Question

If ${\log }_{12}18=\alpha$ and ${\log }_{24}54=\beta$, then the value of $\alpha \beta +(\alpha -\beta )$ will be

• A. $2$
• B. ${\log }_{12}24$
• C. $1$
• D. none of these

Answer 1

The correct option is D none of these

${\log }_{12}18=\alpha$
$\Rightarrow \alpha =\frac{\log 2\left(3^2\right)}{\log 3\left(2^2\right)}=\frac{\log 2+2\log 3}{2\log 2+\log 3}[\because \log (ab)=\log a+\log b\text{\&}\log a^m=m\log a]$
and ${\log }_{24}54=\beta$
$\beta =\frac{\log 2\left(3^3\right)}{\log 3\left(2^3\right)}=\frac{3\log 3+\log 2}{3\log 2+\log 3}$
Now $\alpha \beta +\alpha -\beta =(\alpha -1)(\beta +1)+1$
$=\left(\frac{\log 3-\log 2}{2\log 2+\log 3}\right)\left(\frac{4(\log 3+\log 2)}{3\log 2+\log 3}\right)+1$
$\ne 1,2,{\log }_{12}\left(24\right)$
Ans: D