Đáp án: $x=2$
Giải thích các bước giải:
$2^x+2^{x+1}+2^{x+2}+2^{x+3}+.....+2^{x+2020}=2^{2023}-4$
$⇔2^x+2^x.2+2^x.2^2+2^x.2^3+.....+2^x+2^{2020}=2^2.2^{2021}-2^2$
$⇔2^x.(1+2+2^2+2^3+.....+2^{2020})=2^2(2^{2021}-1)(*)$
Đặt $S=1+2+2^2+2^3+.....+2^{2020}$
$⇔2S=2+2^2+2^3+.....+2^{2020}+2^{2021}$
$⇒2S-S=(2+2^2+2^3+.....+2^{2020}+2^{2021})-(1+2+2^2+2^3+.....+2^{2020})$
$⇔S=2^{2021}-1$
Ta có: $(*)⇔2^x.(2^{2021}-1)=2^2(2^{2021}-1)$
$⇔2^x=2^2⇔x=2$
