Hello.
I'm looking at this equation:
f(x) = e^(2x) - 4e^(x) + 3
I know that in order to find the zero points of this equation, I must write it in this format:
f(x) = u^2 -4u + 3 and then solve for u using the quadratic formula.
While working on this type of problem, my brain came up with an alternate method. This alternate method gives the wrong answer, but it looks to me like I'm not violating any basic rules. Since I'm getting wrong answers, I MUST be doing something wrong, and I need to understand what it is so I don't make that type of mistake again.
Here's my alternate method. Can someone please tell me why this does not work? Thx.
f(x) = e^(2x) - 4e^(x) + 3
e^(2x) - 4e^(x) + 3 = 0
e^(2x) +3 = 4e^(x)
ln( e^(2x) ) + ln3 = ln ( 4e^(x) )
(2x)ln(e) + ln3 = ln4 + ln(e^x)
2x + ln3 = ln4 + (x)ln(e)
2x + ln3 = ln4 +x
x = ln4/ln3
I'm looking at this equation:
f(x) = e^(2x) - 4e^(x) + 3
I know that in order to find the zero points of this equation, I must write it in this format:
f(x) = u^2 -4u + 3 and then solve for u using the quadratic formula.
While working on this type of problem, my brain came up with an alternate method. This alternate method gives the wrong answer, but it looks to me like I'm not violating any basic rules. Since I'm getting wrong answers, I MUST be doing something wrong, and I need to understand what it is so I don't make that type of mistake again.
Here's my alternate method. Can someone please tell me why this does not work? Thx.
f(x) = e^(2x) - 4e^(x) + 3
e^(2x) - 4e^(x) + 3 = 0
e^(2x) +3 = 4e^(x)
ln( e^(2x) ) + ln3 = ln ( 4e^(x) )
(2x)ln(e) + ln3 = ln4 + ln(e^x)
2x + ln3 = ln4 + (x)ln(e)
2x + ln3 = ln4 +x
x = ln4/ln3