Question

Given that ${\log }_{2}3=a,{\log }_{3}5=b,{\log }_{7}2=c$, express the logarithm of a number $63$ to the base $140$ in terms of $a,b$ and $c$.

Answer 1

$\because {\log }_{2}3=a$ ............$\left(1\right)$
$\Rightarrow b={\log }_{3}5=\frac{{\log }_{2}5}{{\log }_{2}3}=\frac{{\log }_{2}5}{a}$ ...... (from $\left(1\right)$)
$\Rightarrow {\log }_{2}5=ab$ .......... $\left(2\right)$
$\Rightarrow \frac{1}{{\log }_{2}7}=c$ or ${\log }_{2}7=\frac{1}{c}$ .....$\left(3\right)$
Now, ${\log }_{140}63=\frac{{\log }_{2}63}{{\log }_{2}140}=\frac{{\log }_2(3^2\times 7)}{{\log }_2(2^2\times 5\times 7)}$
Using product law of logarithms, we get
$=\frac{{\log }_2{3}^2+{\log }_{2}7}{{\log }_2{2}^2+{\log }_{2}5+{\log }_{2}7}=\frac{2a+\frac{1}{c}}{2+ab+\frac{1}{c}}$
On simplification we get
${\log }_{140}63=\frac{2ac+1}{2c+abc+1}$