Question

rezolva ecuatia 2^x=1+2^0+2^1+....+2^1968

Answer 1

Termenii 2^0, 2^1....2^1968 sunt in progresie geometrica cu ratia q=2 si primul termen b1=2^0.
S=b1*(q^n - 1)/q-1
2^0+2^1+...+2^1968=2^0*(2^1969-1)/2-1=2^0(2^1969-1)=2^1969-1
2^x=1+2^0+2^1+....+2^1968
2^x=1+2^1969-1
2^x=2^1969
Functia exponentiala este injectiva, deci putem egala puterile.
x=1969