and study them before the Proofs and Explanations for the next five Formulas, which are dealt with here.

Proof and Explanation of Formula 6 : Logarithm Rules

Logarithm of a Quotient:

log_{a} (m⁄n) = log_{a}m - log_{a}n

Proof: Let log_{a}m = P ⇒ a^{P} = m ............(i) Let log_{a}n = Q ⇒ a^{Q} = n .............(ii) You might have observed that Equations (i) and (ii) are obtained by changing the Logarithmic form to Exponential form. (i) ÷ (ii) gives a^{P}⁄a^{Q} = m⁄n ⇒ a^{P - Q} = m⁄n ( Since a^{P}⁄a^{Q} = a^{P - Q} From laws of Exponents) ⇒ log_{a} (m⁄n) = P - Q ( by changing Exponential form to Logarithmic form) By Replacing the values of P and Q, we get log_{a} (m⁄n) = log_{a}m - log_{a}n (Proved.) In proving this, see how we made use of changing Logarithmic to Exponential form andExponential to Logarithmic form. Remember that Logarithm of a Fraction is the difference of the Logarithms oftheNumerator and Denominator. Remember that the Formula is not for log(m - n) nor forlog m⁄log n. We have Formula for log(m⁄n) and log m - log n. We should be able to apply the formula from L.H.S. to R.H.S. and from R.H.S. to L.H.S.

Proof and Explanation of Formula 7 : Logarithm Rules

Logarithm of a Power: Logarithm Rules

log_{a} (m^{n}) = n log_{a}m

Proof: Let log_{a} (m^{n}) = P ⇒ a^{P} = m^{n}........(i)Let log_{a}m = Q ⇒ a^{Q} = m.............(ii)You might have observed that Equations (i) and (ii) are obtained by changing the Logarithmic form to Exponential form. Using the value of m from (ii) in (i), we get a^{P} = (a^{Q})^{n} = a^{nQ} [Since (a^{Q})^{n} = a^{nQ} from Laws of Exponents] Since the Bases are equal, the Exponents have to be equal. ∴ P = nQ By Replacing the values of P and Q, we get log_{a} (m^{n}) = n log_{a}m (Proved.) Remember that the Logarithm of a Power is equal to the Exponent times the Logarithm of the base of the Power. Depending on the need of the problem, you should be able toapply the formula from L.H.S. to R.H.S. and from R.H.S. to L.H.S.

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Proof and Explanation of Formula 8 : Logarithm Rules

Change of Base of a Logarithm: Logarithm Rules

log_{a}b = (log_{k}b)⁄(log_{k}a)

Proof: Let log_{k}b = P ⇒ k^{P} = b.......(i) Let log_{k}a = Q ⇒ k^{Q} = a ⇒k = a^{(1⁄Q)}.......(ii) Substituting the value of k from (ii) in (i), we get {a^{(1⁄Q)}}^{P} = b ⇒a^{{(1⁄Q) x P}} = b ⇒a^{(P⁄Q)} = b ⇒ P⁄Q = log_{a}b Substituting the values of P and Q, we get (log_{k}b)⁄(log_{k}a) = log_{a}b ⇒ log_{a}b = (log_{k}b)⁄(log_{k}a) (Proved.) While remembering the formula, forget about k. Any variable can take its place. This Formula is very useful in solving many a problem. If we cross multiply the fomula, we get log_{a}b x log_{k}a = log_{k}b In this form also, we use this Formula for change of the Base.

Proof and Explanation of Formula 9 : Logarithm Rules

Logarithm of a Power to the Base of another Power:

log_{(an)} (b^{m}) = (m⁄n) log_{a}b

Proof: L.H.S. = log_{(an)} (b^{m}) Applying Formula 8, we get L.H.S. = {log_{k} (b^{m})}⁄{log_{k} (a^{n})} Applying Formula 7, we get L.H.S. = {m (log_{k}b)}⁄{n (log_{k}a)} = (m⁄n)(log_{k}b)⁄(log_{k}a) Applying Formula 8 again, we get L.H.S. = (m⁄n) log_{a}b = R.H.S. (Proved.) As you can see, this formula is a combination of Formulas 7 and 8. If you remember this formula, it will help you in solving some problems.

Proof and Explanation of Formula 10 : Logarithm Rules

Logarithm of b to the base a; b and a interchanged:

log_{a}b = 1⁄(log_{b}a)

Proof: This formula is a special case of Formula 8. We know, in formula 8, k can be replaced by any variable. If we replace k by b in Formula 8, we get log_{a}b = (log_{b}b)⁄(log_{b}a) Numerator of R.H.S.= (log_{b}b) = 1 (See Formula 4) ∴ log_{a}b = 1⁄(log_{b}a) (Proved.) Remembering this Formula also will help you in solving some problems.

Solved Examples : Logarithm Rules

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