Question

wie rechne ich e^{2x+1}=0,5 mit dem natürlichen logarithmus?

Answer 1

--> 2x+1 = ln0,5

x= (ln0,5-1)/2

Answer 2

 e^2x+1=0,5 | ln

(ln(e))^{2x+1} = ln(0.5)

(2x+1) *ln (e)= ln(1/2)

ln(e)=1

2x+1  =ln(1/2) |

2x+1  =ln (1) -ln(2)

ln(1)=0

2x+1  = -ln(2) |-1

2x  = -ln(2) |-1 |:2

x  = (-ln(2) |-1 )/2

x≈ -0.847