Question

Find $lo{g}_{24}$ 48 in terms of $\alpha$ if ${\log }_{12}36=\alpha$

Answer 1

${\log }_{24}48$

$={\log }_{24}(24\times 2)$

$={\log }_{24}24+{\log }_{24}2$

$=1+\frac{1}{{\log }_{2}24}$

$=1+\frac{1}{{\log }_2(8\times 3)}$

$=1+\frac{1}{{\log }_2{2}^3+{\log }_{2}3}$

$=1+\frac{1}{3{\log }_{2}2+{\log }_{2}3}$

$=1+\frac{1}{3+{\log }_{2}3}$ .........$\left(1\right)$

Given:${\log }_{12}36=a$

$\Rightarrow {\log }_{12}{6}^2=a$

$\Rightarrow 2{\log }_{12}6=a$

$\Rightarrow {\log }_{12}6=\frac{a}{2}$

$\Rightarrow \frac{1}{{\log }_{6}12}=\frac{a}{2}$

$\Rightarrow \frac{1}{{\log }_6(6\times 2)}=\frac{a}{2}$

$\Rightarrow \frac{1}{{\log }_{6}6+{\log }_{6}2}=\frac{a}{2}$

$\Rightarrow \frac{1}{1+\frac{1}{{\log }_{2}6}}=\frac{a}{2}$

$\Rightarrow \frac{{\log }_{2}6}{{\log }_{2}6+1}=\frac{a}{2}$

$\Rightarrow \frac{{\log }_{2}2+{\log }_{2}3}{{\log }_{2}2+{\log }_{2}3+1}=\frac{a}{2}$

$\Rightarrow \frac{1+{\log }_{2}3}{2+{\log }_{2}3}=\frac{a}{2}$

$\Rightarrow 2+2{\log }_{2}3=2a+a{\log }_{2}3$

$\Rightarrow (2-a){\log }_{2}3=2a-2$

$\Rightarrow {\log }_{2}3=\frac{2a-2}{2-a}$

Put this in $\left(1\right)$ we get

${\log }_{24}48=1+\frac{1}{3+{\log }_{2}3}$

$=1+\frac{1}{3+\frac{2a-2}{2-a}}$

$=1+\frac{2-a}{6-3a+2a-2}$

$=1+\frac{2-a}{-a+4}$

$=\frac{4-a+2-a}{4-a}=\frac{6-2a}{4-a}$

$\therefore \alpha =\frac{2(a-3)}{a-4}$