If ${\log _x}a,{a^{x/2}}$ and ${\log _b}x$ are in G.P., then $x$is equal to

${\log _a}\left( {{{\log }_e}a} \right) + {\log _a}\left( {{{\log }_e}b} \right)$

$ - {\log _a}\left( {{{\log }_a}b} \right)$

${\log _a}\left( {{{\log }_e}b} \right) - {\log _a}\left( {{{\log }_e}a} \right)$

JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

1 of 1

Analyzing +4 | -1

{\log _x}a,{a^{x/2}}

and

{\log _b}x

are in G.P., then

x

is equal to

View answer and explanation

Correct option: A

{\log _a}\left( {{{\log }_b}a} \right)

Option A (

{\log _a}({{\log }_b}a)

) is correct because it is the solution for

x

when

{\log _x}a,{a^{x/2}}

and

{\log _b}x

are in G.P.

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}