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Exponential functions are of the form $\,y = b^t\,$ for $\,b > 0\,$ and $\,b\ne 1\,.$
For $\,b > 1\,$ they are increasing, and are called exponential growth functions.
For $\,0 < b < 1\,$ they are decreasing, and are called exponential decay functions.

All exponential functions share common behavior:
when the input changes by a fixed amount, the output gets multiplied by a scaling factor.

It might be that every hour (the fixed input change), something doubles (gets multiplied by $\,2\,$).
Or, it might be that every three seconds (the fixed input change), something gets halved (multiplied by $\,\frac 12\,$).
In general, increasing exponential functions get big fast, and decreasing exponential functions get small fast.

A typical exponential growth/decay problem often has this flavor:

A population has a given initial size.
Later, it is bigger (or smaller).
Assuming exponential behavior, what is its size at some future time?

There are a couple things that can make exponential growth/decay problems seem tricky.
First of all, a ‘basic’ exponential function may need to be generalized, using both vertical and horizontal stretches.
Secondly, the generalized exponential functions go by lots of different names.
You might see a problem solved in different ways, and wonder ‘which is right’.
They're all right! Different people choose different names to work with!
These issues are discussed in this section.

 

Choose $\,t = 0\,$

For exponential growth/decay problems, we call the input $\,t\,$ (for ‘time’) instead of $\,x\,.$
Populations change with time. Populations are a function of time.
The notation $\,P(t)\,$ denotes the size of a population at time $\,t\,.$

You always have an initial choice to make—what you call ‘time zero’—the ‘origin’ for time.
Some people choose the earliest time when a population is known to correspond to $\,t = 0\,.$
It doesn't matter, as long as you are consistent throughout the entire problem.

Suppose $\,t=0\,$ is chosen to correspond to time $\,Y\,$ (think ‘year’):

Then, for $\,n > 0\,$: $\,t = n\,$ corresponds to $\,F = Y + n\,$ (future) $\,t = -n\,$ corresponds to $\,P = Y - n\,$ (past)

Going the other direction, for $\,F > Y > P\,$: a future $\,F\,$ corresponds to positive time $\,t = F - Y\,$ a past $\,P\,$ corresponds to negative time $\,t = P - Y\,$

Generalizing $\,y = b^t\,$ to Handle Typical Exponential Growth/Decay Problems

Pick any allowable value of $\,b\,.$ The sketches below assume $\,b > 1\,,$ but the discussion holds for any $\,b > 0\,,$ $\,b\ne 1\,.$

A typical exponential function, $\,y = b^t\,.$ It may not have the correct initial size (the blue point). It may not have the correct second known size (the red point).

Do a vertical scaling to get the correct initial size: $y = P_{\,0}\,b^t$

Do a horizontal scaling to get the correct second known size: $y = P_{\,0}\,b^{st}$

The exponential function $\,P(t) = b^t\,$ is too restrictive for exponential growth/decay problems.
We need to allow for any initial population size. (The ‘initial population’ is the size at $\,t = 0\,.$ )
However, $\,P(0) = b^0 \ \overset{\text{always}}{=}\ 1\,.$ We don't always have an initial population of size $\,1\,$!

To solve this problem, do a vertical scaling of the basic exponential function, giving $\,P(t) = P_{\,0}\,b^t\,.$
With this generalization, $\,P(0) = P_{\,0}\, b^0 = P_{\,0}(1) = P_{\,0}\,,$ so $\,P_{\,0}\,$ is the desired initial population size.
(You can read ‘$\,P_{\,0}\,$’ as ‘$\,P\,$ sub zero’ or ‘$\,P\,$ naught’.)

