Exponential and Logarithmic functions

Exponential functions
- Definition
- Graph and properties
Logarithmic functions
The number e
Differentiation of logarithmic functions
Derivative of an exponential function
- Important formulas
- Derivatives of exponential functions; examples
Derivative of a real power of x
Derivative of u^v
- Derivatives of u^v ; examples
Limits and de l'Hospital
Solved problems

Exponential functions

Definition

Take a > 0 and not equal to 1 . Then, the function defined by

 
f : R -> R : x -> a^x

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

The domain is R
The range is the set of strictly positive real numbers
The function is continuous in its domain
The function is increasing if a > 1 and decreasing if 0 < a < 1
The x-axis is a horizontal asymptote

Examples:

 
y = 3^x  ; y = 0.5^x ; y = 10^0.2x-1

Logarithmic functions

Definition and basic properties

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

 

f : R -> R : x -> a^x

are either increasing or decreasing, the inverse functions are defined. The inverse function is called the logarithmic function with base a. We write

 
        log_a(x)

        log₁₀(x) is written as log(x)

So,

log_a(x) = y <=> a^y = 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 a^log_a(x) = x and for all x we have log_a(a^x) = x

Example: log(10^2x+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

The range is R
The domain is the set of strictly positive real numbers
The function is continuous in its domain
The function is increasing if a > 1 and decreasing if 0 < a < 1
The y-axis is a vertical asymptote

Examples:

 
log₂(x) ; log(2x+4) ; log_0.5(x)

Properties

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

log(x.y) = log(x) + log(y)
Proof :

 
Let     log(x.y) = u then  a^u  = x.y   (1)

Let     log(x) = v  then   a^v  = x     (2)

Let     log(y) = w  then   a^w  = y     (3)

From (1) , (2) and (3)

        a^u  = a^v  . a^w

=>     a^u  = a^{v + w}

=>      u = v + w

So,

log_a(x.y) = log_a(x) + log_a(y)

In the same way you can prove that

log_a(x/y) = log_a(x) - log_a(y)

For each real number r we have :

log_a(x^r ) = r.log_a(x)

Examples:

 
  log(x² y³) = 2 log(x) + 3 log(y)

  log(x² / y³) = 2 log(x) - 3 log(y)

  log( x^y )= y log(x)

       1
  log ------ = log(1) - log((xy)³) = -3log(xy) = -3 (log(x)+log(y))
      (xy)³

  log(2x) + log(3x) = log(6x²)

  3log(x) + a log(x) = (3+a) log(x) = log(x^3+a)

  0.5 log(x) = log(sqrt(x))

Show that x^log(y) / y^log(x) = 1

The statement implies that x and y are positive, so we can write

 
   x^log(y) / y^log(x) = 1
<=>
   x^log(y) = y^log(x)
<=>
   log( x^log(y) ) = log( y^log(x) )
<=>
   log(y).log(x) = log(x).log(y)

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 log_a(x) =( -------) . log_b(x) log_b(a)

Proof:
We'll prove that

 

 log_b(a) . log_a(x) =  log_b(x)

Let  log_b(a) = u  then b^u  = a        (1)

Let  log_a(x) = v  then a^v  = x        (2)

Let  log_b(x) = w  then b^w  = x        (3)

From (2) and (3) we have

        a^v  = b^w

Using (1)

         b^u.v = b^w
So,
        u.v  = w


=>   log_b(a) . log_a(x) =  log_b(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

Take a function with an equation of the form y = 10^ax+b . If you plot this function with a logarithmic scale on the y-axis, then the graph is a straight line with slope a.
Take a function with an equation of the form y = log(a.x^p) . If you plot this function with a logarithmic scale on the x-axis, then the graph is a straight line with slope p.
Take a function with an equation of the form y = a.x^p. If you plot this function with a logarithmic scale on the x-axis and on the y-axis, then the graph is a straight line with slope p.

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) = a^x

Appealing on the definition of the derivative, we can write

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

                     a^x+h - a^x
              = lim ------------
                h->0     h

                     a^x (a^h  - 1)
              = lim -----------
                h->0     h

                        (since a^x  is constant relative to h )

                          (a^h  - 1)
              = a^x  . lim -----------
                     h->0     h

Now,
             (a^h  - 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,

 
             (e^h  - 1)
        lim ----------- = 1
        h->0     h

The number e as a limit

The expression

 
             (e^h  - 1)
        lim ----------- = 1
         0       h

means that for very very small values of h

 
        e^h  - 1  is approximately  h

<=>     e^h   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 e^x

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

Definition of ln(x)

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

log_e(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.

 
  a^k  = e^r  <=> ln(a^k) = r <=> r = k.ln(a)

Then

 (a^k)^x  = (e^r)^x  <=>   r = k.ln(a)  for all x

<=>

  a^kx  = e^rx  <=>   r = k.ln(a)   for all x

The functions A.a^kx and A.e^rx are identical if and only if r = k.ln(a)

