Exponential and Logarithmic functions




Exponential functions

Definition

Take a > 0 and not equal to 1 . Then, the function defined by
 
f : R -> R : x -> ax
is called an exponential function with base a.

Graph and properties

Let f(x) = an exponential function with a > 1.
Let g(x) = an exponential function with 0 < a < 1.

From the graphs we see that

Examples:
 
y = 3x  ; y = 0.5x ; y = 100.2x-1

Logarithmic functions

Definition and basic properties

Take a > 0 and not equal to 1 . Since the exponential functions
 

f : R -> R : x -> ax
are either increasing or decreasing, the inverse functions are defined. The inverse function is called the logarithmic function with base a. We write
 
        loga(x)

        log10(x) is written as log(x)
So,

loga(x) = y <=> ay = x


From this we see that the domain of the logarithmic function is the set of strictly positive real numbers, and the range is R.
Examples:

 
         log2(8) = 3 ;  log3(sqrt(3)) = 0.5 ; log(0.01) = -2
From the definition it follows immediately that

 
         for x > 0  we have  aloga(x) = x

and
         for all x  we have  loga(ax) = x

Example: log(102x+1) = 2x+1

Graph

Let f(x) = a logarithmic function with a > 1.
Let g(x) = a logarithmic function with 0 < a < 1.

From the graphs we see that

Examples:
 
log2(x) ; log(2x+4) ; log0.5(x)

Properties

In the next 3 properties, all logarithmic functions have base a > 0. For convenience, I don't write this base a.

Change the base of a logarithmic function

Sometimes it is very useful to change the base of a logarithmic function.
Theorem: for each strictly positive real number a and b, different from 1, we have

 
                       1
         loga(x) =( -------) .  logb(x)
                    logb(a)

Proof:
We'll prove that

 

 logb(a) . loga(x) =  logb(x)

Let  logb(a) = u  then bu  = a        (1)

Let  loga(x) = v  then av  = x        (2)

Let  logb(x) = w  then bw  = x        (3)

From (2) and (3) we have

        av  = bw

Using (1)

         bu.v = bw
So,
        u.v  = w


=>   logb(a) . loga(x) =  logb(x)

Calculating images of a logarithmic function with a calculator

Examples:
 
   log(12.5) = 1.0969

   log2(12) = log(12)/log(2) = 3.58496

   log(1/154)= -log(154) = -2.1875

   log7(0.514) = 14 log(0.5)/log(7) = -4.9869

   log(-12.4) is not defined

Logarithmic scale

The number e

A special limit concerning the derivative of an exponential function

We try to calculate the derivative of the exponential function
 
        f(x) = ax
Appealing on the definition of the derivative, we can write
 
                     (f(x+h)-f(x))
        f'(x) = lim ---------------
               h->0        h

                     ax+h - ax
              = lim ------------
                h->0     h

                     ax (ah  - 1)
              = lim -----------
                h->0     h

                        (since ax  is constant relative to h )

                          (ah  - 1)
              = ax  . lim -----------
                     h->0     h

Now,
             (ah  - 1)
        lim ----------- is a constant depending on the value of the base a.
        h->0     h
It can be proved that there is a unique value of a, such that this limit is 1. This very special value of a is called e.
So,
 
             (eh  - 1)
        lim ----------- = 1
        h->0     h

The number e as a limit

The expression
 
             (eh  - 1)
        lim ----------- = 1
         0       h
means that for very very small values of h
 
        eh  - 1  is approximately  h

<=>     eh   is approximately  h +1

<=>     e is approximately (1 + h)1/h
So,

 
        e = lim (1 + h)1/h   = 2.718 28...
             0

Or, if we say that t = 1/h

 
        e = lim (1 + 1/t)t  = 2.718 28...
           infty

The function ex

The number e plays a very important role in mathematics. We'll show the importance of the exponential function ex in the following sections

Definition of ln(x)

The logarithmic function with base number e is noted as ln(x). So,

loge(x) = ln(x)


Transformation of an exponential function with base a, to an exponential function with base e

In applied mathematics, the exponential functions with a base different from e are underused.
This is because an exponential function with base a can be converted to an exponential function with base e.
We now show the procedure.

