darbyhouston
New member
- Joined
- Oct 3, 2020
- Messages
- 4
Hi, I'm having trouble rearranging the following function to solve for x
1-0.8*e^(-(x/3)^3)
where e is Euler's number.
1-0.8*e^(-(x/3)^3)
where e is Euler's number.
You call it a function, so I suppose you want the inverse. That is, if y = 1-0.8*e^(-(x/3)^3), you want to find x as a function of y.Hi, I'm having trouble rearranging the following function to solve for x
1-0.8*e^(-(x/3)^3)
where e is Euler's number.
You call it a function, so I suppose you want the inverse. That is, if y = 1-0.8*e^(-(x/3)^3), you want to find x as a function of y.
The general technique for solving an equation with the variable in only one place is to undo one operation at a time. First undo the subtraction, then the multiplication by 0.8, then the exponentiation (for which logarithms are a good tool!), and so on.
Give it a try and show us where you get stuck.
Ah! That helps.It'll help me if I write it like this:
y=c-d.*e^((-x/a)^b)
(y-c)/-d = e^((-x/a)^b)
ln((y-c)/-d) = (-x/a)^b
This is where I get something weird that I don't think is right, see below.
log(-x/a)(ln(y-c)/-d)=b
Ah! That helps.
The trouble is that you have logs stuck in your mind after using the first one.
When you have something like A = e^(f(x)), you need to take a log and get ln(A) = f(x), as you did, to extract the x from the base.
But now you have something like A = (f(x))^b, with x inside the base rather than in the exponent. Here, you need to raise both sides to the1/b: A^(1/b) = f(x). Now you can solve for x, because x is in a good place.
As you sensed, having x in the base of the logarithm is not a good place to be!
Yes, that will do. For clarity, when typing, I'd add more parentheses, while removing an unnecessary pair: x = -a*(ln((y-c)/-d))^(1/b). That is, with a minor simplification: [MATH]x = -a\left(\ln\left(\frac{c-y}{d}\right)\right)^\frac{1}{b}[/MATH]