^y+z + Log b^z+x + Log c^x+y  = k(y² – z²) + k(z² – x²) + k(x² – y²)

∴ Log (a^y+z b^z+x c^x+y) = k(y² – z² + z² – x² + x² – y²)

∴ Log (a^y+z b^z+x c^x+y) = 0

∴ Log (a^y+z b^z+x c^x+y) = log (1)

∴ a^y+z b^z+x c^x+y = 1

(Proved as required)

Example 04:

If then Show that xyz = 1

Solution:

Given

∴ Log x = k(ab – ac) ……………. (1)

∴ Log y = k(bc – ab) ……………. (2)

∴ Log z = k(ac – bc) ……………. (3)

Adding equations (1), (2) and (3)

Log x + Log y + Log z  = k(ab – ac) + k(bc – ab) + k(ac – bc)

∴ Log (xyz) = k(ab – ac + bc – ab + ac – bc)

∴ Log (xyz) = 0

∴ Log (xyz ) = log (1)

∴ xyz= 1

(Proved as required)

Example 05:

If Show that abc = 1

Solution:

Given

∴ Log a= k(x + y – 2z) ……………. (1)

∴ Log b = k(y + z – 2x) ……………. (2)

∴ Log c= k(z + x -2y) ……………. (3)

Adding equations (1), (2) and (3)

Log a + Log b + Log c= k(x + y – 2z) + k(y + z – 2x) + k(z + x -2y)

∴ Log (abc) = k(x + y – 2z + y + z – 2x + z + x -2y)

∴ Log (abc) = 0

∴ Log (abc) = log (1)

∴ abc = 1

(Proved as required)

Example 06:

If and a³b²c = 1, find value of k

Solution:

Given

∴ Log₂a³ = 12x ……………. (1)

∴ Log₂b² = 12x ……………. (2)

∴ Log₂c= 3kx ……………. (3)

Adding equations (1), (2) and (3)

Log₂a³ + Log₂b² + Log₂c  = 12x + 12x + 3kx

∴ Log₂ (a³ b²c) = 24x + 3kx

∴ Log₂ (1) = x(24 + 3k)

∴ 0 = x(24 + 3k)

∴ 24 + 3k = 0

∴ 3k = – 24

∴ K = – 8

Example 07:

If and a⁴b³c^-2 = 1, find value of x

Solution:

Given

∴ Loga⁴ = 4kx ……………. (1)

∴ Logb³ = 6k ……………. (2)

∴ Logc^-2=  – 10k ……………. (3)

Adding equations (1), (2) and (3)

Loga⁴ + Logb³ + Logc^-2  = 4kx + 6k – 10k

∴ Log(a⁴b³c^-2) = k(4x + 6 – 10)

∴ Log₂ (1) = = k(4x  – 4)

∴ 0 = k(4x  – 4)

∴ 4x – 4 = 0

∴ 4x = 4

∴ x = 1

Example 08:

If a² = b³ = c⁵ = d⁶, then show that log_dabc = 31/5

Solution:

Given

a² = b³ = c⁵ = d⁶

∴ a = d^6/2 = d³

∴ b = d^6/3 = d²

∴ c = d^6/5

∴ L.H.S. = log_dabc

∴ L.H.S. = log_d(d³ d² d^6/5)

∴ L.H.S. = log_d(d^31/5)

∴ L.H.S. = (31/5) log_dd

∴ L.H.S. = (31/5)(1)

∴ L.H.S. = 31/5

(Proved as required)

Example 09:

If a² = b³ = c⁴ = d⁵, then show that log_abcd = 47/30

Solution:

Given

a² = b³ = c⁴ = d⁵

∴ b = a^2/3

∴ c = a^2/4 = d^1/2

∴ d = a^2/5

∴ L.H.S. = log_abcd

∴ L.H.S. = log_a(a^2/3  a^1/2 a^2/5)

∴ L.H.S. = log_a(d^47/30)

∴ L.H.S. = (47/30) log_aa

∴ L.H.S. = (47/30)(1)

∴ L.H.S. = 47/30

(Proved as required)

The post Logarithms appeared first on The Fact Factor.

In the next article, we shall study to solve more problems on the laws of logarithms to prove given logarithmic relation. In this article, we shall study problems based on the change of base rule.