Example:

 
    14 . 3^2.7x  is identical with  14 . e^2.97x

    6 . (0.25)^-x  is  identical with  6 . e^1.39x

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 e^{2.97 x} 6 e^{1.39 x} and 14 3^2.7x 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 -- log_a(x) = ---------- dx x. ln(a) d 1 -- log_a(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) = a^x, then   log_a(a^x ) and x are identical functions.
Hence, the derivative of both functions is the same.
So,

           1
        ---------- .(a^x )' = 1
        a^x .ln(a)


        d
<=>    ---(a^x ) = a^x .ln(a)
        dx

Important formulas

Let u be a differentiable function of x.

d ---(a^x ) = a^x .ln(a) dx d --(e^x ) = e^x dx d --(a^u ) = a^u .ln(a).u' dx d --(e^u ) = e^u .u' dx

Derivatives of exponential functions; examples

 
  y = x.e^-5x

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

  y = 3^2x

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

  y = e^x/x²

       x² e^x - e^x 2x
  y'= ------------------
           x⁴
       e^x (x-2)
    = ------------
          x³
---------------------------

  y = (3e)^x

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

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

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

Derivative of a real power of x

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

   x^r = e^r.ln(x)
=>
   d
   --(x^r) = e^r.ln(x).(r.ln(x))'
   dx

           = x^r.r.(1/x)

           = r.x^r-1

Thus,

For any real number r, we have

 
   d
   --(u^r) = r.u^r-1.u'
   dx

Derivative of u^v

Let u = f(x) and v = g(x), then

 
   u^v = e^v.ln(u)

   d
   --(u^v) = e^v.ln(u).(v.ln(u))'
   dx

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

           = v u^v-1 u' + u^v.ln(u).v'

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

Derivatives of u^v ; examples

 
  y = x^x

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

  y = (ln(x))^2x

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

  y = (e^x)^x

  we rewrite y:  y = e^x²

  y'=  e^x² 2x

Limits and de l'Hospital

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

 
             ln(x)           1/x
        lim ------- =  lim ------- = 0
       infty   x      infty   1

 
              xⁿ         n.x^n-1
        lim ------ = lim ---------
       infty  e^x           e^x

             n.(n-1)x^n-2
      = lim --------------
               e^x

      = ...

              n!
      = lim ------  = 0
              e^x

 
                            ln(x)        1/x
        lim xln(x)   = lim ------ = lim ------ = - lim x = 0
         0                   1/x        -1/x²

 
        lim x^x = lim e^x.ln(x) = e^{lim x.ln(x)}
         0        0

        and from previous example

        = e⁰ = 1

 
        lim (cos(x))^1/x
         0

        = lim e^{(1/x).ln(cos(x))}
           0

        = e^{lim (1/x).ln(cos(x))}


        but  lim (1/x).ln(cos(x))
             0
                 ln(cos(x))
        = lim ---------------
           0        x
               -sin(x)/cos(x)
        = lim ---------------- = 0
           0       1

So,      lim (cos(x))^1/x = e⁰ = 1
         0

 
      lim ( x ln(e + 1/x) -x )
    infty


    = lim  x ( ln(e + 1/x) - 1)


            ln(e + 1/x) - 1
    = lim  -----------------
                 1/x

              -1/x²
    = lim -------------------
          (e + 1/x) . (-1/x²)

    = 1/e

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

We calculate the following limit for x --> + infinity

 
      lim  x^1/x

First we calculate

 
      lim ln( x^1/x ) = lim (1/x) ln(x)  = lim  ln(x)/x    ( case infinity / infinity)

           de l'Hospital

     = lim 1/x = 0

The initial limit is e⁰ = 1

We calculate the following right limit for x --> 0

 
 lim ( tan(x) )^x

First we calculate

 

  lim ln ( tan(x) )^x = lim x.ln(tan(x))   ( case 0. infinity )

         ln (tan(x))
  = lim --------------    ( case infinity / infinity)
             1/x

           de l'Hospital

              - x²
  = lim --------------------
         tan(x) cos²(x)

             x²
  = - lim ----------
           tan(x)

             x
  = - lim --------- . x  = 1.0 = 0
            tan(x)

The initial limit is e⁰ = 1

We calculate the following right limit for x --> 0

 
  lim sin(x)^sin(x)

First we calculate

 
  lim ln sin(x)^sin(x)


= lim sin(x) ln(sin(x))    ( case 0.infinity )

          ln(sin(x)
= lim ----------------       ( case infinity / infinity)
        ( 1/ sin(x))


           de l'Hospital


       ( 1/ sin(x)).cos(x)
= lim -------------------------
       (-1/ sin²(x)) . cos(x)


       ( 1/ sin(x))
= lim -----------------
       (-1/ sin²(x))


= - lim sin(x)  = 0

The initial limit is e⁰ = 1

Solved problems

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

Topics and Problems

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