Question

If $P=3^{\sqrt{{\log }_{3}2}}-2^{\sqrt{{\log }_{2}3}}$ and $Q={\log }_2{\log }_3{\log }_2{\log }_{\sqrt{3}}81$ then:

• A. $P=1,Q=0$
• B. $P=0,Q=1$
• C. $P=Q=0$
• D. $P=Q=1$

Answer 1

Answer: (C)

$P=Q=0$

${\log }_{\sqrt{3}}81={\log }_{3^{1/2}}{3}^4=\frac{4}{\frac{1}{2}}{\log }_{3}3=8$

${\log }_{2}8={\log }_2{2}^3=3$

${\log }_{3}3=1$

${\log }_{2}1=0$

$\therefore Q=0$

Let $a=3^{\sqrt{{\log }_{3}2}}$ and $b=2^{\sqrt{{\log }_{2}3}}$

$\Rightarrow {\log }_{2}a={\log }_2{3}^{\sqrt{{\log }_{3}2}}=\sqrt{{\log }_{3}2}.{\log }_{2}3$

$=\frac{1}{\sqrt{{\log }_{2}3}}.{\log }_{2}3=\sqrt{{\log }_{2}3}$

$\therefore a=2^{\sqrt{{\log }_{2}3}}=b$

$\Rightarrow P=a-b=0$