Exponentials and Percentage Growth

by
K. D.  Stroyan
University of Iowa

copyright 1997 by Academic Press, Inc. - All rights reserved.

Special Mathematica Functions used in this NoteBook

Needs["Graphics`Colors`"];

Notebook Overview

This NoteBook studies the function limit

(Underscript[Limit, Δx -> 0])[
        ((f[x + Δx] - f[x])/Δ)x] = f'[x]

that defines the derivative of an exponential function.

The Moral: Constant Percentage Rate of Change

The point of the whole NoteBook is that exponential functions have  f'[x] proportional to f[x],  or, to say it another way, f[x]  grows at a constant PERCENTAGE rate.  We can write this:

y = (a^x)  =>  (1/y)·(dy/dx) = (Subscript[r, a])

for  (Subscript[r, a])  constant, or

y = (a^x)  =>  (dy/dx) = (Subscript[r, a])·y

with the important special case

y = (e^x)  =>   (dy/dx) = y    (or   (Subscript[r, e])  =  1)

The Official Definition of the Natural Exponential

This is so important that we take it as the "official definition" of the natural exponential fuction by specifying the initial condition:

(y[0] = 1 & (dy/dx) = y)   ⟺   y = (e^x)

The Converse: Constant Percentage Growth => Exponential

The statement that the derivative of an unknown function is proportional to the function (or equivalently, that the instantaneous percentage rate of change of an unknown function is constant)

(dy/dx) ∝ y  or  (dy/dx) =  r y

is equivalent to saying the function has the form  y = y[x] = c (a^x)  for constants  c  and  a.

Linear Growth Comparison

Notice that a linear function  y = f[x] = m x + b  grows at a constant rate, NOT a constant PERCENTAGE rate:

\(Clear[m, b, x, Δx]; 
f[x_] := m\ x\  + \ b\)rate = ((f[x + Δx] - f[x])/Δ)xPrint["Rate of change in y"];  
    Print["for fixed Δx-change in x = "]; Factor[rate]\(Print["Percent rate of change in y"]; 

    Print["for fixed Δx-change in x = ", 
      Simplify[rate\/f[x]], " × 100%"]; 

    Print["Fractional change ="]; 
Simplify[rate\/f[x]]\)

References to the Text

This NoteBook goes with Section 5.4 and Chapter 8 of the text.

Laws of Exponents and
the Percentage Difference Quotient

First we observe (as in Section 5.4) that the rate of change of an exponential is a constant FRACTION of the starting amount  f[x]  when the change in input, Δx,  is constant.

\(Clear[a, x, Δx]; 
f[x_] := a\^x\)\(xChange\  = \ Δx; 

    fctChange\  = \ f[x + Δx] - f[x]; 

    rateOfChange = fctChange\/xChange\)Simplify[rateOfChange]Simplify[(rateOfChange/f)[x]]

The expression in  x  for the percent rate of change of  y = f[x],

(1/f)[x]·((f[x + Δx] - 
            f[x])/Δ)x

actually does not depend on  x; it has the same value for all  x  (as long as  Δx  is fixed.)

Computation Details?

Here's the calculation again with some printed explanation:

\(Clear[a, x, Δx]; 
f[x_] := a\^x\)\(Print["Change in y over Δx-change in x ="]; 

    Print[" rate of change in y for fixed Δx-change ="]; 

    rate = \(f[x + Δx] - f[x]\)\/Δx\)Print["Simplified rate of change in y"];  
    Print["for fixed Δx-change in x = "]; Factor[rate]\(Print[
      \*""Change as a fraction of y,(1/y)Δ(y/Δ)x=""]; 

    Print[\*"" (1/f)[x]·((f[x + Δx] - f[x])/Δ)x=""]; 

Simplify[rate\/f[x]]\)Print[
      "Percent rate of change for fixed Δx-change = "]; 
    Print[Simplify[(rate/f)[x]], " × 100%"];

A Plot of the Difference Quotient

Constant percentage INSTANTANEOUS rate of change for exponentials means that a plot of  f'[x]/f[x]  is constant.  Try this on some functions:

\(Clear[f, x, a]; 
f[x_] := a\^x; 
a = 3\/2; 
f[x]\)f'[x]f'[x]/f[x]Plot[Evaluate[{f[x],f'[x],f'[x]/f[x]}],{x, -1, 2},AxesLabel -> {"x", "y"},PlotStyle 
            -> {Blue, Red, {AbsoluteThickness[2], DarkGreen}}];

Here's another wacky way to see if f[x] and f'[x]  are proportional.  (This might be good if you had data of both  f[x]  and  f'[x].)

