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Exponential functions - converting a linear equation

  • Thread starter LizzieL
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  • #1
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Homework Statement


Ok, so I have an unknown exponential function:
y1 = f1(t)

By measuring the values of t and yi, a linear connection is generated between Yi (=log (yi)) and t:

Yi = Ait + Bi.
A1 = -2.12
B1 = 1.96

Problem 1: Describe f1 in the following matter:
y1 = f1(t) = Ceλt


The Attempt at a Solution


I've been juggling with numbers and letters without ending up with anything reasonable.
Ok, let's ramble down some thoughts:
Adding log function to both sides of the functions gives

log(y1) = log t + log -2.12 + log 1.96 = log (C) + log (e) λt

And this is where it all ends. Any help out there? Would be much appreciated!
 

Answers and Replies

  • #2
ehild
Homework Helper
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Problem 1: Describe f1 in the following matter:
y1 = f1(t) = Ceλt


The Attempt at a Solution



Ok, let's ramble down some thoughts:
Adding log function to both sides of the functions gives

log(y1) = log t + log -2.12 + log 1.96 = log (C) + log (e) λt
The red part has no sense...

The logarithm of y is given: log(y)=At+B=-2.12 t + 1.96.

And -2.12 t + 1.96=log(e)(λt)+log(C)

What is the base of the logarithm?

Compare the terms. The constant terms have to be equal on both sides, and the coefficients of the variable t also have to be equal.

ehild
 
  • #3
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Oh, of course the logarithm of y is given! Why didn't I see that before. Thanks for your reply.

The base of the logarithm is 10, forgot to say.
But, how do I solve it from here? This is my first go at a problem like this, so any help is appreciated. How can I solve this when I have no values for neither t, C and λ? and the log gets in the way.
 
  • #4
I like Serena
Homework Helper
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Welcome to PF, LizzieL! :smile:

ehild's equation must hold for any t.
This means that -2.12=log(e)(λ) and that 1.96=log(C).

For instance -2.12=log(e)(λ) can be rewritten as λ=-2.12/log(e).
 
  • #5
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Oh my, you guys really are angels sent from up above. Thanks a whole lot! Also for the welcome :smile:

So, I have a similar problem:

log(y2) = 1.85t + 1.56 = log(e)μ(t-a)

I cannot use the same principle as before, right? Or can I?

If I try again...

1.85 = log(e)μ
1.56 = -log(e)μa

Is this way off?
 
  • #6
I like Serena
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I'd say it is right on!! :approve:
 

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