Question

If ${\log }_{12}27=a$ then ${\log }_{6}16$=

• A. $4\left(\frac{3-a}{3+a}\right)$
• B. $4\left(\frac{3+a}{3-a}\right)$
• C. $3\left(\frac{4+a}{4-a}\right)$
• D. $3\left(\frac{4-a}{4+a}\right)$

Answer 1

The correct option is A $4\left(\frac{3-a}{3+a}\right)$

$a=lo{g}_{12}27=lo{g}_{12}{3}^3$

$a=\frac{3}{lo{g}_{3}12}$

$a=\frac{3}{lo{g}_3(3\times 2^2)}$

$a=\frac{3}{lo{g}_{3}3+2lo{g}_{3}2}$

$a=\frac{3}{1+2lo{g}_{3}2}$

$(1+2lo{g}_{3}2)=\frac{3}{a}$

$lo{g}_{3}2=\frac{1}{2}(\frac{3}{a}-1)=\frac{3-a}{2a}$

$lo{g}_{2}3=\frac{2a}{3-a}$

Now $lo{g}_{6}16=lo{g}_6{2}^4=4lo{g}_{6}2=\frac{4}{lo{g}_{2}6}$

$=\frac{4}{lo{g}_2(2\times 3)}=\frac{4}{lo{g}_{2}2+lo{g}_{2}3}$

$=\frac{4}{1+lo{g}_{2}3}=\frac{4}{1+\frac{2a}{3-a}}$

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