we have the expression
[tex]\log _912[/tex]
Change the base of log
The formula to change the base is equal to
[tex]\log _bM=\frac{\log _nM}{\log _nb}[/tex]
the new base will be 10
so
n=10
b=9
M=12
substitute
[tex]\log _912=\frac{\log_{10}12}{\log_{10}9}=\frac{\log 12}{\log 9}[/tex]
therefore
[tex]\log _912=\frac{\log12}{\log9}=1.131[/tex]
the answer is the option C
