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# If the terms ${\log _a}x,{\log _b}x,{\log _c}x$ are in A.P where $x \ne 1$ , then show that ${c^2} = {\left( {ac} \right)^{{{\log }_a}b}}$

Last updated date: 23rd Jul 2024
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Hint: The question is taken from the sequence and series from arithmetic progression and it is an arrangement of numbers in instruction in which the difference of any two consecutive numbers always remains constant value and in this given problem which we have to prove the given statement and if the three-term are in A.P. mean arithmetic progression we follow the following step consider a, b, c are in A.P then
$b = \dfrac{{a + c}}{2}$
And since the given terms are logarithmic so we can use the logarithmic formula to obtain the statement.

Complete step-by-step solution:
Step 1:
First we write the given data or ${\log _a}x,{\log _b}x,{\log _c}x$ in A.P. so it follow
a, b, c in A.P.
$\Rightarrow$$b =\dfrac{{a + c}}{2}$
${\log _a}x,{\log _b}x,{\log _c}x$ in A.P. then
$\Rightarrow$${\log _b}x =\dfrac{{{{\log }_a}x + {{\log }_c}x}}{2}$
Now use the logarithmic formula
$\Rightarrow$${\log _p}q =\dfrac{{{{\log }_w}q}}{{{{\log }_w}p}}$
Apply the formula
$\Rightarrow$${\log _b}x =\dfrac{{{{\log }_a}x + {{\log }_c}x}}{2}$
$\Rightarrow$$\dfrac{{{{\log }_c}x}}{{{{\log }_c}b}} =\dfrac{{\dfrac{{{{\log }_c}x}}{{{{\log }_c}a}} +\dfrac{{{{\log }_c}x}}{{{{\log }_c}c}}}}{2}$
$\Rightarrow$$\dfrac{{{{\log }_c}x}}{{{{\log }_c}b}} =\dfrac{{{{\log }_c}x\left[ {\dfrac{1}{{{{\log }_c}a}} +\dfrac{1}{{{{\log }_c}c}}} \right]}}{2}$
$\Rightarrow$$\dfrac{1}{{{{\log }_c}b}} =\dfrac{{\left[ {\dfrac{1}{{{{\log }_c}a}} +\dfrac{1}{{{{\log }_c}c}}} \right]}}{2}$
$\Rightarrow$$\dfrac{1}{{{{\log }_c}b}} =\dfrac{1}{2}\left[ {\dfrac{{{{\log }_c}c + {{\log }_c}a}}{{{{\log }_c}a{{\log }_c}c}}} \right]$
Apply the formula
$\Rightarrow$${\log _q}p + {\log _q}q = {\log _q}pq,\,\,\,\,{\log _q}q = 1$
$\Rightarrow$$\dfrac{1}{{{{\log }_c}b}} =\dfrac{1}{2}\left[ {\dfrac{{{{\log }_c}ac}}{{{{\log }_c}a}}} \right]$
$\Rightarrow$$\dfrac{{{{\log }_c}a}}{{{{\log }_c}b}} = \left[ {\dfrac{{{{\log }_c}ac}}{2}} \right]$
Now again apply the formulas
$\Rightarrow$${\log _p}q =\dfrac{{{{\log }_w}q}}{{{{\log }_w}p}},\,\,{\log _p}q =\dfrac{1}{{{{\log }_q}p}},\,\,\,{\log _p}{q^n} = n{\log _p}q$
$\Rightarrow$${\log _b}a = \left[ {\dfrac{{{{\log }_c}ac}}{2}} \right]$
$\Rightarrow$$\dfrac{1}{{{{\log }_a}b}} = \left[ {\dfrac{1}{{2{{\log }_{ac}}c}}} \right]$
$\Rightarrow$${\log _a}b = 2{\log _{ac}}c$
$\Rightarrow$${\log _a}b = {\log _{ac}}{c^2}$
Now we know that
$\Rightarrow$${\log _q}p = k \Rightarrow p = {q^k}$ Then
$\Rightarrow$${\log _{ac}}{c^2} = {\log _a}b$
$\Rightarrow$${c^2} = {\left( {ac} \right)^{{{\log }_a}b}}$
We proved the given statement

Note: The best part about basic logarithmic is that it can be fit anywhere and the given problem is complex but the basic formula of the logarithmic can be solved easily hence we proved the given statement in the above solution that is provided step by step. Since the given question from the sequence and series of arithmetic progression and we only apply the single concept that is considered a, b, c are in A.P then
$b = \dfrac{{a + c}}{2}$
And after we applied the concept that it becomes a logarithmic-based problem and we have solved the problem easily and proved the given statement as mentioned above.