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Solve for $x\in R$,
$2^x+4^x=8^x$

Using the substitution $2^x=t$, we have

$t+t^2=t^3$

$ \Rightarrow {t^3} - {t^2} - t = 0$

Or, $t({t^2} - t - 1) = 0$

$t \ne 0$ as $t=2^x>0$

So, ${t^2} - t - 1 = 0$

Or, $t = \frac{{1 \pm \sqrt {1 + 4} }}{2} = \frac{{1 \pm \sqrt 5 }}{2}$

Since t > 0, $t = \frac{{1 + \sqrt 5 }}{2}=2^x$

So, $x\log 2 = \log \left( {\frac{{1 + \sqrt 5 }}{2}} \right)$

$ \Rightarrow x = \frac{{\log \left( {\frac{{1 + \sqrt 5 }}{2}} \right)}}{{\log 2}} = \frac{{\log (1 + \sqrt 5 )}}{{\log 2}} - 1$

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${\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$

The range of the function $f(x) = {\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$ is: (A) $[ - 2,2]$ (B) $\left[ {\frac{1}{{\sqrt 5 }},\sqrt 5 } \right]$ (C) $(0,\sqrt 5 )$ (D) $[ 0,2]$ Solution We have, $f(x) = {\log _{\sqrt 5 }}\left( {3 - 2\sin \frac{{3\pi }}{4}\sin x + 2\cos \frac{\pi }{4}\cos x} \right)$ $ \Rightarrow f(x) = {\log _{\sqrt 5 }}\left[ {3 + \sqrt 2 (\cos x - \sin x)} \right]$ Now, $ - \sqrt 2  \le \cos x - \sin x \le \sqrt 2 $ $\therefore - 2 \le \sqrt 2 (\cos x - \sin x) \le 2$ $\therefore 1 \le 3 + \sqrt 2 (\cos x - \sin x) \le 5$ $\therefore{\log _{\sqrt 5 }}1 \le {\log _{\sqrt 5 }}[3 + \sqrt 2 (\cos x - \sin x)] \le {\log _{\sqrt 5 }}5$ $ \Rightarrow 0 \le f(x) \le 2$ Answer: (D)
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