PROPERTIES OF LOGARITHMS -- PROOFS AND EXPLANATIONS OF FORMULAS, LINKS FOR FURTHER STUDY
Please study
Introduction to Logarithms before Properties of Logarithms
if you have not already done so.
There we discussed the need for extension
of Exponents and introduction to the new
branch of study called Logarithms.
It is a prerequisite here.
Proofs and Explanations for
Formulas in Logarithms :
Proof and Explanation of Formula 1 :
Properties of Logarithms
Formula from definition of Logarithm:
ax = n ⇔ loga n = x This formula is directly from
definition of Logarithm.
This formula from
exponential form to logarithmic form
and from
logarithmic form to exponential form
is useful in solving many a problem.
The student should be
thoroughly familiar with it
in applying in both directions.
In remembering this formula,
the following two points are helpful.
- 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.
Proof and Explanation of Formula 2 :
Properties of Logarithms
Logarithm as an Exponent to its Base:
a(loga n) = n Proof:
From the definition of logarithm, we have
if ax = n..............(i)
then x = loga n ................(ii)
Substituting the value of x from (ii) in (i), we get
a(loga n) = n (Proved.)
In remembering this formula,
you have to observe that the
base of the exponential form
is same as the base of the
logarithm in the exponent.
Proof and Explanation of Formula 3 :
Properties of Logarithms
Logarithm of 1 to any Base:
loga 1 = 0 Proof:
From Laws of Exponents, we know a0 = 1
⇒ loga 1 = 0 (Proved.)
It is easy to remember Logarithm of 1 to any Base is zero.
Proof and Explanation of Formula 4 :
Properties of Logarithms
Logarithm of any number to the same Base:
loga a = 1 Proof:
From Laws of Exponents, we know a1 = a
⇒ loga a = 1 (Proved.)
It is easy to remember Logarithm of any number to the same Base is one.
Proof and Explanation of Formula 5 :
Properties of Logarithms
Logarithm of a Product:
loga (mn) = 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) x (ii) gives aP x aQ = mn
⇒ aP + Q = mn ( Since aP x aQ = aP + Q From laws of Exponents)
⇒ loga (mn) = P + Q ( by changing Exponential form to Logarithmic form)
By Replacing the values of P and Q, we get
loga (mn) = loga m + loga n (Proved.)
In proving this, see how we made use of
changing Logarithmic to Exponential form
and Exponential to Logarithmic form.
Remember that Logarithm of a Product
is the sum of the Logarithms of
the Factors of the Product.
Remember that the Formula is not for log(m + n) nor for log m x log n.
We have Formula for log (mn) 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.
Proofs and Explanations of Formula 6 to 10 :
Properties of Logarithms
The following Link takes you to the
Proofs and Explanations of Formula 6 to 10.
Proofs and Explanations of Formula 6 to 10
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