$2^{x}$ + $2^{x+1}$ + $2^{x+2}$ +.....+ $2^{x+2015}$ = $2^{2019}$ - $8$
⇔$2^{x}$.($1$+$2$ + $2^{2}$ + .... + $2^{2015}$) = $2^{2019}$ - $2^{3}$
⇔$2^{x}$.($1$+$2$ + $2^{2}$ + .... + $2^{2015}$) =$2^{3}$($2^{2016-1}$-$1$) ($1$)
Đặt $A$=$1$+$2$ + $2^{2}$ + .... + $2^{2015}$
⇔$2A$=$2$ + $2^{2}$ +$2^{3}$+ .... + $2^{2016}$
⇔$2A-A$=($2$ + $2^{2}$ +$2^{3}$+ .... + $2^{2016}$)-($1$+$2$ + $2^{2}$ + .... + $2^{2015}$)
⇔$A$ = $2^{2016-1}$-$1$
Thay vào ($1$),ta có:
$2^{x}$.($2^{2016-1}$-$1$) =$2^{3}$($2^{2016-1}$-$1$)
⇒$2^{x}$ = $2^{3}$
⇔ $x=3$
Vậy $x=3$
