Question

Let $a =2010^{\log _{2011} 2012}, b =2010^{\log _{2012} 2011}$ and $c =(2010)^2$ then which one of the following is the most appropriate altemative?

• A. $a b=c$
• B. $a b >c$
• C. $a b \geq c$
• D. $a b \leq c$

Answer 1

Answer: (B)

$ a \cdot b=2010^{\left(\log _{2011} 2012+\log _{2012} 2011\right)} \left(y+\frac{1}{y} \geq 2, y>0\right)$
Now, $\log _{2011} 2012+\log _{2012} 2011>2 \left(\Theta \log _{2011} 2012>0, \neq 1\right)$
$\therefore ab > c$