# Thread: confusing exponential increase using variables

1. ## confusing exponential increase using variables

I'm not even sure if this is the right board to post this issue, but I worked this out with variables and eventually confused myself to insanity. Help is very much appreciated!

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The athlete was afraid because the coach's anger increased exponentially. Only 20 minutes after practice began, the coach was twice as angry as when the practice began. How many time more angry will the coach be at the end of the three hour practice?

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(use the At = Aoe^(kt) exponential formula)

2. Hello,
Originally Posted by Savior_Self
I'm not even sure if this is the right board to post this issue, but I worked this out with variables and eventually confused myself to insanity. Help is very much appreciated!

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The athlete was afraid because the coach's anger increased exponentially. Only 20 minutes after practice began, the coach was twice as angry as when the practice began. How many time more angry will the coach be at the end of the three hour practice?

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(use the At = Aoe^(kt) exponential formula)
You're looking for k in $\displaystyle A(t)=A(0)e^{kt}$

But you know that at $\displaystyle t=20$, he's twice as angry as in the beginning.
So $\displaystyle A(20)=2A(0)$

--> $\displaystyle 2A(0)=A(0)e^{20k} \Rightarrow e^{20k}=2\Rightarrow 20k=\ln(2)$

There's a similar notion in physics, called the half-life

3. okay, how would I answer the question of how angry the coach is after three hours (180 min)?

Plugging in ( k = ln2/20 ), would the equation ( A(180) = A(0)e^[(ln2/20)(180)] ) solve for how angry he was after three hours?