\(Clear[f, x, a]; 
f[x_] := a\^x; 
a = 3\/2; 

    ParametricPlot[{f[x], \(f'\)[x]}, {x, \(-1\), 2}, 
      AxesLabel -> {"f[x]", "f'[x]"}, 
      PlotStyle -> {AbsoluteThickness[2], Brown}]; \)

The slope of this parametric line is the constant of proportionality between  f[x]  and  f'[x].

Linear Functions Have Constant Rate of Change
	NOT Constant Percentge Rate of Change

Clear[f, x]; f[x_] := (x/2) + 1f[x]f'[x]((f'[x])/(f[x]))Plot[Evaluate[{f[x],f'[x],((f'[x])/(f[x]))}],{x, -1, 2},AxesLabel -> {"x", "y"},PlotStyle 
            -> {Blue, Red, {AbsoluteThickness[2], DarkGreen}}];

Finally, here's the parametric plot:

Clear[f, x]; f[x_] := (x/2) + 1ParametricPlot[{f[x], f'[x]}, {x, -1, 2}, 
      AxesLabel -> {"x", "y"}, 
      PlotStyle -> {AbsoluteThickness[2], Brown}];

Zero slope tells us that  f[x]  and  f'[x]  are NOT proportional.

The Percent Change in Some Obvious Cases

(2^x) changes by 100% for unit change in  x:

Suppose the base  a=2  of our exponential function  f[x] = (a^x).  When

x = 0  =>  f[0] = (2^0)  = 1

x = 1  =>  f[1] = (2^1)  = 2

The change in  x  is from 0 to 1, or a change of 1.  The change in  y  is from  1 to  2, a change of  1, or percent change of (1/2)×100% = 50%.

Next,

x = 1  =>  f[1] = (2^1)  = 2

x = 2  =>  f[2] = (2^2)  = 4

The change in  x  is from 1 to 2, or a change of 1.  The change in  y  is from  2  to  4, a change of  2, or (2/4)×100% = 50%.

In general,

x   =>  f[x] = (2^x)

x+1  =>  f[x+1] = (2^(x + 1))

The change in  x  is from x to x+1, or a change of 1.  The change in  y  is from  (2^x)  to  (2^(x + 1))= (2^x)·(2^1), a percent change of  \(TraditionalForm
      \`\(\(2\^x·2\^1 - 2\^x\)\/2\^x×\)\)100% = 50%.

\(Clear[f, x, a, Δx]; 
f[x_] := a\^x; 
a = 2; 

    \(f[x + Δx] - f[x]\)\/Δx\)\(Δx = 1; 

    Simplify[\(1\/f[x]\) \(f[x + Δx] - f[x]\)\/Δx]
      \)

(4^x) changes at the rate of 200% for x-change Δx  =  (1/2):

Suppose the base  a=4  of our exponential function  f[x] = (a^x).  When

x = 0  =>  f[0] = (4^0)  = 1

x = (1/2)  =>  f[(1/2)] = (4^(1/2))  = 2

The change in  x  is from 0 to (1/2), or a change of (1/2).  The change in  y  is from  1 to  2, a change of  1, or percent change of (1/1) (1/((1/2)))×100% = 200%.

Next,

x = (1/2)  =>  f[(1/2)] = (4^(1/2))  = 2

x = 1  =>  f[1] = (4^1)  = 4

The change in  x  is from (1/2) to 1, or a change of (1/2).  The change in  y  is from  2  to  4, a change of  2, or (1/2) (2/((1/2)))×100% = 200%.

In general,

x   =>  f[x] = (4^x)

x+(1/2)  =>  f[x+(1/2)] = (4^(x + 1/2))= (4^x)·(4^(1/2))= 2·(4^x)

The change in  x  is from x to x+(1/2), or a change of (1/2).  The change in  y  is from  (4^x)  to  (4^(x + 1/2))= (4^x)·2, a percent change of  \(TraditionalForm
      \`\(1\/4\^x·\(4\^x·2 - 4\^x\)\/\(1/2
              \)×\)\)100% = 200%.

