Can someone help me algebraically solve this?

[darbyhouston]

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.

 

[Oliver Curtis]

What is this equal to? 0 To find a value for X the equation youve given should be equal to something i.e. 1-0.8 ... = 0

 

[skeeter]

what is [math]1-0.8e^{-x^3/27}[/math] equal to ?

 

[Dr.Peterson]

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.

 

[darbyhouston]

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.

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

 

[Dr.Peterson]

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!

 

[darbyhouston]

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!

Thank you! So how about this:

ln((y-c)/-d) = (-x/a)^b

ln((y-c)/-d)^1/b = -x/a

-a*(ln((y-c)/-d)^1/b) = x

 

[Dr.Peterson]

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]

 

[darbyhouston]

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]

Thank you for your help!