Hallo,
a) \( e^{x}=2 \) | ln(..)
x ln (e)= ln(2) -->ln(e)=1
x=ln(2)
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d) \( \mathrm{e}^{2 x+1}=0,5 \) | ln(..)
(2x+1) ln(e)= ln (0.5)
2x+1 = ln (0.5) =ln (1/2)= ln(1) -ln(2) ->ln(1)=0
2x+1 = -ln(2) |-1
2x= - ln(2)-1 |:2
x= (- ln(2)-1) /2
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g) \( \mathrm{e}^{-2 \mathrm{x}}=0,5 \) =1/2
-2x= ln(1) -ln(2) ---->ln(1)=0
-2x= -ln(2)
x= ln(2)/2
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b) \( e^{x}=\pi \)
x= ln(π)
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e) \( e^{x-7}=1 \) |ln(..)
x-7= ln(1) =0
x=7
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h) \( e^{1,5 x}-2=0 \) |+2
e^(1.5x)= 2 |ln(..)
1.5x= ln(2)
x= ln(2)/1.5
x= (2 ln(2))/3
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c) \( e^{0,5 x}=3 \) |ln(...)
0.5x= ln(3)
x= 2 ln(3)
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f) \( e^{-3 x}=5 \) |ln(..)
-3x= ln(5)
x=(-ln(5)/3
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\( e^{\frac{1}{2} x}-5=0 \)
e^(x/2)=5 |ln(..)
x/2= ln(5)
x= 2 ln(5)