\(Clear[f, x, a, Δx]; 
f[x_] := a\^x; 
a = 4; 

    \(f[x + Δx] - f[x]\)\/Δx\)\(Δx = 1/2; 

    Simplify[\(1\/f[x]\) \(f[x + Δx] - f[x]\)\/Δx]
      \)

((((1/8)))^x) decreases at the rate of 150% for x-change Δx  =  (1/3):

Suppose the base  a=(1/8)  of our exponential function  f[x] = (a^x).  When

x = 0  =>  f[0] = ((((1/8)))^0)  = 1

x = (1/3)  =>  f[(1/3)] = ((((1/8)))^(1/3))  = (1/2)

The change in  x  is from 0 to (1/3), or a change of (1/3).  The change in  y  is from  1 to  (1/2), a change of  -(1/2), or percent decrease of ((-(1/2))/1) (1/((1/3)))×100% = -150%.

Next,

x = (1/3)  =>  f[(1/3)] = ((((1/8)))^(1/3))  = (1/2)

x = (2/3)  =>  f[(2/3)] = ((((1/8)))^(2/3))  = (1/4)

The change in  x  is from (1/3) to (2/3), or a change of (1/3).  The change in  y  is from  (1/2)  to  (1/4), a decrease of  (1/4), or ((-(1/4))/((1/2))) (1/((1/3)))×100% = -150%.

In general,

x   =>  f[x] = ((((1/8)))^x)

x + (1/3)  =>  f[x + (1/3)] = ((((1/8)))^(x + 1/3))

The change in  x  is from (1/3) to (2/3), or a change of (1/3).  The change in  y  is  is from  ((((1/8)))^x)  to  ((((1/8)))^(x + 1/3))= ((((1/8)))^x)·(1/2), a percent change of  \(TraditionalForm
      \`\(1\/\((1\/8)\)\^x·\(\(\((1\/8)\)\^x\) 1\/2 - 
              \((1\/8)\)\^x\)\/\(1\/3\)×\)\)100% = -(3/2)×100% = -150%

\(Clear[f, x, a, Δx]; 
f[x_] := a\^x; 
a = 1\/8; 

    \(f[x + Δx] - f[x]\)\/Δx\)\(Δx = 1/3; 

    Simplify[\(1\/f[x]\) \(f[x + Δx] - f[x]\)\/Δx]
      \)

Finding the Base  to Give a Percent Rate of Change

Suppose we want an exponential function that changes at the rate of  "3%  per quarter."  That is, the rate of change of  f[x]  will be  3/100  when  Δx = 1/4.  We know in general that if  f[x] = (a^x), the percentage rate is given by

(((a^Δ)x - 1)/Δ)x

so we simply solve the equation:

\(Clear[a, Δx]; 
Δx = 1\/4; 

    Solve[\(a\^Δx - 1\)\/Δx == 3\/100, a]\)N[%]

Percent Rate of Change for (e^(r*x))

Suppose the base  a=(e^r)  of our exponential function  f[x] = (a^x).  When

x  =>  f[x] = (e^(r*x))

x +Δx  =>  f[x +Δx] = (e^(r*(x + Δx)))  = (e^(r*x)) (e^(r*Δx))

The change in  x  is Δx.  The change in  y  is from  (e^(r*x)) to (e^(r*x)) 
        (e^(r*Δx)), a  percent change of  (1/(e^(r*x))) 
        (((e^(r*x)) (e^(r*Δx)) - 
            (e^(r*x)))/( Δx))×100% = (((e^(r Δx)) - 1)/( Δx
        ))×100%.

