Question

Area of a square ABCD is 36 and side AB is parallel to the X-axis. Vertices A, B and C lie on the graphs of $y=lo{g}_{a}x,y=2lo{g}_{a}x$ and $y=3lo{g}_{a}x$ respectively. Then a =

• A. $3^{\frac{1}{6}}$
• B. $\sqrt{3}$
  
• C. $6^{\frac{1}{3}}$
  
• D. $\sqrt{6}$

Answer 1

The correct option is A $3^{\frac{1}{6}}$

Let $A=(p,lo{g}_{a}p);B=(q,2lo{g}_{a}q)p,q>0$ & $a>0,a\ne 1$, then $C=(q,3lo{g}_{a}q)$
$AB=X-axis\Rightarrow p=q^2.\left|AB\right|=6\Rightarrow |p-q|=6also\left|lo{g}_{a}q\right|=6$.