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)

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