Step 1: Simplify the right side of the equation using the exponent rule $a^m \cdot a^n = a^{m+n}$. $$x^x = 2^x \cdot 2^{320}$$ $$x^x = 2^{x+320}$$ Step 2: Divide both sides by $2^x$. $$\frac{x^x}{2^x} = \frac{2^{x+320}}{2^x}$$ Step 3: Simplify both sides using the exponent rules $\frac{a^m}{b^m} = \left(\frac{a}{b}\right)^m$ and $\frac{a^m}{a^n} = a^{m-n}$. $$\left(\frac{x}{2}\right)^x = 2^{x+320-x}$$ $$\left(\frac{x}{2}\right)^x = 2^{320}$$ Step 4: To make the base and exponent on the left side equal, raise both sides to the power of $\frac{1}{2}$. This uses the rule $(a^b)^c = a^{bc}$. $$\left(\left(\frac{x}{2}\right)^x\right)^{\frac{1}{2}} = \left(2^{320}\right)^{\frac{1}{2}}$$ $$\left(\frac{x}{2}\right)^{\frac{x}{2}} = 2^{\frac{320}{2}}$$ $$\left(\frac{x}{2}\right)^{\frac{x}{2}} = 2^{160}$$ Step 5: Express the right side in the form $k^k$. We need to find a number $k$ such that $k^k = 2^{160}$. We can rewrite $2^{160}$ as $(2^a)^b$ where $a \cdot b = 160$. We want $b = 2^a$. Let's try $a=5$. Then $2^a = 2^5 = 32$. If $a=5$, then $b = \frac{160}{5} = 32$. So, $2^{160} = (2^5)^{32} = 32^{32}$. The equation becomes: $$\left(\frac{x}{2}\right)^{\frac{x}{2}} = 32^{32}$$ Step 6: Equate the bases (and exponents) since they are in the form $A^A = B^B$. $$\frac{x}{2} = 32$$ Step 7: Solve for $x$. $$x = 32 \times 2$$ $$x = 64$$ The final answer is $\boxed{64}$.