Solve : 4^x+1/4^x=16(1/16)

Solve : $4^x + \frac{1}{4^x} = 16\frac{1}{16}$

तपाईं प्रतिभाशाली हुनुहुन्छ । यस्ता समस्या त कति समाधान गर्नु भयो कति, हाेइन र?
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If you want to be a genius in Mathematics, solve the followint problem.
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Mathematics Injection
$\large \textbf{Solve:}$

Solve : $4^x + \frac{1}{4^x} = 16\frac{1}{16}$

Here,

$4^x + \frac{1}{4^x} = 16\frac{1}{16}$

= $4^x + \frac{1}{4^x} = \frac{257}{16}$

let $4^x = a$ .............(i)

So that,$a + \frac{1}{a} = \frac{257}{16}$

or, $\frac{a^2+1}{a} = \frac{257}{16}$

or, $16a^2+16 = 257a$

or, $16a^2 - 257a + 16=0$

or, $16a^2 - 256a - a + 16=0$

or, $16a(a - 16) - 1(a - 16)=0$

or, $(a-16)(16a-1)=0$

Either, $a-16=0$ $\qquad \qquad$ or, $16a-1=0$

$\quad \quad$or, $a=16$ $\qquad \qquad$ or, $16a=1$

$\quad \quad $or, $a=16$ $\qquad \qquad$ or, $a=\frac{1}{16}$

Puting the value of $a$ in the equ. (i), we get

If, $a= 16$, then $4^x=16$ $\Rightarrow$ $4^x=4^2$ $\Rightarrow$ $x=2$

If, $a$= $\frac{1}{16}$, then $4^x=\frac{1}{16} \Rightarrow 4^x=4^{-2} \Rightarrow x=-2$

Therefor, $x= 2$ and $-2$