Graphs of Logarithmic Functions

Graphing and sketching logarithmic functions: a step by step tutorial. The properties such as domain, range, vertical asymptotes and intercepts of the graphs of these functions are also examined in details. Free graph paper is available.

Review

We first start with the properties of the graph of the basic logarithmic function of base a,

f (x) = log a (x) , a > 0 and a not equal to 1.

The domain of function f is the interval (0 , +inf). The range of f is the interval (-inf , +inf).

The symbol inf means infinity.

Function f has a vertical asymptote given by x = 0. This function has an x intercept at (1 , 0). f increases as x increases.

You may want to review all the above properties of the logarithmic function interactively .


Example 1: f is a function given by

f (x) = log2(x + 2)

  1. Find the domain of f and range of f.
  2. Find the vertical asymptote of the graph of f.
  3. Find the x and y intercepts of the graph of f if there are any.
  4. Sketch the graph of f.

Answer to Example 1

a - The domain of f is the set of all x values such that 

x + 2 > 0

or x > -2

The range of f is the interval (-inf , +inf).

b - The vertical asymptote is obtained by solving

 x + 2 = 0

which gives

 x = -2

As x approaches -2 from the right (x > -2) , f(x) decreases without bound. How do we know this?

Let us take some values:

f(-1) = log2 (-1 + 2) = log2 (1) = 0

f(-1.5) = log2 (-1.5 + 2) = log2 (1/2) = -1

f(-1.99) = log2 (-1.99 + 2) = log2 (0.01) which is approximately equal to  -6.64

f(-1.999999) = log2 (-1.999999 + 2) = log2 (0.000001) which is approximately equal to  -19.93.

c - To find the x intercept we need to solve the equation f(x) = 0

log2 (x + 2) = 0

Use properties of logarithmic and exponential functions to write the above equation as

2log2 (x + 2) = 20

Then simplify

 x + 2 = 1

x = -1

The x intercept is at (-1 , 0).

The y intercept is given by (0 , f(0)) = (0,log2 (0 + 2)) = (0 , 1).

d - So far we have the domain, range, x and y intercepts and the vertical asymptote. We need more points. Let us consider a point at x = -3/2 (half way between the x intercept and the vertical asymptote) and another point at x = 2.

f(-3/2) = log2 (-3/2 + 2) = log2 (1/2) = log2 (2-1) = -1.

f(2) = log2 (2 + 2) = log2 (22) = 2.

We now have more information on how to graph f. The graph increases as x increases. Close to the vertical asymptote x = -2, the graph of f decreases without bound as x approaches -2 from the right. The graph never cuts the vertical asymptote. We now join the different points by a smooth curve.


Matched Problem to Example 1: f is a function given by

f (x) = log2 (x + 3)

 

  1. Find the domain of f and range of f.
  2. Find the vertical asymptote of the graph of f.
  3. Find the x and y intercepts of the graph of f if there are any.
  4. Sketch the graph of f.


Example 2: f is a function given by

f (x) = -3ln(x - 4)

 

  1. Find the domain of f and range of f.
  2. Find the vertical asymptote of the graph of f.
  3. Find the x and y intercepts of the graph of f if there are any.
  4. Sketch the graph of f.

Answer to Example 2

a - The domain of f is the set of all x values such that 

x - 4 > 0

or x > 4

The range of f is the interval (-inf , +inf).

b - The vertical asymptote is obtained by solving

 x - 4 = 0

or x = 4

As x approaches 4 from the right (x > 4) , f(x) increases without bound. How do we know this?

Let us take some values:

f(5) = ln(5-4) = -3ln(1) = 0

f(4.001) = -3ln(0.001) which is approximately equal to  20.72.

f(4.000001) = -3ln(0.000001) which is approximately equal to  41.45.

c - To find the x intercept we need to solve the equation f(x) = 0

-3ln(x - 4) = 0

Divide both sides by -3 to obtain

ln(x - 4) = 0

Use properties of logarithmic and exponential functions to write the above equation as

eln(x - 4) = e0

Then simplify

 x - 4 = 1

x = 5

The x intercept is at (5 , 0).

