Logarithm Rules or Log Rules


In mathematics logarithm rules or log rules we have discussed mainly on logarithm laws along with their proof. If students understand the basic proof on general laws of logarithm then it will be easier to solve any types of questions on logarithm like ………

Logarithm Rules or Log Rules

  • How to change exponential form to logarithm form?

  • How to change logarithmic form to exponential form?

  • How to add logarithm?

  • How to subtract logarithm?

  • How to multiply logarithm?

  • How to divide logarithm?

  • How to write as a single logarithm?

  • Write the expression as a single logarithm?

  • 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 = loga 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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