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e^x=t, t>0
t-1/t=1
t²-t-1=0
Δ=5
√Δ=√5
t1=(1+√5)/2 ∨ t=(1-√5)/2<0 odrzucamy
e^x=(1+√5)/2
lne^x=ln[(1+√5)/2]
x=ln[(1+√5)/2]
1.
e^x>2^x⇔x<0 (rys.1)
e(-x)≤x^5+1⇔x≥0 (rys.2)
rys.3