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Solve for x, $4^x+6^x=9^x$

We have, ${({2^x})^2} + {2^x}{.3^x} = {({3^x})^2}$

Dividing by $2^x.3^x$,

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

Let, ${\left( {\frac{3}{2}} \right)^x} = t$

We have, $\frac{1}{t} + 1 = t$

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

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

Since, $t = {\left( {\frac{3}{2}} \right)^x} > 0,t = \frac{{1 + \sqrt 5 }}{2}$

We have, ${\left( {\frac{3}{2}} \right)^x} = \frac{{1 + \sqrt 5 }}{2}$

So, $x\log \left( {\frac{3}{2}} \right) = \log \left( {\frac{{1 + \sqrt 5 }}{2}} \right)$

Or, $x = \frac{{\log \left( {\frac{{1 + \sqrt 5 }}{2}} \right)}}{{\log \left( {\frac{3}{2}} \right)}}$

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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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