a. factor [ 0,2^{22}] [ 0,04+0,2+1]=1,24·[( 0,2)² la n]
0,2^(22) [1,24]= 1,24 · (0,2)^(2n)
termeni identici , egalam exponentii 22=2n n=11
b. 1,1^(3n+6) =(1,1)^([2(n+9)]
1,21=1,1²
egalam exponentii 3n+6=2(n+9)
3n-2n=18-6
n=12
2. 1,2²=1,44
1,2^[1+2+3+..... +n] = 1,2^[2 · 105]
stim 1+2+3+.....+n=n(n+1) /2
n(n+1) /2 = 210
n(n+1)=420
n² -n - 420=0
Δ=4+4 ·420=1681 √Δ=41
n=21 ∈N ( celalat numar este negativ )
