LOGARITHMS -- DEFINITION, FORMULAS WITH PROOFS, SOLVED EXAMPLES, EXERCISES

Before you proceed , Look at the need of Logarithms in Exponents, explained at the end, by means of an example,if you have not already done so. Also knowledge of Exponents is a prerequisite here.

Look at a question in exponents: what is x if 7^x = 343. The same question can be asked here by using a symbol.The question is " what is log₇ 343 ? " If you understand this algebraic symbol, which you will, a little later, you will understand the question as find x such that 7^x = 343. More over while solving the question, you will write 7^x = 343 = 7³ ⇒ x = 3. Since 343 is 7³, you could easily answer it. If I give you find x such that 7^x = 100, what will you do? With the symbolism you learn here, you will answer it as x = log₇ 100. It is not only the symbolism for finding the exponents, but a useful body of knowledge, you will learn here.

Definition of Logarithm:

If a(≠1) and n are any positive real numbers and for some real x, a^x = n, then x is said to be the logarithm of n to the base a.
It is written as log_a n = x.

For real values of a, x, n and a≠1, a > 0, n > 0
a^x = n ⇒ log_a n = x

The logarithms of the same number to different bases are different.

Some more examples:
64^1⁄3 = 4 ⇒ log₆₄ 4 = 1⁄3
5^-2 = 1⁄25 ⇒ log₅ (1⁄25) = -2

You may note the following points.

• Base of the logarithm (a in definition) is positive and not equal to 1.
  
• Here logarithms are defined only for positive real numbers (n in definition).
  
• The value of the logarithm (x in definition) can be negative.
  
• a^x = n is called exponential form.
  
• log_a n = x is called logarithmic form.
  
• Both in exponential form and logarithmic form, the base is same.
  
• The exponent in exponential form is the value of the logarithm (to the same base to which the exponent is raised) in logarithmic form.
  
• Also there exists a unique x which satisfies the equation a^x = n. So log_a n is unique.
  

Nature of Logarithmic Function:

Look at the following two sets of examples
1st set:
2³ = 8 ⇒ log₂ 8 = 3
2⁴ = 16 ⇒ log₂ 16 = 4
2⁵ = 32 ⇒ log₂ 32 = 5
Here as the number (8, 16, 32) increases, the log of the number (3, 4, 5) increases. Note that the base (2) is more than 1.
What is seen to be true here in this example, is infact true in general.
The general statement is

If a > 1, n₁ > n₂ ⇒ log_a n₁ > log_a n₂
i.e. when the base is more than 1, the logarithmic function is an increasing function.

2nd set:
(1⁄2)⁵ = 1⁄32 ⇒ log_(1⁄2) (1⁄32) = 5
(1⁄2)⁴ = 1⁄16 ⇒ log_(1⁄2) (1⁄16) = 4
(1⁄2)³ = 1⁄8 ⇒ log_(1⁄2) (1⁄8) = 3
Here as the number (1⁄32, 1⁄16, 1⁄8) increases, the logarithm of the number (5, 4, 3) decreases. Note that the base (1⁄2) is less than 1.
What is seen to be true here in this example, is infact true in general.
The general statement is

If 0 < a < 1, n₁ > n₂ ⇒ log_a n₁ < log_a n₂ i.e. when the base is less than 1 (and positive), the logarithmic function is a decreasing function.

Algebra Formulas in Logarithms:

In the following formulas, the base of the logarithm is positive real number (≠ 1). When we write logarithm of a literal, that literal is a positive real number.

Algebra Formula 1 in Logarithms:
Formula from definition of Logarithm:

a^x = n ⇔ log_a n = x

Algebra Formula 2 in Logarithms:
Logarithm as an Exponent to its Base:

a^{log_a n} = n

Algebra Formula 3 in Logarithms: Logarithm of 1 to any Base:

log_a 1 = 0

Algebra Formula 4 in Logarithms:
Logarithm of any number to the same Base:

log_a a = 1

Algebra Formula 5 in Logarithms:
Logarithm of a Product:

log_a (mn) = log_a m + log_a n

Algebra Formula 6 in Logarithms:
Logarithm of a Quotient:

log_a (m⁄n) = log_a m - log_a n

Algebra Formula 7 in Logarithms:
Logarithm of a Power:

log_a (mⁿ) = n log_a m

Algebra Formula 8 in Logarithms:
Change of Base of a Logarithm:

log_a b = (log_k b)⁄(log_k a)

