Question
Question: Prove that \(\dfrac{{{{\log }_a}x}}{{{{\log }_{ab}}x}} = 1 + {\log _a}b\)...
Prove that logabxlogax=1+logab
Solution
In case of changing the bases, there can be any number at the position of base in case of logarithmic except 1. Any real number different to 1 is possible.
Logarithmic of a number x in the base a and in the base b are related as follows:
logbx=logablogax
If we know the values of logarithms in base a, we can find the values of logarithms in base b.
Logarithmic functions can be solved by using properties;
logba=logbloga
logab=loga+logb
Complete step-by-step answer:
taking Left hand side
L.H.S = logabxlogax=logablogxlogalogx
Simplifying the fraction;
= logxlogx×logalogab
= 1×[logaloga+logb]
Separating the denominator;
= logaloga+logalogb
=1+logalogb
Using base changing property;
=1+logab= R.H.S.
Hence, L.H.S. = R.H.S.
Note: In the above equation, two of the logarithmic properties have been used i.e.
logab=loga+logb and
logba=logbloga but there are many more properties of logarithm i.e. logba=loga−logb
Mainly, the properties of sum of logarithm, multiplication of the logarithmic function, divisions of logarithmic functions are used. In this equation, logarithm with the different bases is used.
There are two types of logarithmic functions
Common log
Natural log
Common log is having 10 as the base in the logarithmic function. It is represented as
logx=∣!n10∣!nA.
Natural log is having ‘e’ at the base of the logarithmic function. It is represented as
x=logaA which is also read as ;
“ x is equal to the logarithmic of A at the base a “
So logaA=x or ax=A
The most commonly used properties of logarithmic functions are :
logmn=nlogm
logmn=logm+logn
lognm=logm−logn
lognm=lognlogm