Power getting some power !!

Take $A=(7+4\sqrt{3})$ . Then, note that we have $A^{-1}=(7-4\sqrt{3})$ . Using this, the equation transforms to,

$A^{-x}+A^x=14\implies \left(A^{x}\right)^2-14A^x+1=0$

This is a quadratic in terms of $A^x$ and so we proceed to solve for $A^x$ using the quadratic formula. Then, we get,

$A^x=\frac{14\pm\sqrt{196-4}}{2}=\frac{14\pm 8\sqrt{3}}{2}=7\pm 4\sqrt{3}\\ \implies (7+4\sqrt{3})^x=7\pm 4\sqrt{3}=(7+4\sqrt{3})^{\pm 1}\implies x=\pm 1$

Since the question mentioned that $x\lt 0$ , we take the negative solution $x=(-1)$ and conclude our answer as $(-1)^5+\frac{1}{(-1)^5}=-1-1=(-2)$

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A quicker approach to the problem would be to make the following two observations:

• The equation has symmetric solutions, i.e., if $x=\alpha$ is a solution, then $x=(-\alpha)$ will also be a solution. To prove the symmetry, take $m=(-x)$ in the equation. You'll see that the same equation appears with all $x$ 's replaced by $m$ .

• It is quite trivial to note that $x=1$ satisfies the equation. Hence, by virtue of the previous point, $x=(-1)$ is also a solution and hence is our required value of $x$ .

A sure hit is by plotting the function $f(x) = (7-4\sqrt{3})^x+(7+4\sqrt{3})^x-14$ and check for the root. If it is not considered as cheating. I have done it with an Excel spreadsheet and the root is clearly $x = \boxed{-1}$ .

Tricked me at first. THEN got it.