7.208. $2^{x-1}+2^{x-4}+2^{x-2}=6.5+3.25+1.625+\ldots$ (the expression on the right side is an infinite geometric progression).
Solution.

On the right side, we have the sum of the terms of an infinitely decreasing geometric progression $S$, where $b_{1}=6.5 ; q=\frac{3.25}{6.5}=0.5 \Rightarrow S=\frac{b_{1}}{1-q}=\frac{6.5}{1-0.5}=13$.

Rewrite the equation as $\frac{2^{x}}{2}+\frac{2^{x}}{16}+\frac{2^{x}}{4}=13 \Leftrightarrow \frac{13}{16} \cdot 2^{x}=13$, $2^{x}=16$, from which $x=4$.

Answer: 4.