The number a is a strictly positive number.

 
  ak  = er  <=> ln(ak) = r <=> r = k.ln(a)

Then

 (ak)x  = (er)x  <=>   r = k.ln(a)  for all x

<=>

  akx  = erx  <=>   r = k.ln(a)   for all x
The functions A.akx and A.erx are identical if and only if r = k.ln(a)

Example:

 
    14 . 32.7x  is identical with  14 . e2.97x

    6 . (0.25)-x  is  identical with  6 . e1.39x
Working with the functions with base e, has many benefits for algebraic calculations.

The product of the functions with base e, from the previous example, is much easier to work with.

Compare the expressions: 14 e2.97 x 6 e1.39 x and 14 32.7x 6 (0.25)-x.

Half Value of an exponential function

Take the function e-rx with r > 0. It is a decreasing function.
We specify, for x, a fixed initial value a. Then the image is y1 = e-ra.
For higher x values, we find smaller images.
We let x go from a to a+h. Let y2 = e-r(a+h).

We'll find the value of h, such that y2 = y1/2.

 
      y2 = (1/2) y1

<=>   e-r(a+h) = (1/2)  e-ra

<=>   e-ra  e-rh = (1/2)  e-ra

<=>          e-rh = (1/2)

<=>      - rh = ln(1/2)

<=>      - rh = ln(1) -ln(2)

<=>       - rh = -ln(2)

<=>        h = ln(2)/r
If we increase x with ln(2)/r, then the image of e-rx reduces to half the initial value. Remarkable is that ln(2)/r does not depend on our initial value a.

This means that the image is halved, when a random initial value of x increases with ln(2)/r.

That's why ln(2)/r is called the 'half value' of the function e-rx.

Example : Take the function y = e-0.5x.

The half value is ln(2)/0.5 = 1.386

Exercise: plot the graph and check all previous properties.

Differentiation of logarithmic functions

Derivative of a logarithmic function

In this section, all logarithmic functions have base a. For convenience, I don't write this base number.
 
Let f(x) = log(x) , then

                     (f(x+h)-f(x))
        f'(x) = lim ---------------
               h->0        h

                     (log(x+h)-log(x))
<=>     f'(x) = lim -------------------
               h->0        h

                     log( (x+h)/x )
<=>     f'(x) = lim -------------------
               h->0        h

                     1
<=>     f'(x) = lim --- . log( (x+h)/x )
               h->0  h

<=>     f'(x) = lim  log( (x+h)/x )1/h
               h->0

<=>     f'(x) = lim  log( (x+h)/x )1/h
               h->0

<=>     f'(x) = lim  log(1 + h/x)1/h
               h->0

<=>     f'(x) = lim  log((1 + h/x)x/h )1/x
               h->0

<=>     f'(x) = lim  (1/x).log(1 + h/x)x/h
               h->0

<=>     f'(x) =(1/x). lim  log(1 + h/x)x/h
                      h->0

<=>     f'(x) =(1/x). lim  log(1 + h/x)x/h
                     h/x->0


<=>     f'(x) =(1/x).log  lim  (1 + h/x)x/h
                         h/x->0


<=>     f'(x) =(1/x).log(e)


<=>     f'(x) =(1/x).ln(e)/ln(a)

<=>     f'(x) =(1/x)/ln(a)

                   1
<=>     f'(x) = ----------
                 x. ln(a)

Important Formulas

Let u be a differentiable function of x.

