Logarithm Rules Or Log Rules | Laws Of Logarithm - Math Only Math

How to solve logarithm equations?

There are four following math logarithm formulas:

● Product Rule Law:

loga (MN) = loga M + loga N

● Quotient Rule Law:

loga (M/N) = loga M - loga N

● Power Rule Law:

IogaMn = n Ioga M

● Change of base Rule Law:

loga M = logb M × loga b

Let’s observe the detailed step-by-step explanation of mathematical proof of logarithm rules or log rules.

1. Proof of Product Rule Law:

loga (MN) = loga M + loga N Let loga M = x ⇒ a sup>x = M and Ioga N= y ⇒ ay = N Now ax ∙ ay = MN or, ax + y = MN Therefore from definition, we have, loga (MN) = x + y = loga M + loga N [putting the values of x and y] Corollary: The law is true for more than two positive factors i.e., loga (MNP) = loga M + loga N + loga P since, loga (MNP) = 1oga (MN) + loga P = loga M+ loga N+ loga P Therefore in general, loga (MNP ….... )= loga M + loga N + loga P + ……. . Hence, the logarithm of the product of two or more positive factors to any positive base other than 1 is equal to the sum of the logarithms of the factors to the same base.

2. Proof of Quotient Rule Law:

loga (M/N) = loga M - loga N Let loga M = x ⇒ ax = M and loga N = y ⇒ ay = N Now ax/ay = M/N or, ax - y = M/N Therefore from definition we have, loga (M/N) = x - y = loga M- loga N [putting the values of x and y] Corollary: loga [(M × N × P)/R × S × T)] = loga (M × N × P) - loga (R × S × T) = loga M + Ioga N + loga P - (loga R + loga S + loga T) The formula of quotient rule [loga (M/N) = loga M - loga N] is stated as follows: The logarithm of the quotient of two factors to any positive base other than I is equal to the difference of the logarithms of the factors to the same base. Logarithm Rules or Log Rules

3. Proof of Power Rule Law:

IogaMn = n Ioga M Let loga Mn = x ⇒ ax = Mn and loga M = y ⇒ ay = M Now, ax = Mn = (ay)n = any Therefore, x = ny or, loga Mn = n loga M [putting the values of x and y].

4. Proof of Change of base Rule Law:

loga M = logb M × loga b Let Ioga M = x ⇒ ax = M, logb M = y ⇒ by = M, and loga b = z ⇒ az = b. Now, ax = M= by - (az)y = ayz Therefore x = yz or, loga M = Iogb M × loga b [putting the values of x, y, and z]. Corollary: (i) Putting M = a on both sides of the change of base rule formula [loga M = logb M × loga b] we get, loga a = logb a × loga b or, logb a × loga b = 1 [since, loga a = 1] or, logb a = 1/loga b i.e., the logarithm of a positive number a with respect to a positive base b (≠ 1) is equal to the reciprocal of logarithm of b with respect to the base a. (ii) From the log change of base rule formula we get, logb M = loga M/loga b i.e., the logarithm of a positive number M with respect to a positive base b (≠ 1) is equal to the quotient of the logarithm of the number M and the logarithm of the number b both with respect to any positive base a (≠ 1). Note: (i) The logarithm formula loga M = logb M × loga b is called the formula for the change of base. (ii) If bases are not stated in the logarithms of a problem, assume same bases for all the logarithms.Logarithm Rules or Log Rules

Summarisation of logarithm rules or log rules:

If M > 0, N > 0, a > 0, b > 0 and a ≠ 1, b ≠ 1 and n is any real number, then     (i) loga 1 = 0     (ii) loga a = 1     (iii) a Ioga M = M     (iv) loga (MN) = loga M + loga N     (v) loga (M/N) = loga M - loga N     (vi) loga Mn = n loga M     (vii) loga M = logb M × loga b     (viii) logb a × loga b = 1     (ix) 10gb a = 1/loga b     (x) logb M = 1oga M/loga b

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