Negative Exponential Growth and Decay  TOPIC_SOLVED

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Negative Exponential Growth and Decay

Postby Whist34 on Tue Nov 01, 2011 12:19 am

I am trying to wrap my mind around the concept of Exponential Growth and Decay, but when the initial value is negative I cannot seem to find a confident answer.

Let us use: y = abx
a is the initial value
b (If b > 1 then we have growth, if 0 < b < 1 then we have decay)(b is not equal to 1 and is greater than 0 by definition)
x > 0 (assume b is adjusted to remove negative exponent)

My confusion comes from the different resources on the net and the lack of a definitive answer. Some websites restrict initial values (a > 0) for growth and decay problems, others allow it to be any real number, but then they do not demonstrate the negative initial value case. I have only found one website where examples with negative initial values are presented (http://mathonweb.com/help_ebook/html/expoapps.htm) - this site demonstrates my understanding of growth and decay. The site uses an example for exponential decay of drinks cooling or warming to room temperature. But this site is only one in 100 and I want to be sure about it.

This is my current understanding of Exponential Growth and Decay:
Exponential Growth - the rate at which y changes increases (note - y could increase or decrease)
a is any real number, b > 1
Examples:
A) y = 3(1.2)x initial value is 3 and rate of change increases (The y value would increase as input increases)
B) y = -3(1.2)x initial value is -3 and rate of change increases (The y value would decrease as input increases)

Exponential Decay - the rate at which y changes decreases (note - y could increase or decrease)
a is any real number, 0 < b < 1
A) y = 3(0.5)x initial value is 3 and rate of change decreases (The y value would decrease as input increases)
B) y = -3(0.5)x initial value is 3 and rate of change decreases (The y value would increase as input increases)

Please either confirm my understanding or correct it.
The following questions sum up my confusion:
1) It seems like many sites regard growth as an increase in y instead of an increase in the rate of change. Does y have to increase for there to be exponential growth? Alternatively does y have to decrease for there to be exponential decay?

2) Should a be limited to a > 0 by definition for growth and decay?

3) If I am wrong and exponential growth is indeed as x increases so does y, then would y = -3(0.5)x be growth and y = -3(1.2)x be decay?

Thanks
Whist34
 
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Re: Negative Exponential Growth and Decay  TOPIC_SOLVED

Postby maggiemagnet on Tue Nov 01, 2011 3:11 pm

Whist34 wrote:1) It seems like many sites regard growth as an increase in y instead of an increase in the rate of change. Does y have to increase for there to be exponential growth? Alternatively does y have to decrease for there to be exponential decay?

As far as I know, "growth" means "getting bigger" while "getting smaller" is called "decay" (or at least "negative growth").

Whist34 wrote:2) Should a be limited to a > 0 by definition for growth and decay?

As far as I know, "a" can be any non-zero value (a = 0 would be pretty dull!), but, for "real life" stuff, positive is what makes sense. So it's more of a word-problem thing than a math thing.

Whist34 wrote:3) If I am wrong and exponential growth is indeed as x increases so does y, then would y = -3(0.5)x be growth and y = -3(1.2)x be decay?

I think the "growth" and "decay" thing is more related to the exponential part, bx, and whether it is getting larger or getting closer to zero.
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Re: Negative Exponential Growth and Decay

Postby Whist34 on Tue Nov 01, 2011 11:54 pm

Thank you for the reply.

Here are some more questions to illustrate the point.

Identify the following equations as exponential growth, exponential decay, or neither and explain why.

1) y = -2(5)x

2) y = -2(1/5)x


Describe the following real life situations as representing exponential growth, exponential decay, or neither.

3) A cup of hot coffee as it cools to room temperature.

4) A cup of iced coffee as it warms to room temperature.

Thanks again for any help.
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