If {log _{12}}27 = a,then find the value of { | vedclass

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(C) Given that ${\log _{12}}27 = a$. Using the change of base formula,$\frac{{\log _3}27}{{\log _3}12} = a$. Since $27 = 3^3$ and $12 = 3 	imes 2^2$,

Vedclass · Accepted Answer

$4\frac{3-a}{3+a}$

Vedclass · Answer

(C) Given that ${\log _{12}}27 = a$. Using the change of base formula,$\frac{{\log _3}27}{{\log _3}12} = a$. Since $27 = 3^3$ and $12 = 3 	imes 2^2$,we have $\frac{3}{{\log _3}3 + {\log _3}2^2} = a$. This simplifies to $\frac{3}{1 + 2{\log _3}2} = a$. Rearranging gives $3 = a(1 + 2{\log _3}2)$,so $3 = a + 2a{\log _3}2$. Thus,$2a{\log _3}2 = 3 - a$,which means ${\log _3}2 = \frac{3-a}{2a}$. Now,we need to find ${\log _6}16 = \frac{{\log _3}16}{{\log _3}6} = \frac{{\log _3}2^4}{{\log _3}(2 	imes 3)} = \frac{4{\log _3}2}{{\log _3}2 + 1}$. Substituting ${\log _3}2 = \frac{3-a}{2a}$: ${\log _6}16 = \frac{4(\frac{3-a}{2a})}{\frac{3-a}{2a} + 1} = \frac{\frac{2(3-a)}{a}}{\frac{3-a+2a}{2a}} = \frac{2(3-a)}{a} 	imes \frac{2a}{3+a} = \frac{4(3-a)}{3+a}$. Therefore,the correct option is $C$.

If ${\log _{12}}27 = a$,then find the value of ${\log _6}16$.

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