Algebra Formula 9 in Logarithms:
Logarithm of a Power to the Base of another Power:

log_aⁿ (b^m) = (m⁄n) log_a b

Algebra Formula 10 in Logarithms:
Logarithm of b to the base a; b and a interchanged:

log_a b = 1⁄(log_b a)

All the above 10 Formulas are to be remembered and to be applied to solve various problems. What we said about Formulas and problems in Exponents holds good here also. We repeat that here.

The best way to remember various Algebra Formulas (or Math Formulas) is to apply them to a number of problems. The best way to solve Algebra Problems (or Math Problems) is to remember various formulas.

Proofs and Explanations for Formulas in Logarithms:

Proof and Explanation of Formula 1 in Logarithms:
Formula from definition of Logarithm:

a^x = n ⇔ log_a n = x

• Both in exponential form and logarithmic form, the base is same.
  
• The exponent in exponential form is the value of the logarithm in logarithmic form.

a^{(log_a n)} = n

Proof and Explanation of Formula 3 in Logarithms:
Logarithm of 1 to any Base:

log_a 1 = 0

Proof and Explanation of Formula 4 in Logarithms:
Logarithm of any number to the same Base:

log_a a = 1

Proof and Explanation of Formula 5 in Logarithms:
Logarithm of a Product:

log_a (mn) = log_a m + log_a n

Proof and Explanation of Formula 6 in Logarithms:
Logarithm of a Quotient:

log_a (m⁄n) = log_a m - log_a n

Proof and Explanation of Formula 7 in Logarithms:
Logarithm of a Power:

log_a (mⁿ) = n log_a m

Proof and Explanation of Formula 8 in Logarithms:
Change of Base of a Logarithm:

log_a b = (log_k b)⁄(log_k a)

Proof and Explanation of Formula 9 in Logarithms:
Logarithm of a Power to the Base of another Power:

log_(aⁿ) (b^m) = (m⁄n) log_a b

Proof and Explanation of Formula 10 in Logarithms:
Logarithm of b to the base a; b and a interchanged:

log_a b = 1⁄(log_b a)

Solved Examples : Logarithms

Let us see solved Examples on
Application of these Formulas.

Solved Example 1 of Logarithms:

If log_x y = log_y z = log_z x, prove that x = y = z

Solution:
Let log_x y = log_y z = log_z x = K
By converting from Logarithmic Form to Exponential Form (See Formula 1), we get
x^K = y........(i)
y^K = z.............(ii)
z^K = x.............(iii)
Using the value of y from (i) in (ii), we get
z = y^K = (x^K)^K = x^{K x K} = x^K2.....(iv)
Using the value of x from (iii) in (iv), we get
z = x^K2 = (z^K)^K2
= z^{K x K2} = z^K3
Thus we have z¹ = z^K3
Since the bases are equal, the exponents have to be equal.
∴ K³ = 1 ⇒ K = 1
Using the value of K in (i) and (ii), we get x¹ = y and y¹ = z.
⇒ x = y = z (Proved.)

Solved Example 2 of Logarithms:

If (log x)⁄(b - c) = (log y)⁄(c - a) = (log z)⁄(a - b), then prove that
xyz = x^ay^bz^c = x^{(b + c)}y^{(c + a)}z^{(a + b)} = 1

Solution:
Let (log x)⁄(b - c) = (log y)⁄(c - a) = (log z)⁄(a - b) = K
⇒ log x = K(b - c); log y = K(c - a); log z = K(a - b);
Let the base of the logarithms be p.
Then the above three equations become
log_p x = K(b - c); log_p y = K(c - a); log_p z = K(a - b)
By converting from logarithmic form to exponential form (See Formula 1), we get
p^{K(b - c)} = x; p^{K(c - a)} = y; p^{K(a - b)} = z;
⇒ xyz = p^{K(b - c)} x p^{K(c - a)} x p^{K(a - b)}
= p^{{K(b - c) + K(c - a) + K(a - b)}} = p^{{K(b - c + c - a + a - b)}}
= p^{K(0)} = p⁰ = 1. (Proved.)

