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# If ${\log _a}x,\,\,{\log _b}x,\,\,{\log _c}x\,$ are in A.P. where x$\ne$1, then show that ${c^2} = {\left( {ac} \right)^{\left( {{{\log }_a}b} \right)}}$. Verified
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Hint: Here in the given question ${\log _a}x,\,\,{\log _b}x,\,\,{\log _c}x\,$ are in Arithmetic progression, then the middle term will be the arithmetic mean of the other two terms. The terms given in A.P are in logarithmic form so while solving the problem we apply logarithm rules such as product rule, quotient rule and power rule.

Complete step-by-step solution:
The logarithm is the inverse function to exponent. The base b logarithm of a number is the exponent that we need to raise in order to get the number.
Given ${\log _a}x,\,\,{\log _b}x,\,\,{\log _c}x\,$ are in arithmetic progression then the common difference between the consecutive terms will be equal
$\Rightarrow$$d = \,{\log _b}x - {\log _a}x$-----------(1)
$\Rightarrow d = {\log _c}x\, - {\log _b}x$-----------(2)
Equating (1) and (2) we get
$\Rightarrow {\log _b}x - {\log _a}x = {\log _c}x\, - {\log _b}x$
On rearranging the terms
$\Rightarrow {\log _b}x\, + {\log _b}x = {\log _c}x\, + {\log _a}x$
$\Rightarrow 2{\log _b}x = {\log _a}x + {\log _c}x$
Using change of base rule, we get
$\Rightarrow$$2\left( {\dfrac{{\log x}}{{\log b}}} \right) = \dfrac{{\log x}}{{\log a}} + \dfrac{{\log x}}{{\log c}}$
On taking log x common on both the sides we get
$\Rightarrow 2\left( {\log x} \right)\left( {\dfrac{1}{{\log b}}} \right) = \log x\left( {\dfrac{1}{{\log a}} + \dfrac{1}{{\log c}}} \right)$
Cancelling log x on both the sides the equation reduces as following
$\Rightarrow 2\left( {\dfrac{1}{{\log b}}} \right) = \left( {\dfrac{1}{{\log a}} + \dfrac{1}{{\log c}}} \right)$-------(3)
On simplifying equation (3) we get
$\Rightarrow \dfrac{2}{{\log b}} = \dfrac{{\log c + \log a}}{{\log a\log c}}$-------(4)
On rearranging the equation (4)
$\Rightarrow 2\log c = \left( {\dfrac{{\log b}}{{\log a}}} \right)\left( {\log c + \log a} \right)\,\,\,\,$--------(5)
Using product rule $\log m + \log n = \log mn$ in equation (5)
$\Rightarrow 2\log c = \left( {{{\log }_a}b} \right)\left( {\log ac} \right)$-------(6)
Using power rule $n\log m = \log {m^n}$ in equation (6) we get
$\Rightarrow \log {c^2} = \log {\left( {ac} \right)^{\left( {{{\log }_a}b} \right)}}$-------(7)
Now cancelling log on both the sides we get
$\Rightarrow {c^2} = {\left( {ac} \right)^{\left( {{{\log }_a}b} \right)}}$
Hence proved
Note: Arithmetic progression is a sequence of numbers in which each successive term is a sum of its preceding term and a fixed term. We use ${T_n} = a + \left( {n - 1} \right)d$to find any term ${T_n}$ in the given arithmetic progression, where ‘a’ is the first term of the A.P, nth position of the term, d is the common difference. Similarly ${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$ is used to find the sum of n terms of the A.P.