Question

If $(\sqrt{5\sqrt{2}-7}{)}^x+6(\sqrt{5\sqrt{2}+7}{)}^x=7$, then the value of $x$ can be equal to :

• A. 0
• B. ${\log }_{(5\sqrt{2}-7)}36$
• C. $\frac{-2}{{\log }_6(5\sqrt{2}+7)}$
• D. ${\log }_{\sqrt{5\sqrt{2}-7}}6$

Answer 1

Answer: (A), (B), (C), (D)

Let $(\sqrt{5\sqrt{2}-7}{)}^x=y$
$\therefore y+\frac{6}{y}=7\Rightarrow y^2-7y+6=0\Rightarrow (y-6)(y-1)=0\Rightarrow y=6\text{ or }1$
$\text{ If }y=6$
$(\sqrt{5\sqrt{2}-7}{)}^x=6$ ......(1)
Taking log on both sides
$x{\log }_6(\sqrt{5\sqrt{2}-7})={\log }_{6}6=1$
$x=\frac{1}{{\log }_6(5\sqrt{2}-7)}=\frac{1}{{\log }_{36}(5\sqrt{2}-7)}$
$x={\log }_{(5\sqrt{2}-7)}36(B)$
$\text{ If }y=1$
$\frac{x}{2}{\log }_6(5\sqrt{2}-7)=1$
$x=\frac{2}{{\log }_6(5\sqrt{2}-7)}\Rightarrow (D)$
$x=\frac{2}{{\log }_6\left(\frac{1}{5\sqrt{2}+7}\right)}=\frac{-2}{{\log }_6(5\sqrt{2}+7)}\Rightarrow (C)$
$\text{ If }(5\sqrt{2}-7{)}^x=1\Rightarrow x{\log }_6(5\sqrt{2}-7)=0$ $\Rightarrow x=0\Rightarrow \text{ (A) }$