We have p^{K(b - c)} = x; p^{K(c - a)} = y; p^{K(a - b)} = z;
Raising the powers a, b, c to x, y, z respectively, we get
x^a = (p^{K(b - c)})^a
= p^{Ka(b - c)};
Similarly, y^b = p^{Kb(c - a)}
and z^c = p^{Kc(a - b)}
Multiplying these three, we get
x^ay^bz^c = p^{Ka(b - c)}p^{Kb(c - a)}p^{Kc(a - b)} = p^{{Ka(b - c) + Kb(c - a) + Kc(a - b)}} = p^{K{ab - ac + bc - ba + ca - cb}}
= p^K{0} = p⁰ = 1. (Proved.)

Raising the powers (b + c), (c + a), (a + b) to x, y, z respectively, we get
x^{(b + c)} = {p^{K(b - c)}}^{(b + c)}
= p^{K(b + c)(b - c)} = p^{K(b2 - c2)};
Similarly, y^{(c + a)} = p^{K(c2 - a2)}
and z^{(a + b)} = p^{K(a2 - b2)}
Multiplying these three, we get
x^{(b + c)}y^{(c + a)}z^{(a + b)}
= p^{K(b2 - c2)}p^{K(c2 - a2)}p^{K(a2 - b2)}
= p^{K(b2 - c2) + K(c2 - a2) + K(a2 - b2)}
= p^{K(b2 - c2 + c2 - a2 + a2 - b2)}
= p^K(0) = p⁰ = 1.(Proved.)

Solved Example 3 of Logarithms:

Express the following as single logarithms to any base.
(i) (1⁄2) log (a + b) - (1⁄2) log (a - b)
(ii) log (a + b) + log (a² - ab + b²) - log (x - y) - log (x² + xy + y²)
(iii) (3⁄2) log x - (1⁄3) log y + (2⁄3) log z - (1⁄5) log a

Solution:
(i) Let P = (1⁄2) log (a + b) - (1⁄2) log (a - b)
We know in log of a power (see Formula 7) , the exponent multiplies the log. Considering the reverse, If a quantity multiplies the log, it becomes exponent.
∴ P = log {(a + b)^(1⁄2)} - log {(a - b)^(1⁄2)} We know, log of a quotient can be written as the difference of the logs of the numerator and denominator of the quotient (See Formula 6)
Considering the reverse, the difference of the logs of different quantities can be written as log of a quotient (the quantity whose log has negative sign, in the denominator) of the quantities.
= log {(a + b)^(1⁄2)}⁄{(a - b)^(1⁄2)}
= log {(a + b)⁄(a - b)}^(1⁄2). Ans.

(ii) Let P = log (a + b) + log (a² - ab + b²) - log (x - y) - log (x² + xy + y²)
= log (a + b) + log (a² - ab + b²) - {log (x - y) + log (x² + xy + y²)}
We know, log of a product can be written as the sum of the logs of the factors of the product (See Formula 5)
Considering the reverse, the sum of the logs of different quantities can be written as log of product of the quantities.
∴ P = log {(a + b)(a² - ab + b²)} - log {(x - y)(x² + xy + y²)}
We know from Polynomials, (a + b)(a² - ab + b²) = a³ + b³;
and (x - y)(x² + xy + y²) = x³ - y³
∴ P = log (a³ + b³) - log (x³ - y³)
We know, log of a quotient can be written as the difference of the logs of the numerator and denominator of the quotient (See Formula 6)
Considering the reverse, the difference of the logs of different quantities can be written as log of a quotient (the quantity whose log has negative sign, in the denominator) of the quantities.
∴ P= log {(a³ + b³)⁄(x³ - y³)} Ans.

