Question

Let ${\log }_{a}3=2$ and ${\log }_{b}8=3.$ If $\alpha =\left[{\log }_{a}b\right]+1,$ where $[.]$ denotes the greatest integer function and $\beta$ is the integral part of ${\log }_{\sqrt{2}}\left(\sqrt{\alpha +\sqrt{\alpha +\sqrt{\alpha +\sqrt{\alpha +\cdots \text{upto}\infty }}}}\right),$ then

• A. $\alpha =2$
• B. $\alpha =3$
• C. $\beta =3$
• D. $\beta =2$

Answer 1

Answer: (A), (D)

$\beta =2$

${\log }_{a}3=2,{\log }_{b}8=3$
$\Rightarrow a=\sqrt{3},b=2$
${\log }_{a}b={\log }_{\sqrt{3}}2={\log }_{3}4$
We know that,
${\log }_{3}3<{\log }_{3}4<{\log }_{3}9$
$\Rightarrow 1<{\log }_{3}4<2$
$\Rightarrow \left[{\log }_{3}4\right]=1$
So, $\alpha =\left[{\log }_{a}b\right]+1=\left[{\log }_{3}4\right]+1=2$

$\beta$ is the integral part of ${\log }_{\sqrt{2}}\left(\sqrt{\alpha +\sqrt{\alpha +\sqrt{\alpha +\sqrt{\alpha +\cdots \text{upto}\infty }}}}\right)$
Let $\sqrt{\alpha +\sqrt{\alpha +\sqrt{\alpha +\sqrt{\alpha +\cdots }}}}=t$
$\Rightarrow \sqrt{\alpha +t}=t$
$\Rightarrow t^2-t-\alpha =0$
$\Rightarrow t^2-t-2=0$ $(\because \alpha =2)$
$\Rightarrow t=2,-1$
For log to be defined, $t>0$
So, $t=2$
${\log }_{\sqrt{2}}t={\log }_{\sqrt{2}}2=2$
$\therefore \beta =2$