The y intercept is given by (0 , f(0)). f(0) is undefined since x = 0 is not a value in the domain of f. There is no y intercept.

d - So far we have the domain, range, x intercept and the vertical asymptote. We need extra points to be able to graph f.

f(4.5) = -3ln(4.5 - 4) approximately equal to 2.08

f(8) = -3ln(8 - 4) approximately equal to - 4.16

f(14) = -3ln(14 - 4) approximately equal to - 6.91

Let us now sketch all the points and the vertical asymptote. Join the points by a smooth curve and f increases as x approaches 4 from the right.

 


Matched Problem to Example 2:  f is a function given by

f (x) = 2ln(x + 5)

 

  1. Find the domain of f and range of f.
  2. Find the vertical asymptote of the graph of f.
  3. Find the x and y intercepts of the graph of f if there are any.
  4. Sketch the graph of f.


Example 3: f is a function given by

f (x) = 2ln(| x |)

 

  1. Find the domain of f and range of f.
  2. Find the vertical asymptote of the graph of f.
  3. Find the x and y intercepts of the graph of f if there are any.
  4. Sketch the graph of f.

Answer to Example 3

a - The domain of f is the set of all x values such that 

| x | > 0

The domain is the set of all real numbers except 0.

The range of f is the interval (-inf , +inf).

b - The vertical asymptote is obtained by solving

| x | = 0

which gives

x = 0

As x approaches 0 from the right (x > 0) , f(x) decreases  without bound. How do we know this?

Let us take some values:

f(1) =2 ln(| 1 |) = 0

f(0.1) = 2ln(0.1)  which is approximately equal to  -4.61.

f(0.0001) = 2ln(0.0001) which is approximately equal to -18.42.

f(0.0000001) = 2ln(0.0000001) which is approximately equal to -32.24.

As x approaches 0 from the left (x < 0) , f(x) decreases  without bound. How do we know this?

Let us take some values:

f(-1) =2 ln(| -1 |) = 0

f(-0.1) = 2ln(| -0.1 |)  which is approximately equal to  -4.61.

f(-0.0001) = 2ln(| -0.0001 |) which is approximately equal to -18.42.

f(-0.0000001) = 2ln(| -0.0000001 |) which is approximately equal to -32.24.

c - To find the x intercept we need to solve the equation f(x) = 0

2ln(| x |) = 0

Divide both sides by 2 to obtain

ln(| x |) = 0

Use properties of logarithmic and exponential functions to write the above equation as

eln(| x |) = e0

Then simplify

 | x |  = 1

Two x intercepts at (1 , 0) and (-1 , 0).

The y intercept is given by (0 , f(0)). f(0) is undefined since x = 0 is not a value in the domain of f. There is no y intercept.

d - So far we have the domain, range, x intercept and the vertical asymptote. By examining function f it is easy to show that this is an even function and its graph is symmetric with respect to the y axis.

f( -x) = 2 ln(| -x |)

but

| -x | = | x |

and therefore

f( -x) = 2 ln(| x |) = f( x ),  this shows that f is an even function.

Let us find extra points.

f(4) = 2ln(| 4 |) approximately equal to 2.77.

f(0.5) = 2ln(| 0.5 |) approximately equal to - 1.39.

Since f is even f(-4) = f(4) and f(-0.5) = f(0.5).

Let us now sketch all the points, the vertical asymptote and Join the points by a smooth curve.


Matched Problem to Example 3:  f is a function given by

f (x) = -2ln(x2)

  1. Find the domain of f and range of f.
  2. Find the vertical asymptote of the graph of f.
  3. Find the x and y intercepts of the graph of f if there are any.
  4. Sketch the graph of f.


More references on logarithmic functions and graphing.

  • Graphing Functions


  • Logarithmic Functions.

  • Graphs of Basic Functions.