Question

If $b=\sqrt{3+2\sqrt{2}}+\sqrt{3-2\sqrt{2}}$, then the value of ${\log }_{b}64$ is equal to

• A. 3
• B. 4
• C. 5
• D. 6

Answer 1

Answer: (B)

Given, $b=\sqrt{3+2\sqrt{2}}+\sqrt{3-2\sqrt{2}}=(\sqrt{2}+1)+(\sqrt{2}-1)=2\sqrt{2}$
$\therefore {\log }_{b}64={\log }_{2^{3/2}}{2}^6=\frac{6}{3/2}==4$.