LOGARITHM RULES -- PROOFS AND EXPLANATIONS OF FORMULAS, LINKS FOR FURTHER STUDY
Proofs and Explanations for Logarithm Formulas :
Please go to
The Proofs and Explanations for
the first five Logarithm Rules (Formulas)
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:
loga (m⁄n) = loga m - loga n Proof:
Let loga m = P ⇒ aP = m ............(i)
Let loga n = Q ⇒ aQ = n .............(ii)
You might have observed that Equations (i) and (ii) are obtained by changing the Logarithmic form to Exponential form.
(i) ÷ (ii) gives aP⁄aQ = m⁄n
⇒ aP - Q = m⁄n ( Since aP⁄aQ = aP - Q From laws of Exponents)
⇒ loga (m⁄n) = P - Q ( by changing Exponential form to Logarithmic form)
By Replacing the values of P and Q, we get
loga (m⁄n) = loga m - loga 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
loga (mn) = n loga m Proof:
Let loga (mn) = P ⇒ aP = mn........(i)Let loga m = Q ⇒ aQ = 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
aP = (aQ)n
= anQ [Since (aQ)n = anQ 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
loga (mn) = n loga 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.
Proof and Explanation of Formula 8 :
Logarithm Rules
Change of Base of a Logarithm: Logarithm Rules
loga b = (logk b)⁄(logk a) Proof:
Let logk b = P ⇒ kP = b.......(i)
Let logk a = Q ⇒ kQ = 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 = loga b
Substituting the values of P and Q, we get
(logk b)⁄(logk a) = loga b
⇒ loga b = (logk b)⁄(logk 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
loga b x logk a = logk 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) (bm) = (m⁄n) loga b Proof:
L.H.S. = log(an) (bm)
Applying Formula 8, we get
L.H.S. = {logk (bm)}⁄{logk (an)}
Applying Formula 7, we get
L.H.S. = {m (logk b)}⁄{n (logk a)}
= (m⁄n)(logk b)⁄(logk a)
Applying Formula 8 again, we get
L.H.S. = (m⁄n) loga 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:
loga b = 1⁄(logb 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
loga b = (logb b)⁄(logb a)
Numerator of R.H.S.= (logb b) = 1 (See Formula 4)
∴ loga b = 1⁄(logb a) (Proved.)
Remembering this Formula also will help you in solving some problems.
Solved Examples
The following Link takes you
to the set of Solved Examples.
Set of Solved Examples
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