\(Clear[f, x, a, Δx, r]; 
f[x_] := a\^x; 

    a = E\^r; 

    \(f[x + Δx] - f[x]\)\/Δx\)Simplify[
      (1/f)[x] ((f[x + Δx] - f[x])/Δ)x]

Suppose we want to control the constant fractional change:

k = (((e^(r Δx)) - 1)/( Δx
        ))

What value of  r  makes  k = 1, for example?  This depends on  Δx, of course, but we can solve:

\(Clear[k, r, Δx]; 

    Solve[k\  == \ 
        \(E\^\(r\ Δx\) - 1\)\/\(\ Δx
          \), r]\)

Simplifying this result, we get

r = Log[1 + k Δx]/Δx

Comparison of  Log[1 + k Δx]/Δx  to Known Changes

For base  a = 2  and  Δx = 1, we know that  k = 1.  To compare, we need to make   2  =  a  = (e^r), or  Log[2] = Log[(e^r)] = r:

\(Clear[k, Δx]; 
Δx\  = \ 1; 
k\  = \ 1; 

    Log[1 + k\ Δx]\/Δx\)Log[2]

Finding  r  in (e^(r*x))  for a Given Percentage

Suppose we want an exponential function that changes at the rate of  "3%  per quarter."  That is, the rate of change of  f[x]  will be  3/100  when  Δx = 1/4.  We know in general that if  f[x] = (e^(r*x)), the percentage rate is given by

(((((e^(r*)))^Δ)x - 1
          )/Δ)x

so we simply solve the equation:

\(Clear[r, Δx]; 
Δx = 1\/4; 

    Solve[\(E\^\(r*Δx\) - 1
            \)\/Δx == 3\/100, r]\)N[%]

When we solved this problem for  a  in  f[x] = (a^x), we got  a = 1.03034.  Let's compare:

\(r\  = \ 0.029888059354804227; 
E\^r\)

Expressing Constant Percentage Change as(((1 + k Δx))^(x/Δx))

Suppose we know we have a function  y = f[x]  that changes at the PERCENTAGE RATE of  k×100%  for each  x-change of  Δx  First, we show that this forces  f[x]  to satisfy

f[x] = 
        f[0] (((1 + k Δx))^(x/Δx))

so if  f[0] = 1,

f[x] = (((1 + k Δx))^(x/Δx))

This formula allows us to make the comparisons

(a^x) = (((1 + k Δx))^(x/Δx))

and

(e^(r x)) = 
        (((1 + k Δx))^(x/Δx))

in the cases where we know the relation between  a  and  k  or  r  and  k.

Proof that Percentage Rate of  k×100% Implies
	f[x] = 
        f[0] (((1 + k Δx))^(x/Δx))

Since  y  changes at the percentage rate of  k×100%  as  x changes by  Δx, we have

(1/y)·((y - change)/(x - change)) =  k

(1/y)·((y - change)/Δ)x =  k

y-change = y k Δx

y-change = (Subscript[y, new]) - (Subscript[y, old]) = (Subscript[y, old]) k Δx

(Subscript[y, new]) = (Subscript[y, old]) (1 + k Δx)

or

f[Δx] = f[0] · (1+k Δx)

Also suppose that  f[0] = 1, as is the case when  f[x] = (a^x)or  f[x] = (e^(r x)), so

f[Δx] = (1+k Δx)

When  y = f[x]  changes from  Δx  to 2Δx, we still have percentage rate of growth  k×100%, so

(Subscript[y, new]) = (Subscript[y, old]) (1 + k Δx)

or

f[Δx+Δx] = f[Δx] · (1+k Δx) = (1+k Δx)·(1+k Δx)

f[2Δx](( = (1 + k Δx))^2)

Similarly, when  x  changes from  2Δx  to 3Δx, we still have percentage rate of growth  k×100%, so

(Subscript[y, new]) = (Subscript[y, old]) (1 + k Δx)

or

f[Δx+2Δx] = f[2Δx] · (1+k Δx) = (((1 + k Δx))^2)·(1+k Δx)

f[3Δx](( = (1 + k Δx))^3)

an in general,

f[n Δx](( = (1 + k Δx))^n)

If we take  x = n Δx  so  n = x/Δx, we can write this as

f[x](( = (1 + k Δx))^(x/Δx))

In general,  f[x] = f[0] (((1 + k Δx))^(x/Δx)), for  x = Δx, 2Δx, 3Δx, ⋯

The (Underscript[Limit, Δx -> 0])[(1/f)[x]·((f[x + Δx] - 
            f[x])/Δ)x]  when  f[x] = (e^(r*x))

