Question

If \[{\log _a}x,{\log _b}x,{\log _c}x\] are in A.P. where \[x \ne 1\] then show that \[{c^2} = {(ac)^{{{\log }_a}b}}\].

Answer 1

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 \[{a_1},{a_2},{a_3},...{a_n}\].Nth term of an AP is given by \[{a_n} = {a_1} + (n - 1)d\] Where \[{a_1}\] is the first term, \[{a_n}\] is the last term and \[d\] is the common difference.
Formulas used:
\[\log (ab) = \log a + \log b\]
\[\log \left( {\dfrac{a}{b}} \right) = \log a - \log b\]
\[{\log _a}b = \dfrac{{\log b}}{{\log a}}\]

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 \[{a_1},{a_2},{a_3},...{a_n}\].Nth term of an AP is given by \[{a_n} = {a_1} + (n - 1)d\] Where \[{a_1}\] is the first term, \[{a_n}\] 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 \[{\log _a}x,{\log _b}x,{\log _c}x\] are in A.P.
Therefore we have \[2{\log _b}x = {\log _a}x + {\log _c}x\]
Therefore we get \[2\left( {\dfrac{{\log x}}{{\log b}}} \right) = \dfrac{{\log x}}{{\log a}} + \dfrac{{\log x}}{{\log c}}\]
On simplification we get ,
\[\dfrac{2}{{\log b}} = \dfrac{1}{{\log a}} + \dfrac{1}{{\log c}}\]
Now taking LCM we get ,
\[\dfrac{2}{{\log b}} = \dfrac{{\log c + \log a}}{{\log a\log c}}\]
On cross multiplication we get ,
\[2\log c = \left( {\dfrac{{\log b}}{{\log a}}} \right)\left( {\log c + \log a} \right)\]
On simplification we get ,
\[2\log c = {\log _a}b\left( {\log ac} \right)\]
On simplification we get ,
\[\log {c^2} = \log {(ac)^{{{\log }_a}b}}\]
Hence we get \[{c^2} = {(ac)^{{{\log }_a}b}}\].
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.