We also need the curve to pass through a second point, which describes the population at some known future or past time.
A horizontal scaling accomplishes this:   $\,P(t) = P_{\,0}\, b^{st}\,$

Checking that the Generalized Exponential Function Still Behaves As Expected

It's worthwhile to check that the generalized exponential function, $\,P(t) = P_{\,0}\, b^{st}\,,$ still has the property mentioned above:
when the input changes by a fixed amount, the output gets multiplied by a scaling factor.

$$ \begin{alignat}{2} P(t+\Delta t) &= P_{\,0} \, b^{s(t+\Delta t)}&&\text{definition of function $\,P\,$}\cr\cr &= P_{\,0} \, b^{(st+s\Delta t)}&&\text{distributive law}\cr\cr &= P_{\,0} \, b^{st}\, b^{s\Delta t}&&\text{exponent law}\cr\cr &= \bigl( P_{\,0\,} b^{st}\bigr) \cdot \bigl(b^{s\Delta t}\bigr)\qquad &&\text{re-group}\cr\cr &= P(t) \cdot \bigl(b^{s\Delta t}\bigr)\qquad &&\text{definition of $\,P(t)\,$}\cr\cr \end{alignat} $$

In this case, the scaling factor is $\,b^{s\Delta t}\,,$ which depends on three things:

• the base of the exponential function ($\,b\,$)
• the horizontal scaling factor ($\,s\,$)
• how much the inputs are changing by ($\,\Delta t\,$)

Any Generalized Exponential Function Can Be Written With Any Allowable Base

Let $\,b\,,$ $\,k\,$ and $\,s\,$ be the current base, vertical scaling factor, and horizontal scaling factor for a generalized exponential function: $$y = kb^{st}$$ You've decided you don't want base $\,b\,.$ You want a different base $\,B\,$ instead.
This can be easily done! Let $\,y = KB^{St}\,$ denote the new function, with the desired base.
Notice that we're using lowercase letters for the old function ($\,b\,,$ $\,k\,,$ $\,s\,$), and uppercase letters for the new function ($\,B\,,$ $\,K\,,$ $\,S\,$).

• We want $kb^{st} = KB^{St}$ to be true for all values of $\,t\,.$
• In particular, it must be true when $\,t = 0\,.$
  Substituting $\,0\,$ for $\,t\,$ gives $\,kb^0 = KB^0\,,$ so that $\,k = K\,.$
  The vertical scaling factor stays the same!
• Now we have $\,kb^{st} = kB^{St}\,.$
  Divide both sides by $\,k\,,$ giving $\,b^{st} = B^{St}\,.$
• To get to the exponents, take logs of both sides.
  Any log will do, so choose the natural logarithm:
  $\ln b^{st} = \ln B^{St}$
• Bring down the exponents:
  $st \ln b = St\ln B$
• Divide both sides by $\,t\,$:
  $s\ln b = S\ln B$
• Solve for $\,S\,$:
  $\displaystyle S = s\cdot\frac{\ln b}{\ln B}$

Here's a quick example to show how easy it is!
Initial function: $\,y = 10\cdot 5^{3t}\,$   ($\,k = 10\,,$ $\,b = 5\,,$ $\,s = 3\,$)
New desired base: $\,7\,$   ($\,B = 7\,$).

The $\,10\,$ stays the same ($\,K = k\,$).   The new horizontal scaling factor is $\,\displaystyle S = s\cdot\frac{\ln b}{\ln B} = 3\cdot \frac{\ln 5}{\ln 7}\,.$
The new function is therefore $\,y = K\cdot B^{St} = 10\cdot 7^{3t(\ln 5)/(\ln 7)}\,.$

Don't take my word for it!
Hop up to wolframalpha.com and cut-and-paste:

plot y = 10 * 5^{3t}, y = 10 * 7^{3t ln(5)/ln(7)}
You'll only see one curve, because they're exactly the same!

So, any base can be written as any other base. Exponential functions have lots of different names!
The point is this: when solving exponential growth/decay problems, you can use any (allowable) base that you want.
Having said this, though, there is a strongly-favored base that is usually used $\ldots$

Since Any Base Can Be Used, Most People Use Base $\,\text{e}\,$

Since any base can be used when solving exponential growth/decay problems, most people use base $\,\text{e}\,.$
Why? Why would someone want to use an irrational number as a base, instead of a ‘simple’ number like (say) $\,2\,$?

The answer is that the exponential function with base $\,\text{e}\,$ has calculus properties that make it easier to work with than any other exponential function.
Also, when the base is $\,\text{e}\,,$ the horizontal scaling constant takes on special significance, and is given a special name—the relative growth rate.
More on this in the next section!

On this exercise, you will not key in your answer. However, you can check to see if your answer is correct.

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