Question

Total number of values of $x$, satisfying $(\sqrt{3}+1{)}^{2x}+(\sqrt{3}-1{)}^{2x}=2^{3x}$, is equal to

• A. $1$
• B. $3$
• C. $2$
• D. None

Answer 1

The correct option is A

$1$

$(\sqrt{3}+1{)}^{2x}+(\sqrt{3}-1{)}^{2x}=2^{3x}=(2\sqrt{2}{)}^{2x}$

$\Rightarrow {\left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)}^{2x}+{\left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)}^{2x}=1$

$\Rightarrow {\left(\frac{2+\sqrt{3}}{4}\right)}^x+{\left(\frac{2-\sqrt{3}}{4}\right)}^x=1$

$\Rightarrow x=1$ is a solution.

Also $0<\frac{2+\sqrt{3}}{4}<1,0<\frac{2-\sqrt{3}}{4}<1$

$\therefore$ If $x>1,{\left(\frac{2+\sqrt{3}}{4}\right)}^x<\frac{2+\sqrt{3}}{4},{\left(\frac{2-\sqrt{3}}{4}\right)}^x<\frac{2-\sqrt{3}}{4}$

$\Rightarrow {\left(\frac{2+\sqrt{3}}{4}\right)}^x+{\left(\frac{2-\sqrt{3}}{4}\right)}^x<1$

$\therefore$ if $x<1,{\left(\frac{2+\sqrt{3}}{4}\right)}^x+{\left(\frac{2-\sqrt{3}}{4}\right)}^x>1$

Hence, $x=1$ is the only solution.