(iii) Let P = (3⁄2) log x - (1⁄3) log y + (2⁄3) log z - (1⁄5) log a
We know in log of a power (see Formula 7) , the exponent multiplies the log. Considering the reverse, If a quantity multiplies the log, it becomes exponent.
∴ P = log x^(3⁄2) + log z^(2⁄3) - {log y^(1⁄3) + log a^(1⁄5)}
We know, log of a product can be written as the sum of the logs of the factors of the product (See Formula 5)
Considering the reverse, the sum of the logs of different quantities can be written as log of product of the quantities.
∴ P = log {x^(3⁄2) x z^(2⁄3)} - log {y^(1⁄3) x a^(1⁄5)}
We know, log of a quotient can be written as the difference of the logs of the numerator and denominator of the quotient (See Formula 6)
Considering the reverse, the difference of the logs of different quantities can be written as log of a quotient (the quantity whose log has negative sign, in the denominator) of the quantities.
∴ P = log [{x^(3⁄2) x z^(2⁄3)}⁄{y^(1⁄3) x a^(1⁄5)}] Ans.

Solved Example 4 of Logarithms:

Solve the equations
(i) 3^{(2x + 1)}.4^{(4x - 1)} = 36
(ii) 4^{(log₉ 3)} + 9^{(log₂ 4)} = 10^{(log_x 83)}

Solution:
(i) 3^{(2x + 1)}.4^{(4x - 1)} = 36
Taking logarithms on both sides, we get
log {3^{(2x + 1)}.4^{(4x - 1)}} = log 36
⇒ log {3^{(2x + 1)}} + log {4^{(4x - 1)}} = log 36
We know in log of a power (see Formula 7) , the exponent multiplies the log.
∴ (2x + 1) log 3 + (4x - 1) log 4 = log 36
⇒ 2x log 3 + 4x log 4 + log 3 - log 4 = log 36
⇒ 2x(log 3 + 2 log 4) = log 36 - log 3 + log 4
We know in log of a power (see Formula 7) , the exponent multiplies the log. Considering the reverse, If a quantity multiplies the log, it becomes exponent.
∴ L.H.S. = 2x(log 3 + log 4²)
We know, log of a product can be written as the sum of the logs of the factors of the product (See Formula 5)
Considering the reverse, the sum of the logs of different quantities can be written as log of product of the quantities.
We know, log of a quotient can be written as the difference of the logs of the numerator and denominator of the quotient (See Formula 6)
Considering the reverse, the difference of the logs of different quantities can be written as log of a quotient (the quantity whose log has negative sign, in the denominator) of the quantities.
Applying these to L.H.S. and R.H.S., we get
2x{log (3 x 4²)} = log (36 x 4⁄3) = log 48
⇒ 2x{log 48} = log 48 ⇒ 2x = 1 ⇒ x = 1⁄2. Ans.

Alternative Solution: This problem can be solved from Exponents knowledge without using logarithms as given below.
3^{(2x + 1)}.4^{(4x - 1)} = 36 ⇒ 3^(2x).3¹.4^(4x).4^(-1) = 36
⇒ (3)^2x.(4²)^2x.(3⁄4) = 36
⇒ 3^2x.(16)^2x = 36 x (4⁄3) = 48 ⇒ (3 x 16)^2x = 48
⇒ (48)^2x = 48 ⇒ 2x = 1 ⇒ x = 1⁄2. Ans.

(ii) The given equation is 4^{(log₉ 3)} + 9^{(log₂ 4)} = 10^{(log_x 83)}
We have log₉ 3 = log_3² 3¹
We know, in log of a power to the base of another power (see Formula 9), the quotient of the exponents (exponent of the base in the denominator) multiplies the log.
∴ log₉ 3 = (1⁄2) log₃ 3
We know log of a quantity to the same base is 1. (See Formula 4)
∴ log₉ 3 = (1⁄2) (1) = 1⁄2;
log₂ 4 = log₂ 2²
We know in log of a power (see Formula 7) , the exponent multiplies the log.
∴ log₂ 4 = 2 log₂ 2
We know log of a quantity to the same base is 1. (See Formula 4)
∴ log₂ 4 = 2 (1) = 2
∴ the given equation becomes 4^(1⁄2) + 9² = 10^{(log_x 83)}
⇒ 2 + 81 = 10^{(log_x 83)}
⇒ 83 = 10^{(log_x 83)} .......(i)
By converting the exponential form of (i) to logarithmic form (See Formula 1), we get
log_x 83 = log₁₀ 83
When two logarithms of same quantities are equal, their bases have to be equal.
∴ x = 10. Ans.
You can also apply Formula 2 to equation (i), to conclude x = 10.Ans.

