Question

Find the values of ${\log }_{2-\sqrt{3}}(2+\sqrt{3})$

• A. $1$
• B. $-1$
• C. $-2$
• D. $2$

Answer 1

Answer: (B)

$-1$

Let $a={\log }_{2-\sqrt{3}}(2+\sqrt{3})$

Consider $2+\sqrt{3}=\frac{2+\sqrt{3}}{1}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}=\frac{4-3}{2-\sqrt{3}}={(2-\sqrt{3})}^{-1}$

$a={\log }_{2-\sqrt{3}}{(2-\sqrt{3})}^{-1}$

We know that, $\log a^m=m\log a$

$a=-{\log }_{2-\sqrt{3}}2-\sqrt{3}$

Since ${\log }_{a}a=1$

$a=-1$