B BAC_Mono New member Joined Nov 16, 2015 Messages 2 Nov 16, 2015 #1 e^(-3x) +e(-x - x^2 = 0 All the other questions I've come across on this topic have had all x the values raised to e - never a mixture like this. Thanks in advance.
e^(-3x) +e(-x - x^2 = 0 All the other questions I've come across on this topic have had all x the values raised to e - never a mixture like this. Thanks in advance.
B BAC_Mono New member Joined Nov 16, 2015 Messages 2 Nov 16, 2015 #2 e^(-3x) +e(-x) = x^2 e^(-3x) +e(-x) = x^2 I can't see how to group the x-values, without having Ln(x^2) somewhere. Never come across a question like this. Thanks in advance.
e^(-3x) +e(-x) = x^2 e^(-3x) +e(-x) = x^2 I can't see how to group the x-values, without having Ln(x^2) somewhere. Never come across a question like this. Thanks in advance.
stapel Super Moderator Staff member Joined Feb 4, 2004 Messages 16,550 Nov 16, 2015 #3 BAC_Mono said: e^(-3x) +e(-x) - x^2 = 0 Click to expand... What were the instructions for this? What are you supposed to be doing with this equation? Please be complete, including a clear listing of any attempts you have made, even if you think they're wrong. Thank you!
BAC_Mono said: e^(-3x) +e(-x) - x^2 = 0 Click to expand... What were the instructions for this? What are you supposed to be doing with this equation? Please be complete, including a clear listing of any attempts you have made, even if you think they're wrong. Thank you!
J Jedi Equester New member Joined Nov 14, 2015 Messages 9 Nov 16, 2015 #4 First of all I assume you talk about e^(-3x) + e^(-x) - x^2 = 0 To solve this equation you won't be able to use standard algebraic transformations. Having a look at this kind of equation remember Newton's method or similar methods.
First of all I assume you talk about e^(-3x) + e^(-x) - x^2 = 0 To solve this equation you won't be able to use standard algebraic transformations. Having a look at this kind of equation remember Newton's method or similar methods.