Question

If $B=\frac{12}{3+\sqrt{5}+\sqrt{8}}$ and $A=\sqrt{1}+\sqrt{2}+\sqrt{5}-\sqrt{10}$, then ${\log }_{A}B$ will be

• A. a negative integer.
• B. a prime integer.
• C. a positive integer.
• D. an even-natural number.

Answer 1

The correct option is C a positive integer.

$B=\frac{12}{3+\sqrt{5}+\sqrt{8})}$

$B=\left(\begin{array}{c}\frac{12(3+\sqrt{5}-\sqrt{8})}{(3+\sqrt{5}{)}^2-8}\end{array}\right)=\frac{12(3+\sqrt{5}-\sqrt{8})}{6+6\sqrt{5}}$

$=\left(\begin{array}{c}\frac{2(3+\sqrt{5}-2\sqrt{2})}{1+\sqrt{5}}\end{array}\right)=\frac{6+2\sqrt{5}}{\sqrt{5}+1}-\frac{4\sqrt{2}}{\sqrt{5}+1}$

$=\frac{(\sqrt{5}+1{)}^2}{\sqrt{5}+1}-\frac{4\sqrt{2}(\sqrt{5}-1)}{4}$

$=\sqrt{5}+1-\sqrt{10}+\sqrt{2}=A\Rightarrow {\log }_{A}B=1[\because {\log }_{a}a=1]$
Ans: C