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

Logarithmic functions

Definition and basic properties

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

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

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

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

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

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

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)

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 with respect 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

Definition of ln(x)

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

loge(x) = ln(x)


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 cases

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

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 corollaries

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

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

You can find solved problems about exponential and logarithmic functions here

Topics and Problems

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