Question

if log_{12}27=a .calculate \[\log_{6}16 \] with using a.

Answer 1

\[\log_{27}12=a \]

Answer 2

can you screenshot this question, I;m not quite following it

Answer 3

\[\log_{12}27=a \]
\[\log_{6}16=? \]

Answer 4

\[\log_{12} 27=a\]
\[\log_{12} 3^{3}=a\]
\[3\log_{12} 3=a\]
\[\log_{12} 3=\frac{a}{3}\ ..............(1)\]
\[\log_{12} 1=1\]
\[\log_{12} 4=\log_{12} 12-\log_{12} 3=1-\frac{a}{3}\ .........(2)\]
\[\log_{12} 16=2\log_{12} 4=2(1-\frac{a}{3})\ ..............(3)\]
From equation (2):
\[2\log_{12} 2=1-\frac{a}{3}\]
\[\log_{12} 2=\frac{1-\frac{a}{3}}{2}\]
\[\log_{12} 6=\log_{12} 3+\log_{12} 2=\frac{a}{3}+\frac{1-\frac{a}{3}}{2}\ .......(4)\]
By the change of base formula:
\[\log_{6} 16=\frac{\log_{12} 16}{\log_{12} 6}=\frac{2(1-\frac{a}{3})}{\frac{a}{3}+\frac{1-\frac{a}{3}}{2}}\]
This simplifies to:
\[\log_{6} 16=\frac{4(3-a)}{a+3}\]