Laws of Logarithms

• Log a + Log b = Log (ab) (Law of Product)
• Log a – log b = log (a/b)    b ≠ 0 (law of Quotient)
• Log a^m = m Log a (Law of exponent)
• Log (1) = 0
• Log_aa = 1
• 
• 
• 

Change of Base Rule

Example 01:

Show that

Solution:

(Proved as required)

Example 02:

Show that

Log_ab . log_bc . log_cd = log_ad

Solution:

Log_ab . log_bc . log_cd = log_ad 

(Proved as required)

Example 03:

Show that

Log_ba . log_cb . log_ac = 1

Solution:

Log_ba . log_cb . log_ac = 1

(Proved as required)

Example 04:

Show that

Solution:

(Proved as required)

Example 05:

Show that

Solution:

(Proved as required)

Example 06:

Show that

Solution:

(Proved as required)

Example 07:

Show that

Solution:

(Proved as required)

Example 08:

Show that

Solution:

(Proved as required)

Example 09:

Show that

Solution:

(Proved as required)

Example 10:

Show that

Solution:

(Proved as required)

Example 11:

Show that

Solution:

(Proved as required)

Example 12:

Show that

Solution:

(Proved as required)

Example 13:

Show that

Solution:

(Proved as required)

Example 14:

Show that

Solution:

(Proved as required)

Example 15:

Show that

Solution:

In the next article, we shall study to solve problems on the laws of the logarithm to find the value of x.

Click Here for More Subtopics in Logarithms

Click Here for More Topics in Mathematics

The post Use of the Laws of Logarithms: Set – V appeared first on The Fact Factor.

In the last article, we have studied to solve problems on the laws of logarithms to evaluate or simplify given logarithmic expression. In this article, we shall study to solve more problems on these laws to prove given logarithmic expression.

Laws of Logarithms

• Log a + Log b = Log (ab) (Law of Product)
• Log a – log b = log (a/b)    b ≠ 0 (law of Quotient)
• Log a^m = m Log a (Law of exponent)
• Log (1) = 0
• Log_aa = 1
• 
• 
• 

Example 01:

Show that,

log 360 = 3 log2 + 2 log 3+ log 5

Solution:

L.H.S. = log 360

∴ L.H.S. = log (2 x 2 x 2 x 3 x 3 x 5)

∴ L.H.S. = log (2³ x 3² x 5)

∴L.H.S. = log 2³ + log 3² + log 5

∴L.H.S. = 3 log2 + 2 log 3+ log 5

∴ L.H.S. = R.H.S.

∴Log 360 = 3 log2 + 2 log 3+ log 5 (Proved as required)

Example 02:

Show that,

log 540 = 2 log2 + 3 log 3+ log 5

Solution:

L.H.S. = log 540

∴ L.H.S. = log (2 x 2 x 3 x 3 x 3 x 5)

∴ L.H.S. = log (2² x 3³ x 5)

∴L.H.S. = log 2² + log 3³ + log 5

∴L.H.S. = 2 log2 + 3 log 3+ log 5

∴ L.H.S. = R.H.S.

∴Log 540 = 2 log2 + 3 log 3+ log 5 (Proved as required)

Example 03:

Show that

Solution:

(Proved as required)

Example 04:

Show that

Solution:

(Proved as required)

Example 05:

Show that

Solution:

(Proved as required)

Example 06:

Show that

Solution:

(Proved as required)

Example 07:

Show that

Solution:

(Proved as required)

Example 08:

Show that

Solution:

Example 09:

Show that

Log(log x⁷) – log(logx³) = log(7/3)

Solution:

Log(log x⁷) – log(logx³) = log(7/3)

(Proved as required)

Example 10:

Show that

Log(log x⁴) – log(logx) = log 4

Solution:

Log(log x⁴) – log(log x) = log 4

(Proved as required)

Example 11:

Show that

Log(log x⁴) – log(logx⁶) = log(2/3)

Solution:

Log(log x⁴) – log(logx⁶) = log(2/3)

(Proved as required)

Example 12:

Show that

Solution:

(Proved as required)

Example 13:

Show that

Solution:

(Proved as required)

Example 14:

Show that

log(1 + 2 + 3) = log 1 + log2 + log3

Solution:

L.H.S. = log(1 + 2 + 3)

∴ L.H.S. = log(6)

∴ L.H.S. = log(1 x 2 x 3)

∴ L.H.S. = log 1 + log2 + log3

∴ L.H.S. = R.H.S.

∴ log(1 + 2 + 3) = log 1 + log2 + log3

(Proved as required)

Example 15:

Show that

Solution:

(Proved as required)

In the next article, we shall study to solve more problems on these laws to prove given relation using given logarithmic expression.

Click Here for More Subtopics in Logarithms

Click Here for More Topics in Mathematics

The post Use of the Laws of Logarithms: Set – II appeared first on The Fact Factor.