How Do I Solve Equations with e and ln?

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Discussion Overview

The discussion revolves around solving equations involving the exponential function e and its natural logarithm ln, particularly in the context of finding derivatives and determining concavity of the function f(t) = e^(-9t). Participants explore the steps involved in differentiating the function and the implications of the second derivative.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant expresses difficulty in solving for t after finding the second derivative of f(t) = e^(-9t).
  • Another participant asks for clarification on whether the goal is to find t or something else, indicating confusion about the problem statement.
  • A participant explains that determining concavity requires understanding when the second derivative is positive, negative, or zero, and notes that setting it to zero only identifies inflection points.
  • There is a correction regarding the need to set the first derivative to zero to find stationary points before analyzing the second derivative.
  • One participant mentions that the function e^(-9t) does not have stationary points, as its first derivative is never zero, but acknowledges that other functions involving exponentials can have stationary points.
  • Another participant clarifies that the second derivative of e^(-9t) is always positive, indicating that the function is concave upward for all t.

Areas of Agreement / Disagreement

Participants express differing views on the correct approach to determining concavity and the role of the first and second derivatives. There is no consensus on the initial problem statement or the method for solving it.

Contextual Notes

Some assumptions about the function and its derivatives are not explicitly stated, and the discussion includes various interpretations of the problem, leading to potential misunderstandings.

winston2020
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For example, let's say that f(x) = e^(-9t)
Once I've found the second derivative, I equate it to zero in order to determine concavity, but I don't know how to solve it after that :rolleyes:

I'm just having a hard time solving with the ln or whatever you have to do..:confused:
 
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Are you trying to solve the equation to find t=something? Or are you tring to do something else? Is t a function of x? Could you clarify your problem please because at the moment it makes no sense whatsoever.
 
Last edited:
If you're trying to determine concavity, you need to figure out when the 2nd derivative is positive, negative, or zero. Just setting it to zero will tell you where the inflection points are (neither concave up nor down), and if there are no such points then you'll get "no solution" when you do this.
 
sorry, what i meant was set the first derivative to zero, and solve for t. Then use the t value to solve for the second derivative.

(The function should have been f(t) not f(x))
 
Okay.

Plug the value of t you have (from setting 1st derivative = 0) into the 2nd derivative.

Is the result positive, negative, or zero?
 
Redbelly98 said:
Plug the value of t you have (from setting 1st derivative = 0) into the 2nd derivative.
Err, just one problem with that: et has no roots. A purely exponential function has no stationary points and therefore it's first derviative is never zero. However, functions involving expoentials can have stationary points. For example,

[tex]y(x) = e^x-x[/tex]

[tex]y^\prime(x) = e^x-1[/tex]

Hence, y(x) has a stationary point when,

[tex]e^x = 1 \Rightarrow x=0[/tex]
 
Last edited:
winston2020 said:
For example, let's say that f(x) = e^(-9t)
Once I've found the second derivative, I equate it to zero in order to determine concavity, but I don't know how to solve it after that :rolleyes:

I'm just having a hard time solving with the ln or whatever you have to do..:confused:
No, you don't "equate the second derivative to zero in order to determin concavity".

The concavity of f(t) at a particular t depends upon the sign of f" at that particular t.

The second derivative of e^(-9t) is 81e^(-9t) and , as pointed out by others, that is never equal to 0. At t= 0, it is 81> 0. Since 81e^(-9t) is continuous and never equal to 0, it is always positive and so f(t)= e^(-9t) is concave upward for all t.
 
Thanks for all your help everyone :smile:
 

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