If $16x^4+\frac{1}{16x^4}=14159,$ then find the value of $2x+\frac{1}{2x}.$

The correct answer is 11.

To find the value of $2x+\frac{1}{2x}$, let us define a variable $y=2x+\frac{1}{2x}$.

First, we square this expression to find a relation with higher powers of $x$:
$y^2={\left(2x+\frac{1}{2x}\right)}^2$
Using the algebraic identity ${(a+b)}^2=a^2+b^2+2ab$, we expand:
$y^2=4x^2+\frac{1}{4x^2}+2·\left(2x\right)·\left(\frac{1}{2x}\right)$
$y^2=4x^2+\frac{1}{4x^2}+2$
From this, we get:
$4x^2+\frac{1}{4x^2}=y^2-2$

Next, we square this expression again to relate it to the given term $16x^4+\frac{1}{16x^4}$:
${\left(4x^2+\frac{1}{4x^2}\right)}^2={\left(y^2-2\right)}^2$
Expanding the left side:
$16x^4+\frac{1}{16x^4}+2·\left(4x^2\right)·\left(\frac{1}{4x^2}\right)={\left(y^2-2\right)}^2$
$16x^4+\frac{1}{16x^4}+2={\left(y^2-2\right)}^2$

We are given that $16x^4+\frac{1}{16x^4}=14159$. Substituting this value:
$14159+2={\left(y^2-2\right)}^2$
$14161={\left(y^2-2\right)}^2$

Taking the square root on both sides:
$y^2-2=\sqrt{14161}$
To find $\sqrt{14161}$, we observe that ${120}^2=14400$ and ${119}^2=14161$. Assuming positive values for real $x$:
$y^2-2=119$
$y^2=119+2$
$y^2=121$

Finally, taking the square root:
$y=\sqrt{121}=11$

Therefore, the value of $2x+\frac{1}{2x}$ is 11.