If ${\log _a}x,{\log _b}x,{\log _c}x$ are in A.P., where $x \ne 1,$ then ${c^2} = $

${\left( {ab} \right)^{{{\log }_a}b}}$

${\left( {ab} \right)^{{{\log }_b}a}}$

JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Medium

1 of 1

Analyzing +4 | -1

{\log _a}x,{\log _b}x,{\log _c}x

are in A.P., where

x \ne 1,

then

{c^2} =

View answer and explanation

Correct option: B

{\left( {ac} \right)^{{{\log }_a}b}}

Option D (

{\left( {ac} \right)^{{{\log }_b}a}}

) is correct because by using the properties of logarithms and the given condition that

{\log _a}x,{\log _b}x,{\log _c}x

are in A.P., we can derive the relationship

{c^2} = {\left( {ac} \right)^{{{\log }_a}b}}

Practice All Chapter MCQs →

A=\{1,2,3\}, B=\{3,4\}, C=\{4,5,6\}

{A} \cup({B} \cap {C})

A-(B \cap C)

(B \cup C)-A

a,b,c

a,x,b

b,y,c

{x^2},{b^2},{y^2}

P(x)=1-x+x^2-x^3+\ldots+x^{2012}

y

Q(y)=a_0+a_1 y+\ldots+a_{2012} y^{2012}

y=x-2

\sum\limits_{{i=0}}^{2012} a_i

\log \left(\frac{5 c}{a}\right), \log \left(\frac{3 b}{5 c}\right)

\log \left(\frac{a}{3 b}\right)

A P

a, b

G P

a, b, c

\frac{3}{4}