3×e2x-5=7 |+5
3×e2x= 12 |:3
e2x=4
2x ln(e)= ln(4)
ln(e)=1
2x = ln(4)
x= ln(22)/2
x=2 *ln(2)/2
x=ln(2)
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ex-8e-x=2 |*ex
e2x -8= 2 ex
e2x - 2 ex -8=0
Substituiere z= ex
z2-2z -8=0 ->pq Formel:
z1.2= 1± √( 1+8)
z1.2= 1± 3
z1=4
z2=-2
Resubstitution:
4= ex
ln(4)=x
x= 2 ln(2)
-2 =ex hat keine Lösung