[tex] log_{24}3=c; \frac{1}{ log_{3}24 }=c; \frac{1}{1+3 log_{3}2 } =c; c(1+ log_{3}2)=1;1+ log_{3}2= \frac{1}{c} [/tex]
[tex] log_{3}2= \frac{1-c}{c} [/tex]
[tex] log_{24}2= \frac{1}{ log_{2}24 }= \frac{1}{3+ log_{2}3 }= \frac{1}{3+ \frac{c}{1-c} }= \frac{1-c}{3+2c} [/tex]