Solved Example 5 of Logarithms:

(i) If log₁₂ 27 = a, compute log₆ 32 in terms of a
(ii) If log₃₀ 3 = a and If log₃₀ 5 = b, find log₃₀ 8

Solution:
(i) If log₁₂ 27 = a, compute log₆ 32
We have a = log₁₂ 27 = log₁₂ 3³
We know, in log of a power (See Formula 7), the exponent multiplies the log.
∴ a = 3 log₁₂ 3
We know, log of a quantity to a base can be written as the ratio of log of the quantity and log of the base.(See Formula 8)
∴ a = 3 {(log 3)⁄(log 12)}
We know 12 = 3 x 4 = 3 x 2² ⇒ log 12 = log (3 x 2²)
We know, log of a product is equal to the sum of the logs of the factors of the product (See Formula 5)
∴ log 12 = log 3 + log (2²)
We know, in log of a power (See Formula 7), the exponent multiplies the log.
∴ log (2²) = 2 log 2
Thus log 12 = log 3 + 2 log 2
∴ a = 3 {(log 3)⁄(log 3 + 2 log 2 )}
Dividing numerator and denominator by log 3, we get
a = 3 [(1)⁄{1 + 2 (log 2⁄log 3)}]
= 3⁄(1 + 2p) where p = (log 2⁄log 3)
⇒ 1 + 2p = 3⁄a ⇒ 2p = 3⁄a - 1 = (3 - a)⁄a
⇒ p = (3 - a)⁄2a ...........(i)

Let x = log₆ 32
We know, log of a quantity to a base can be written as the ratio of log of the quantity and log of the base.(See Formula 8)
∴ x = (log 32)⁄(log 6)
We know 32 = 2 x 2 x 2 x 2 x 2 = 2⁵; 6 = 2 x 3;
∴ x = {log (2⁵}⁄{log (2 x 3)}
Numerator is logarithm of a power and Denominator is logarithm of a product.
We know, in log of a power (See Formula 7), the exponent multiplies the log.
We know, log of a product is equal to the sum of the logs of the factors of the product (See Formula 5)
∴ x = {5 log 2}⁄{log 2 + log 3}
Dividing numerator and denominator by log 3, we get
x = 5 (log 2⁄log 3)⁄{(log 2⁄log 3) + 1}
= 5p⁄(p + 1) [Since p = (log 2⁄log 3)]
We have p = (3 - a)⁄2a ...........(i)
∴ p + 1 = (3 - a)⁄2a + 1 = (3 - a + 2a)⁄2a = (3 + a)⁄2a........(ii)
Substituting the value of p from (i) and (p + 1) from (ii) in x, we get
x = 5 {(3 - a)⁄2a}⁄{(3 + a)⁄2a} = 5 {(3 - a)⁄(3 + a)}. Ans.

(ii) we know 30 = 3 x 5 x 2; log₃₀ 30 = 1
∴ log₃₀ 30 = 1 ⇒ log₃₀ (3 x 5 x 2) = 1
We know, log of a product is equal to the sum of the logs of the factors of the product (See Formula 5)
∴ log₃₀ 3 + log₃₀ 5 + log₃₀ 2 = 1
⇒ a + b + log₃₀ 2 = 1 ⇒ log₃₀ 2 = 1 - a - b
log₃₀ 8 = log₃₀ (2³)
We know, in log of a power (See Formula 7), the exponent multiplies the log.
∴ log₃₀ 8 = 3(log₃₀ 2) = 3(1 - a - b). Ans.

Solved Example 6 of Logarithms:

If b = √(ac), prove that
(log_a n)⁄(log_c n) = {(log_a n) - (log_b n)}⁄{(log_b n) - (log_c n)}

