\(2^x+2^{x+1}+...+2^{x+2021}=2^{2025}-16\)
=>\(2^x\left(1+2^1+...+2^{2021}\right)=2^{2025}-16=2^4\left(2^{2021}-1\right)\)
Đặt \(A=1+2^1+...+2^{2021}\)
=>\(2A=2+2^2+...+2^{2022}\)
=>\(2A-A=2+2^2+...+2^{2022}-1-2^1-...-2^{2021}\)
=>\(A=2^{2022}-1\)
Phương trình sẽ trở thành: \(2^x\left(2^{2022}-1\right)=2^4\left(2^{2021}-1\right)\)
=>\(2^x=\dfrac{2^4\left(2^{2021}-1\right)}{2^{2022}-1}\)
=>\(x=log_2\left(\dfrac{2^4\left(2^{2021}-1\right)}{2^{2022}-1}\right)\)
\[
2^x + 2^{x+1} + 2^{x+3} + \ldots + 2^x + 2021 = 2^{2025} - 16
\]
\[
2^x + 2^{x+1} + 2^{x+3} + \ldots + 2^x + 2021
\]
\[
2^x + 2^{x+1} + 2^{x+3} + \ldots
\]
\[
2^x (1 + 2^1 + 2^3 + \ldots) + 2021
\]
\[
2^x (1 + 2^1 + 2^2 + \ldots + 2^k) + 2021
\]
\[
\frac{2^{k+1} - 1}{2 - 1} = 2^{k+1} - 1
\]
\[
2^x (2^{k+1} - 1) + 2021
\]
\[
2^x (2^{k+1} - 1) + 2021 = 2^{2025} - 16
\]
\[
2^x (2^{k+1} - 1) \approx 2^{2025} - 16 - 2021
\]
\[
2^x \approx 2^{2025} - 2021 - 16
\]
\[
2^x = 2^{2021}
\]
\[
x = 2021
\]
`2^x + 2^(x+1) + ... + 2^(x + 2021) = 2^2026 - 16 = 2^4 (2^2022 - 1)`
Đặt `S = 2^x + 2^(x+1) + ... + 2^(x + 2021)`
`2S = 2^(x+1) + 2^(x+2) + ... + 2^(x + 2022) `
`2S - S = (2^(x+1) + 2^(x+2) + ... + 2^(x + 2022)) - (2^x + 2^(x+1) + ... + 2^(x + 2021))`
`S = 2^(x + 2022) - 2^x `
`S = 2^x (2^2022 - 1) `
`=> x = 4`
Vậy `x = 4`