If $\log _{12} 27=a$, then $\log _6 16=$

$2\left(\frac{3-a}{3+a}\right)$

$2\left(\frac{3+a}{3-a}\right)$

JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

1 of 1

Application +4 | -1

\log _{12} 27=a

, then

\log _6 16=

View answer and explanation

Correct option: B

4\left(\frac{3-a}{3+a}\right)

Option B (

4\left(\frac{{3-a}}{{3+a}}\right)

) is correct because it results from expressing

\log_6 16

in terms of

\log_{12} 27 = a

using change of base formulas.

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}