Solution:
By data b = √(ac) ⇒ b² = ac ⇒ b⁄a = c⁄b.........(i)
Let x = log_n a then 1⁄x = 1⁄log_n a;
Let y = log_n b then 1⁄y = 1⁄log_n b;
Let z = log_n c then 1⁄z = 1⁄log_n c;
We know, the reciprocal of the log of a quantity to a base is equal to the log with the quantity and the base interchanged.(See Formula 10)
∴ 1⁄x = log_a n; 1⁄y = log_b n; 1⁄z = log_c n
To prove (log_a n)⁄(log_c n) = {(log_a n) - (log_b n)}⁄{(log_b n) - (log_c n)}
L.H.S. = (log_a n)⁄(log_c n) = (1⁄x)⁄(1⁄z) = z⁄x ......(ii)
R.H.S. = {(log_a n) - (log_b n)}⁄{(log_b n) - (log_c n)}
= {(1⁄x) - (1⁄y)}⁄{(1⁄y) - (1⁄z)} = {(y - x)⁄(xy)}⁄{(z - y)⁄(yz)}
= {(y - x)⁄(xy)} x {(yz)⁄(z - y)} = (z⁄x){(y - x)⁄(z - y)}.........(iii)
(y - x)⁄(z - y) = (log_n b - log_n a)⁄(log_n c - log_n b)
We know, log of a quotient can be written as the difference of the logs of the numerator and denominator of the quotient (See Formula 6)
Considering the reverse, the difference of the logs of different quantities can be written as log of a quotient (the quantity whose log has negative sign, in the denominator) of the quantities.
∴ (y - x)⁄(z - y) = {log_n (b⁄a)}⁄{log_n (c⁄b)} = 1 [Since b⁄a = c⁄b from (i)]
Using this value in (iii), we get
R.H.S. = (z⁄x) (1) = z⁄x
From (ii), L.H.S. = z⁄x
∴ L.H.S. = R.H.S. (Proved.)

Solved Example 7 of Logarithms:

If log_a bc = x, log_b ca = y, log_c ab = z,
then prove that 1⁄(x + 1) + 1⁄(y + 1) + 1⁄(z + 1) = 1.

Solution:
By data, log_a bc = x
By converting from Logarithmic Form to Exponential Form (See Formula 1), we get
a^x = bc
Multiplying both sides with a, we get
a x a^x = a x bc ⇒ a^{x + 1} = abc
By converting from Exponential Form to Logarithmic Form (See Formula 1), we get
x + 1 = log_a abc
Taking reciprocals, we get
Reciprocal of L.H.S. = 1⁄(x + 1);
Reciprocal of R.H.S. = 1⁄(log_a abc)
We know, the reciprocal of the log of a quantity to a base is equal to the log with the quantity and the base interchanged.(See Formula 10)
∴ Reciprocal of R.H.S. = log_abc a
∴ 1⁄(x + 1) = log_abc a.......(i)

Next two paragraphs are repetition of the procedure in the above paragraph with variables changed. You can use the word 'Similarly' and write Equations (ii) and (iii) directly. But I am giving below the repititions.Two more repetitions of the Application of Formula 1 (both ways) and Formula 10 will help you remember them.

By data, log_b ca = y
By converting from logarithmic form to exponential form (See Formula 1), we get
b^y = ca
Multiplying both sides with b, we get
b x b^y = b x ca ⇒ b^{y + 1} = bca = abc
By converting from Exponential Form to Logarithmic Form (See Formula 1), we get
y + 1 = log_b abc
Taking reciprocals, we get
Reciprocal of L.H.S. = 1⁄(y + 1)
Reciprocal of R.H.S. = 1⁄(log_b abc)
We know, the reciprocal of the log of a quantity to a base is equal to the log with the quantity and the base interchanged.(See Formula 10)
∴ Reciprocal of R.H.S. = log_abc b
∴ 1⁄(y + 1) = log_abc b .......(ii)

By data, log_c ab = z
By converting from logarithmic form to exponential form (See Formula 1), we get
c^z = ab
Multiplying both sides with c, we get
c x c^z = c x ab ⇒ c^{z + 1} =ca b = abc
By converting from Exponential Form to Logarithmic Form (See Formula 1), we get
z + 1 = log_c abc
Taking reciprocals, we get
Reciprocal of L.H.S. = 1⁄(z + 1)
Reciprocal of R.H.S. = 1⁄(log_c abc)
We know, the reciprocal of the log of a quantity to a base is equal to the log with the quantity and the base interchanged.(See Formula 10)
∴ Reciprocal of R.H.S. = log_abc c
∴ 1⁄(z + 1) = log_abc c .......(iii)

(i) + (ii) + (iii) gives
1⁄(x + 1) + 1⁄(y + 1) + 1⁄(z + 1) = log_abc a + log_abc b + log_abc c
= log_abc abc
We know log of a quantity to the same base is 1. (See Formula 4)
∴ 1⁄(x + 1) + 1⁄(y + 1) + 1⁄(z + 1) = 1. (proved.)

