Question

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

• A. $\frac{1+a}{a}$
• B. $4\left(\frac{3-a}{3+a}\right)$
• C. $\frac{2a}{3-a}$
• D. $5\left(\frac{2-a}{2+a}\right)$

Answer 1

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

$Given:{\log }_{12}27=a⟹\frac{\log 27}{\log 12}=a⟹\frac{\log 3^3}{\log 3\times 2^2}=a⟹3\log 3=[\log 3+2\log 2]a⟹\frac{\log 3+2\log 2}{3\log 3}=\frac{1}{a}⟹[\frac{1}{3}+\frac{2}{3}\left(\frac{\log 2}{\log 3}\right)]=\frac{1}{a}⟹\frac{2\log 2}{3\log 3}=\frac{1}{a}-\frac{1}{3}⟹\frac{\log 2}{\log 3}=\frac{3}{2}\left[\frac{3-a}{3a}\right]⟹\frac{\log 2}{\log 3}=\frac{3-a}{2a}⟹\log 3=\left(\frac{2a}{3-a}\right)\log 2-------\left(1\right)Now,{\log }_{6}16=\frac{\log 16}{\log 6}=\frac{\log 2^4}{\log (2\times 3)}=\frac{4\log 2}{(\log 2+\log 3)}=\frac{4\log 2}{(\log 2+\left(\frac{2a}{3-a}\right)\log 2)}\left[from\right(1\left)\right]{\log }_{6}16=\frac{4\log 2}{\log 2(1+\left(\frac{2a}{3-a}\right))}=\frac{4(3-a)}{3-a+2a}=\frac{4(3-a)}{(3+a)}$