Question

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

Answer 1

$\alpha =\frac{\log 36}{\log 12}=\frac{\log 3+\log 12}{\log 12}=1+\frac{\log 3}{2\log 2+\log 3}\alpha -1=\frac{\log 3}{2\log 2+\log 3}2(\alpha -1)\log 2+(\alpha -1)\log 3=\log 32(\alpha -1)\log 2=(2-\alpha )\log 3\frac{2(\alpha -1)}{2-\alpha }={\log }_{2}3$

${\log }_{24}48=\frac{\log 2}{\log 24}+\frac{\log 24}{\log 24}=\frac{\log 2}{3\log 2+\log 3}+1=\frac{1}{3+{\log }_{2}3}+1=\frac{1}{3+\frac{2(\alpha -1)}{2-\alpha }}+1=\frac{(2-\alpha )}{4-\alpha }+1=\frac{2-\alpha +4-\alpha }{4-\alpha }=\frac{2(3-\alpha )}{4-\alpha }$