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3^x<10 i x<2
f.rosnąca, bo 3>0
niech 3^x=t, t>0
10-t≥9/t
10t-t²-9≥0
-t²+10t-9≥0
t²-10t+9≤0
(t-9)(t-1)≤0
t∈<1,9>
3^x∈<1,9>
1≤3^x≤9
3^0≤3^x≤3^2
0≤x≤2
uwzględniając dziedzinę
x∈<0,2)