Question

If ${\log }_{6}15=a,{\log }_{12}18=b,$ then find ${\log }_{25}24$ in terms of $a$ and $b$.

Answer 1

${\log }_{6}15=\frac{{\log }_{3}15}{{\log }_{3}6}=\frac{1+{\log }_{3}5}{1+{\log }_{3}2}=a$ ....(i)
${\log }_{12}18=\frac{{\log }_{3}18}{{\log }_{3}12}=\frac{1+{\log }_{3}2}{1+{\log }_{3}2}=b$ .....(ii)
From (i) and (ii),
${\log }_{3}2=\frac{2-b}{2b-1}$ and ${\log }_{3}5=\frac{ab+a-2b+1}{2b-1}$
${\log }_{25}24=\frac{{\log }_{3}24}{{\log }_{3}25}=\frac{1+3{\log }_{3}2}{2{\log }_{3}5}$

$=\frac{1+3\left(\frac{2-b}{2b-1}\right)}{2\left(\frac{ab+a-2b+1}{2b-1}\right)}=\frac{5-b}{2(a-2b+ab+1)}$