 
        d               1
        --  loga(x) = ----------
        dx            x. ln(a)

        d               1
        --  loga(u) = ---------- . u'
        dx            u. ln(a)

        d           1
        -- ln(x) = ---
        dx          x

        d           1
        -- ln(u) = ---.u'
        dx          u

Calculating derivatives of logarithmic functions; examples

 
  y = ln(2x2+6)

  y' = (1/(2x2+6)). 4x
---------------------------

  y = ln2(x)

  y' = 2 ln(x) . (1/x)
---------------------------

  y = ln(1/x)

  y'= x.(-1/x2) = -1/x
---------------------------

  y = ln(ln(x))

  y'= (1/ln(x)) . (1/x)
---------------------------

  y = x3 ln(x)

  y'= (3x2).ln(x) + (x3).(1/x) = (3x2).ln(x) + x2
---------------------------

  y = log3(x2)

  y'= (1/(x2.ln(3)).(2x) = 2/(x ln(3))
---------------------------

  y = ln(x)/x


       x.(1/x) - ln(x)       1 - ln(x)
  y'= ------------------ = ------------
           x2                x2
---------------------------

  y = ln(sqrt(2x2+x))

  first we rewrite y

  y = 0.5 ln(2x2+x)

            4x+1
  y'= 0.5 ----------
           2x2+x
---------------------------

  y = sqrt(ln(x2))

         1
  y'= --------------- .(2/x)
      2 sqrt(ln(x2))

Derivative of an exponential function

 
Let f(x) = ax, then   loga(ax ) and x are identical functions.
Hence, the derivative of both functions is the same.
So,

           1
        ---------- .(ax )' = 1
        ax .ln(a)


        d
<=>    ---(ax ) = ax .ln(a)
        dx

Important formulas

Let u be a differentiable function of x.

 
        d
       ---(ax ) = ax .ln(a)
        dx

        d
        --(ex ) = ex
        dx

        d
        --(au ) = au .ln(a).u'
        dx

        d
        --(eu ) = eu .u'
        dx

Derivatives of exponential functions; examples

 
  y = x.e-5x

  y'= e-5x + x.(-5)e-5x
---------------------------

  y = 32x

  y'=  32x ln(3) 2 =  32x ln(9)
---------------------------

  y = ex/x2

       x2 ex - ex 2x
  y'= ------------------
           x4
       ex (x-2)
    = ------------
          x3
---------------------------

  y = (3e)x

  y'=  (3e)x ln(3e) =  (3e)x (ln3 +lne) = (3e)x (ln3 + 1)

---------------------------
  y = arcsin(2x)

        2x ln(2)
  y'= -----------------
       sqrt(1-22x)

Derivative of a real power of x

 
Let f(x) = xr with r any real number.

   xr = er.ln(x)
=>
   d
   --(xr) = er.ln(x).(r.ln(x))'
   dx

           = xr.r.(1/x)

           = r.xr-1
Thus,
For any real number r, we have
 
   d
   --(ur) = r.ur-1.u'
   dx

Derivative of uv

Let u = f(x) and v = g(x), then
 
   uv = ev.ln(u)

   d
   --(uv) = ev.ln(u).(v.ln(u))'
   dx

           = uv . (v' ln(u) + v.(1/u).u'

           = v uv-1 u' + uv.ln(u).v'
 
   d
   --(uv) = v uv-1 u' + uv.ln(u).v'
   dx

Derivatives of uv ; examples

 
  y = xx

  y'= xx ln(x) + x xx-1 = xx (1 + ln(x))

  y = (ln(x))2x

  y'= (ln(x))2x ln(ln(x)).2 + 2x.(ln(x))2x-1 (1/x)

  y = (ex)x

  we rewrite y:  y = ex2

  y'=  ex2 2x

Limits and de l'Hospital

We give some examples of the use of de l'Hospital's rule

The limit of the function uv, with u and v functions of x, can also be calculated via a detour. We first calculate the limit of ln(uv). If this is done, we apply the reverse process to the outcome.

Solved problems

 
You can find Solved Problems about exponential and logarithmic functions using this link
 





Topics and Problems

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