Solved Example 8 of Logarithms:

If x = log _2a a, y = log _3a 2a, z = log _4a 3a,
then prove that 1 + xyz = 2yz

Solution:
We know, log of a quantity to a base can be written as the ratio of log of the quantity and log of the base.(See Formula 8)
x = log _2a a = (log a)⁄(log 2a)
y = log _3a 2a = (log 2a)⁄(log 3a)
z = log _4a 3a = (log 3a)⁄(log 4a)
xyz = {(log a)⁄(log 2a)}{(log 2a)⁄(log 3a)}{(log 3a)⁄(log 4a)}
= (log a)⁄(log 4a)
We know, log of a quantity to a base can be written as the ratio of log of the quantity and log of the base.(See Formula 8)
Considering the reverse, the ratio of logs of two quantities can be written as a single log with the denominator quantity as base.
∴ xyz = log _4a a
Now to prove 1 + xyz = 2yz
L.H.S. = 1 + xyz = 1 + (log a)⁄(log 4a) = {(log 4a) + (log a)}⁄(log 4a)
We know, log of a product can be written as the sum of the logs of the factors of the product (See Formula 5)
Considering the reverse, the sum of the logs of different quantities can be written as log of product of the quantities.
∴ L.H.S. = {log (4a x a)}⁄(log 4a) = {log (4a²)}⁄(log 4a)
= {log (4a²)}⁄(log 4a) = {log (2a)²}⁄(log 4a)
We know, in log of a power (See Formula 7), the exponent multiplies the log.
∴ L.H.S. = {2 log (2a)}⁄(log 4a)
R.H.S. = 2yz = 2 {(log 2a)⁄(log 3a)}{(log 3a)⁄(log 4a)}
= 2 (log 2a)⁄(log 4a)
∴ L.H.S. = R.H.S. (proved.)

Solved Example 9 of Logarithms:

Find the value of
(i) log₃₄₃ 49 (ii) log_0.01 (0.0001) (iii) log_(2√3) (1728)

Solution:
(i) Let A = log₃₄₃ 49
We know 49 = 7 x 7 = 7²; 343 = 49 x 7 = 7 x 7 x 7 = 7³
∴ A = log₃₄₃ 49 = log_(7³) (7²)
We know, in log of a power to the base of another power (See Formula 9), the quotient of the exponents (exponent of the base in the denominator) multiplies the log.
∴ A = (2⁄3) log₇ 7
But log of any number to the same Base is one (see Formula 4).
∴ A = (2⁄3) (1) = 2⁄3. Ans.

(ii) Let A = log_0.01 (0.0001)
We know 0.0001 = 0.01 x 0.01 = (0.01)²;
∴ A = log_(0.01) (0.0001) = log_(0.01) {(0.01)²}
We know, in log of a power (see Formula 7), the exponent multiplies the log.
∴ A = 2 log_(0.01) (0.01)
But log of any number to the same Base is one (see Formula 4).
∴ A = 2 (1) = 2. Ans.

(iii) Let A = log_(2√3) (1728)
We know 2√3 = 2 x (3)^(1⁄2) = (2²)^(1⁄2) x (3)^(1⁄2)
= 4^(1⁄2) x (3)^(1⁄2) = (4 x 3)^(1⁄2) = 12^(1⁄2)
1728 = 12 x 144 = 12 x 12 x 12 = 12³
See how the base (2√3) and 1728 are made as powers of the same number 12.
∴ A = log_(2√3) (1728) = log_{12^(1⁄2)} (12³)
We know, in log of a power to the base of another power (see Formula 9), the quotient of the exponents (exponent of the base in the denominator) multiplies the log.
∴ A = {3⁄(1⁄2)} log₁₂ 12
But log of any number to the same Base is one (see Formula 4).
∴ A = (3 x 2) (1) = 6.
Thus log_(2√3) (1728) = 6. Ans.

