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^2-7 y+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 $
$x=\frac{2}{\log _6(5 \sqrt{2}-7)} \Rightarrow (D)$
$x=\frac{2}{\log _6\left(\frac{1}{5 \sqrt{2}+7}\right)}=\frac{-2}{\log _6(5 \sqrt{2}+7)} \Rightarrow (C)$
$\text { If } (5 \sqrt{2}-7)^x=1 \Rightarrow x \log _6(5 \sqrt{2}-7)=0$ $\Rightarrow x=0 \Rightarrow \text { (A) }$