Mathematica can compute the limit

\(Clear[f, x, Δx]; 
f[x_]\  := \ E\^\(r*x\); 
    
Limit[\(f[x + Δx] - f[x]\)\/Δx, 
      Δx -> 0]\)Limit[(1/f)[x]*((f[x + Δx] - f[x])/Δ)x, 
      Δx -> 0]

We also know from above that the percentage rate of change  k   for  (e^(r*x))  in an x-change of Δx is related to  r  by the equation:

r = Log[1 + k Δx]/Δx

As  Δx  tends to zero, the Log[.] expression tends to  k:

\(Clear[k, Δx]; 

    Limit[Log[1 + k*Δx]\/Δx, 
      Δx -> 0]\)

In other words, if we know (the fact) that

(Underscript[Limit, Δx -> 0])[
          Log[1 + k*Δx]/Δx] =  k

then the instantaneous percentage rate of change of  y = (e^(r*x)) is

(1/y) (dy/dx)= r

(Underscript[Limit, Δx -> 0])[(1/f)[x]·((f[x + Δx] - 
            f[x])/Δ)x] =  Log[a] , a constant,
when  f[x] = (a^x).  ((Subscript[k, a]) = Log[a])

Mathematica can compute the limit

\(Clear[f, x, a, Δx]; 
f[x_]\  := \ a\^x; 

    Limit[\(f[x + Δx] - f[x]\)\/Δx, 
      Δx -> 0]\)Limit[(1/f)[x]*((f[x + Δx] - f[x])/Δ)x, 
      Δx -> 0]

But we can also write  (a^x)  in the form  (e^(r*x)), solving for  r  by:

a = (e^r)
	Log[a] = Log[(e^r)] = r

The law of repeated exponents shows that if  a = (e^r), then

(a^x) = ((((e^r)))^x) = (e^(r*x))

Since we know from the previous section that the instantaneous percentage rate of change of  y = (e^(r*x)) is

(1/y) (dy/dx)= r

we have y  =  (a^x) = (e^(r*x))  (with  r = Log[a])  implies that

(1/y) (dy/dx)= Log[a]

Exercises

The Exponential Project

Percentage Change of  (3^x)  for x-change  Δx  = 1

Let  f[x] = (3^x)  and compute  f[0], f[1], f[2].

Show that as  x  goes from  0  to  1,  f[x]  changes by  200% of its starting value.

Show that as  x  goes from  1  to  2,  f[x]  changes by  200%  of its starting value.

Show that for any  x  the rate of change in  f[x]  when  x  increases by  1  is  200%  of its starting value.

Find  a  to give a certain percentage rate for (a^x) when  Δx = 1

Find the base  a  so that the percentage rate of change of  (a^x)  is  50%  whenever  x  changes by  1 unit.

Find  b  to give a certain percentage rate for (b^x) when  Δx = 1/2

Find the base  b  so that the percentage rate of change of  (b^x)  is  25%  whenever  x  changes by  1/2 unit.

Comparison of  a  and  b

If   (a^x)  changes at the rate of  50% in  1  unit and  (b^x)  changes at the rate of 25%  in 1/2  unit, is  a = b?  (Look at the previous two solutions and also change  (b^x)  twice to explain why not.)

1.25 *1.25

Find  r  to give a certain percentage rate for  (e^(r*x))  when  Δx = 1/100

Find the number r  so that the percentage rate of change of  (e^(r*x))  is  25%  whenever  x  changes by  1/100 unit.

Compare your solution to the value of

Log[1 + k Δx]/Δx  for  k = 0.25  and  Δx = 1/100

How much is the difference between  r  and  0.25?

Find  r  to give a certain percentage rate for  (e^(r*x))  when  Δx = 1/1000

Find the number r  so that the percentage rate of change of  (e^(r*x))  is  25%  whenever  x  changes by  1/1000 unit.

Compare your solution to the value of

Log[1 + k Δx]/Δx  for  k = 0.25  and  Δx = 1/1000

How much is the difference between  r  and  0.25?