Solved Example 10 of Logarithms:

If (2.3)^x = (0.23)^y = 1000, show that 1⁄x - 1⁄y = 1⁄3

Solution:
By data (2.3)^x = 1000
By converting from Exponential Form to Logarithmic Form (See Formula 1), we get
x = log_(2.3) 1000
Taking reciprocals, we get
Reciprocal of L.H.S. = 1⁄x
Reciprocal of R.H.S. = 1⁄(log_(2.3) 1000)
We know, the reciprocal of the log of a quantity to a base is equal to the log with the quantity and the base interchanged.(See Formula 10)
∴ Reciprocal of R.H.S. = log₁₀₀₀ (2.3)
∴ 1⁄x = log₁₀₀₀ (2.3)

By data (0.23)^y = 1000
By converting from Exponential Form to Logarithmic Form (See Formula 1), we get
y = log_(0.23) 1000
Taking reciprocals, we get
Reciprocal of L.H.S. = 1⁄y
Reciprocal of R.H.S. = 1⁄(log_(0.23) 1000)
We know, the reciprocal of the log of a quantity to a base is equal to the log with the quantity and the base interchanged.(See Formula 10)
∴ Reciprocal of R.H.S. = log₁₀₀₀ (0.23)
∴ 1⁄y = log₁₀₀₀ (0.23)

To prove 1⁄x - 1⁄y = 1⁄3
L.H.S. = 1⁄x - 1⁄y = log₁₀₀₀ (2.3) - log₁₀₀₀ (0.23)
We know, log of a quotient can be written as the difference of the log of the numerator and denominator of the quotient (See Formula 6)
Considering the reverse, the difference of the logs of different quantities can be written as log of a quotient (the quantity whose log has negative sign, in the denominator) of the quantities.
∴ L.H.S. = log₁₀₀₀ {(2.3)⁄(0.23)} = log₁₀₀₀ (10)
= log_10³ (10¹)
We know, in log of a power to the base of another power (see Formula 9), the quotient of the exponents (exponent of the base in the denominator) multiplies the log.
∴ L.H.S. = (1⁄3) log₁₀ 10
We know log of a quantity to the same base is 1. (See Formula 4)
∴ L.H.S. = (1⁄3) (1) = 1⁄3 = R.H.S. (Proved.)

Logarithm Tables : Characteristic and Mantissa

To find the log value of a number using Tables
we need the knowledge of Charateristic and Mantissa
which is covered in Logarithm Tables.

Exercise : Logarithms

1. If a^x = b^y = c^z and y² = zx,
   Prove that log_b a = log_c b.
   
2. Find the value of x from the equation
   a^{(3 - x)}.b^5x = a^3x.b^{(x + 5)}
   
3. Solve 4^x + 2^{(2x - 1)} = 3^{(x + 1⁄2)} + 3^{(x - 1⁄2)}
   (i) by using logarithms (ii) without using logarithms.
   
4. If (log_y x.log_z x - log_x x) + (log_z y.log_x y - log_y y) + (log_x z.log_y z - log_z z) = 0,
   prove that xyz = 1.
   
5. If 1⁄x = 1 + log_a bc, 1⁄y = 1 + log_b ca, 1⁄z = 1 + log_c ab,
   then prove that x + y + z = 1.
   
6. If a = log₂₄ 12, b = log₃₆ 24, c = log₄₈ 36,
   then prove that 1 + abc = 2bc.
   
7. Find the value of (i) log₂₄₃ 9 (ii) log₃ √(243)
   
8. If (3.4)^x = (0.034)^y = 10000,
   show that 1⁄x - 1⁄y = 1⁄2
   
9. If (log₂ a)⁄4 = (log₂ b)⁄6 = (log₂ c)⁄(3p) and a³.b².c = 1,
   find the value of p
   
10. If 2 log_x a + log_ax a + 3 log_a²x a = 0,
    find the value of x.

    For Answers see at the bottom of the page.

    Answers to Exercise : Logarithms

    (2) (5 log b - 3 log a)⁄{4(log b - log a)} (3) 3⁄2 (7) (i) 2⁄5 (ii) 5⁄2 (9) -8 (10) a^-1⁄2, a^-4⁄3