Question

If log 27 base 12 = a then find the value of log16 base 6 in terms of a​

Answer 1

Answer:

log12^27=a

then

log12/log27=a

log12=a×log3^4

log4+log3=4alog3

log4=(4a-1)log3

log2=(4a-1)log3/2

then

log6^16

=log16/log6

=4log2/log2+log3

put value of log2

then

2×2×(4a-1)log3/(4a-1)log3+2log3

4(4a-1)/(4a+1)

the answer is

16a-4/4a+1

ok

follow

me

guy's

Answer 2

${\mathcal{\red{ANSWER}}}$

Given:

$\tt{\log_{12}27=a \implies \dfrac{\log_{27}}{\log_{12}} = a}$

$\tt{\implies \dfrac{\log 12}{\log 27}=\dfrac{1}{a}}$

$\tt{\implies \dfrac{\log 3\times 4}{\log 3^{3}}=\dfrac{1}{a}}$

$\tt{\implies \dfrac{\log 3+\log 4}{3\log 3}=\dfrac{1}{a}}$

$\tt{\implies \dfrac{\log 3}{\log 3}+\dfrac{\log 4}{\log 3}=\dfrac{3}{a}}$

$\tt{\implies 1+ \dfrac{\log 2^{2}}{\log 3}=\dfrac{3}{a}}$

$\tt{\implies \dfrac{2\log 2}{\log 3} = \dfrac{3}{a} -1=\dfrac{3-a}{a}}$

$\tt{\implies \dfrac{\log 2}{\log 3} = \dfrac{3-a}{2a}}$

$\tt{\implies \dfrac{\log 3}{\log 2} = \dfrac{2a}{3-a}}$

$\bf{\log_{2}3=\dfrac{2a}{3-a}\;\;\;\;\;.........(1)}$

Given,

$\bf{\log_{6}16=\dfrac{1}{\log_{16}6}}$

$\tt{\implies \dfrac{1}{\dfrac{\log 6}{\log 16}}}$

$\tt{\implies \dfrac{1}{\dfrac{\log 3\times 2}{\log 2^{4}}}}$

$\tt{\implies \dfrac{1}{\dfrac{\log 3+\log 2}{4\log 2}}}$

$\tt{\implies \dfrac{4}{\dfrac{\log 3}{\log 2}+1}}$

$\tt{\implies \dfrac{4}{\log_{2}3+1} = \dfrac{4}{\dfrac{2a}{3-a}+1}}$

$\tt{\implies \dfrac{4\times (3-a)}{2a+3-a}}$

$\bf{\implies \dfrac{4\times (3-a)}{a+3}}$

_________________________________________[ANSWER]