Question

Let $a =27^{\log _{\sqrt{3}} \sqrt[6]{4^k+6}}, b =\frac{1}{5^{\log _{\sqrt{5}} \sqrt{2^k-1}}}$ and $ab =10$. Then the sum of the squares of the real value(s)

• A. 4
• B. 5
• C. 9
• D. 10

Answer 1

Answer: (D)

$a =4^{ k }+6, b =\frac{1}{2^{ k }-1}$
$\text { Put } 2^k=t $
$\frac{t^2+6}{t-1}=10 $
$\Rightarrow t^2-10 t+16=0$
$(t-8)(t-2)=0$
$t=8, t=2$
$\Rightarrow k=3,1 $
$\Rightarrow 3^2+1^2=10$