Question

Express ${\log }_{25}24$ in terms of ${\log }_{6}15=\alpha$ and ${\log }_{12}18=\beta$

Answer 1

We will change all the logarithms to base 5.

Thus ${\log }_{25}24={\log }_{5^2}(2^3\times 3)$
$={\log }_{5^2}\left(2^3\right)+{\log }_{5^2}\left(3\right)$
${\log }_{25}24=\frac{3}{2}{\log }_{5}2+\frac{1}{2}{\log }_{5}3$ ...(1)

Now $\alpha ={\log }_{6}15=\frac{{\log }_{5}15}{{\log }_{5}6}$

$\because {\log }_{5}5=1$

$=\Rightarrow \alpha =\frac{{\log }_{5}3+1}{{\log }_{5}2+{\log }_{5}3}$ ...(2)

$\beta ={\log }_{12}18=\frac{{\log }_{5}18}{{\log }_{5}12}$

$=\frac{{\log }_5\left(3^2.2\right)}{{\log }_5\left(2^2.3\right)}$
$=\frac{2{\log }_{5}3+{\log }_{5}2}{2{\log }_{5}2+{\log }_{5}3}$ ...(3)

Now putting ${\log }_{5}2=x$ and ${\log }_{5}3=y,$ we get from (2) and (3)
$\alpha =\frac{y+1}{x+y}$
or $\alpha x+(\alpha -1)y-1=0$ ...(4)
$\beta =\frac{2y+x}{2x+y}$
$(2\beta -1)x-(2-\beta )y+0=0$ ...(5)

Solving (4) and (5), we get
$\frac{x}{-(2-\beta )}=\frac{y}{-(2\beta -1)}=\frac{1}{-[\alpha (2-\beta )+(\alpha -1)(2\beta -1)]}$
or $\frac{x}{2-\beta }=\frac{y}{2\beta -1}=\frac{1}{\alpha \beta +\alpha -2\beta +1}.$
$\Rightarrow {\log }_{5}2=x=\frac{2-\beta }{\alpha \beta +\alpha -2\beta +1}$
and ${\log }_{5}3=y=\frac{2\beta -1}{\alpha \beta +\alpha -2\beta +1}$
Substituting these values of ${\log }_{5}2$ and ${\log }_{5}3,$ in (1), we get

${\log }_{25}24=\frac{3}{2}\left(\frac{2-\beta }{\alpha \beta +\alpha -2\beta +1}\right)+\frac{1}{2}\left(\frac{2\beta -1}{\alpha \beta +\alpha -2\beta +1}\right)$

${\log }_{25}24=\frac{5-\beta }{2\alpha \beta +2\alpha -4\beta +2}.$