Question

Tìm số tự nhiên x, biết:

a) \(2^0\)+\(2^1+2^2+...+2^{2017}=4^x-1\)

b) \(3^x+3^{x+1}+3^{x+2}+3^{x+3}=360\)

Answer 1

a.

\(2^0+2^1+2^2+...+2^{2017}=4^x-1\) (1)

\(\Rightarrow2^1+2^2+2^3+...+2^{2018}=2.\left(4^x-1\right)\) (2)

Trừ vế (2) cho (1):

\(\Rightarrow2^{2018}-2^0=2.\left(4^x-1\right)-\left(4^x-1\right)\)

\(\Rightarrow2^{2018}-1=4^x-1\)

\(\Rightarrow2^{2018}=4^x\)

\(\Rightarrow4^{1009}=4^x\)

\(\Rightarrow x=1009\)

b.

\(3^x+3^{x+1}+3^{x+2}+3^{x+3}=360\)

\(\Rightarrow3^x\left(1+3+3^2+3^3\right)=360\)

\(\Rightarrow3^x.40=360\)

\(\Rightarrow3^x=9\)

\(\Rightarrow3^x=3^2\)

\(\Rightarrow x=2\)