Question
Question: If \[{\log _a}x,{\log _b}x,{\log _c}x\] are in A.P. where \[x \ne 1\] then show that \[{c^2} = {(ac)...
If logax,logbx,logcx are in A.P. where x=1 then show that c2=(ac)logab.
Solution
Hint : Arithmetic Progression (AP) is a sequence of numbers in order in which the difference of any two consecutive numbers is a constant value. General form of an AP is a1,a2,a3,...an.Nth term of an AP is given by an=a1+(n−1)d Where a1 is the first term, an is the last term and d is the common difference.
Formulas used:
log(ab)=loga+logb
log(ba)=loga−logb
logab=logalogb
Complete step-by-step answer :
Arithmetic Progression (AP) is a sequence of numbers in order in which the difference of any two consecutive numbers is a constant value. General form of an AP is a1,a2,a3,...an.Nth term of an AP is given by an=a1+(n−1)d Where a1 is the first term, an is the last term and d is the common difference.
Logarithmic functions are the inverses of exponential functions, and any exponential function can be expressed in logarithmic form. Similarly, all logarithmic functions can be rewritten in exponential form. The domain of a logarithmic function is real numbers greater than zero, and the range is real numbers.
We are given that logax,logbx,logcx are in A.P.
Therefore we have 2logbx=logax+logcx
Therefore we get 2(logblogx)=logalogx+logclogx
On simplification we get ,
logb2=loga1+logc1
Now taking LCM we get ,
logb2=logalogclogc+loga
On cross multiplication we get ,
2logc=(logalogb)(logc+loga)
On simplification we get ,
2logc=logab(logac)
On simplification we get ,
logc2=log(ac)logab
Hence we get c2=(ac)logab.
Hence showed.
Note : Logarithmic functions are the inverses of exponential functions, and any exponential function can be expressed in logarithmic form. Similarly, all logarithmic functions can be rewritten in exponential form. The domain of a logarithmic function is real numbers greater than zero, and the range is real numbers.