The Numerical Approximations  a = 2, 3; 2.5,2.75; 2.7,⋯

We know that the derivative

((d ((a^x)))/dx)  =  
      (lim_(Δx -> 0))(((a^(x + Δx)) - 
              (a^x))/Δ)x =  
        (a^x)*(lim
              _(Δx -> 0))(((a^Δ)x - 1
                )/Δ)x

This means that the constant  (Subscript[r, a]) in the proportionality

(dy/dx) = (Subscript[r, a])·y

satisfies

(Subscript[r, a])  =  
        (lim_(Δx -> 0))(((a^Δ)x - 1
              )/Δ)x

This part of the project asks you to estimate (Subscript[r, a])  numerically for various choices of  a.  Here is a start:

\(Clear[a, Δx]; 
a\  = \ 2; 
Print["a = ", a]; 

    Δx\  = \ 2; 

    Do[Δx\  = \ Δx/2; 
	
      \(Print[\*""(Subscript[r, a]) ≈ (((a^Δ)x - 1)/Δ)x= "", 
        N[\(a\^Δx - 1\)\/Δx], 
        " for Δx = ", N[Δx]]; \)

      	, {n, 1, 6}]; 

    Print["Log[", a, "] ≈ ", \ N[Log[a]]]\)

\(Clear[a, Δx]; 
a\  = \ 3; 
Print["a = ", a]; 

    Δx\  = \ 2; 

    Do[Δx\  = \ Δx/2; 
	
      \(Print[\*""(Subscript[r, a]) ≈ (((a^Δ)x - 1)/Δ)x= "", 
        N[\(a\^Δx - 1\)\/Δx], 
        " for Δx = ", N[Δx]]; \)

      	, {n, 1, 6}]; 

    Print["Log[", a, "] ≈ ", \ N[Log[a]]]\)

First, make the numerical calculation more accurate by computing to smaller values of  Δx.  Notice the numbers converging to the natural Log?

Once you have accurate numerical data for the cases  a = 2  and  a = 3, notice that  (Subscript[r, 2]) < 1  and  (Subscript[r, 2]) > 1.  Use your numerical program to approximate  (Subscript[r, a]) for the nbumbers:

a = 2.5

a = 2.75

Continue using  a = the average of the last  a  with  (Subscript[r, a]) < 1 and the last  a  with  (Subscript[r, a]) > 1 until you have approximated the base  a  which makes  (Subscript[r, a])  ≈  1  to the accuracy of your computations.

Compare your answer to:

N[E, 40]

Plots of Instantaneous Percentage Rates

For each of the functions below plot  f[x], the instantaneous rate of change of  f[x]  and the percentage instantaneous rate of change of  f[x]:

y = f[x] = (2^x)

y = f[x] = (3^x)

y = f[x] = (e^x)

y = f[x] = ((((1/2)))^x)

y = f[x] = 2 x  -  1

y = f[x] = Cos[x]

y = f[x] = Log[x]

Exercise CD5.4.5

Let  f[x] = (e^(r x)) and solve the specific equation

(1/f)[x] ((f[x + Δx] - f[x])/Δ)x = k

for  r, showing that

r = (1/Δ)x Log[1 + k Δx]

This problem asks you to show that  k ≈ r, when  Δx≈0, by using a property of the natural logarithm.

Step 1: The Increment Approximation: Log[ 1 + δx] = δx + ϵ·δx

Assume that it is given that

((d (Log[x]))/dx) =  (1/x)

and write the specific increment approximation at  x = 1,

F[x+δx] - F[x] = F'[x]·δx + ϵ·δx

for the function  F[x] = Log[x].  This gives

Log[1+δx] = δx + ϵ·δx

with  ϵ≈0  when  δx ≈ 0.

Step 2: Compute (1/Δ)x Log[1 + k Δx]

Use the result of part 1, Log[ 1 + δx] = δx + ϵ·δx, with  ϵ≈0, when δx≈0, to show

(1/Δ)x Log[1 + k Δx]≈ k

when  Δx ≈ 0.  Conclude that if   f[x] = (e^(r x)) changes at the percentage fraction  k  in x-change  Δx ≈ 0, then

r ≈ k

Step 3:  y  =  (e^(r x))  =>  (dy/dx)  =  r*(e^(r x))

Conclude that if  f[x] = (e^(r x)), then

(1/f)[x] f'[x]  =  r

(HINT:  f'[x] ≈ \(\(?? \/?\)\),  when  Δx≈0.)