THE LOGARITHMIC FUNCTION

The logarithmic function, with base ‘ a’ is represented by the expression . To determine the shape of its graph we start by constructing a table of values for the function

and comparing it with the table of values for :

x 0 1 2 4 8 16

---From the table of values we observe that the x and y values have interchanged! Plotting these results on the same set of axes, we observe that the graph of the logarithmic function is a reflection of the exponential function about the line y = x. So, whereas for the exponential function, the asymptote is the x–axis (i.e., y = 0), for the logarithmic function, the asymptote is the y–axis (i.e., x = 0).

So, how do the graphs of compare to . The best

way to see this is to sketch the graphs on the same set of axes:

Notice that for all a > 0 and .

As is the case for the exponential functions, the base ‘ e’ also plays an important role when dealing with logarithmic functions. When using the number ‘ e’ as the base for the logarithmic function, we refer to it as the natural logarithmic function and can write it in one of two ways:

(1,0) (0,1)

(2,1) (1,2)

( ,–1)1 2 (–1, )1

2 y

x

y = log2x x, >0 y = 2^x

The implied domain of the basic logarithmic function is ]0, ∞[ and the y–axis is its asymptote.

y = x

y = log_ex,y = log₁₀x,y = log₃x y = log₂x

(1,0) (2,1) y

x y = log2x x, >0

The implied domain of the basic logarithmic function with any positive base is ]0, ∞[ and has the y–axis as its asymptote.

y = log3x x, >0

y = log10x x, >0 (3,1)

(10,1)

Observe that each of the logarithmic functions is a reflection about the line y = x of its corresponding exponential function (of the same base).

a1

log = 0 log_aa = 1

f x( ) = logex x, >0 or f x( ) = lnx x, >0

Sketch the following, specifying the implied domain in each case.

(a) f x( ) = log3(x–2) (b) g x( ) = log2(2x+3)

E

^XAMPLE5.25

(a) We begin by looking at the domain: As we cannot have the logarithm of a negative number

we must have that .

Therefore, domain of f = ]2, ∞[.

Note also then, that by default we have obtained the equation of the vertical asymptote, in this case it is x = 2.

When x = 3, and

when x = 5, .

(b) This time we need to have .

Therefore, the implied domain of g = . The vertical asymptote has the equation .

When x = –1, .

When x = 0, .

We make the following general observations:

Notice that in part (b) of example 5.25, we have = , where

we can still ‘see’ the equation of the asymptote as . The extra factor of ‘2’ can be viewed as either a dilation or a translation – we will leave further discussion of this to Chapters 6 and 7.

(a) The implied domain in this case is x > 0.

So, the vertical asymptote has the equation x = 0.

We note that the negative sign in front of the will have the effect of reversing the sign of the values.

That is, the graph of is a reflection

the right and has a vertical asymptote at x = k.

2. The graph of is identical to but moved ‘ k’ units to

the left and has a vertical asymptote at x = –k.

y = loga(x–k),k>0 y = log_ax

Sketch the following, specifying the implied domain in each case.

(a) f x( ) = –2log₃x (b) f x( ) 1

The factor of 2 will have the effect of ‘stretching’ the graph of by a factor of 2 along the y–axis.

Also, we have that and .

(b) This time the implied domain is ]0, ∞[.

Therefore, the equation of the asymptote is x = 0. The one third factor in front of will have the effect of

‘shrinking’ the graph of by a factor of 3.

Then, and .

Again, we have the following observations:

(a) The effect of adding 3 to the graph of

will result in being moved up

3 units.

Its implied domain is ]0, ∞[ and its asymptote has equation x = 0.

(b) The effect of subtracting 2 from the graph of

will result in being

moved down 2 units.

Its implied domain is ]0, ∞[ and its asymptote has equation x = 0.

(c) As we cannot have the logarithm of a negative

number we must have that .

This means that the vertical asymptote is given by x = 3 and the graph must be drawn to the left of the asymptote.

y = log₃x

i. Stretched along the y–axis if k > 1.

ii. Shrunk along the y–axis if 0 < k < 1.

2. The graph of is identical to but

i. Reflected about the x–axis and stretched along the y–axis if k < –1.

ii. Reflected about the x–axis and shrunk along the y–axis if –1 < k < 0.

y = k×log_ax,k>0 y = log_ax

y = k×log_ax,k<0 y = log_ax

Sketch the following, specifying the implied domain in each case.

(a) g x( ) = log₂x+3 (b) g x( ) = log₂x–2 (c) g x( ) = log₂(3–x)

We make the following general observations:

1. Sketch the graph of the following functions, clearly stating domains and labelling asymptotes.

(a) (b)

(c) (d)

(e) (f)

(g) (h)

2. Sketch the graph of the following functions, clearly stating domains and labelling asymptotes.

(a) (b)

(c) (d)

(e) (f)

3. Sketch the graph of the following functions, clearly stating domains and labelling asymptotes.

(a) (b) (c)

(d) (e) (f)

4. Sketch the graph of the following functions, clearly stating domains and labelling asymptotes.

(a) (b) (c)

(d) (e) (f)

1. The graph of is identical to but moved ‘ k’ units

vertically down and has a vertical asymptote at x = 0.

2. The graph of is identical to but moved ‘ k’ units

vertically up and has a vertical asymptote at x = 0.

y = log_ax–k,k>0 y = log_ax y = log_ax+k,k>0 y = log_ax

1. The graph of is identical to but reflected about the y–axis and has a vertical asymptote at x = 0.

2. The graph of is identical to but reflected about

the y–axis, moved ‘k’ units to the right and has a vertical asymptote at x = k.

y = loga( )–x y = log_ax

y = loga(k–x) k 0, > y = log_ax

E

^XERCISES

^5.3.4

f x( ) = log4(x–2) f x( ) = log2(x+3) h x( ) = log₁₀x+2 g x( ) = –3+log₃x

f x( ) = log5(2x–1) h x( ) = log2(2–x) g x( ) = 2log₁₀x f x( ) = –log₁₀x+1

f x( ) = 2log₂x+3 f x( ) = 10–2log₁₀x h x( ) = 2log22 x( –1) g x( ) 1

2---log10(1–x) –

=

f x( ) = log2(3x+2)–1 h x( ) 3 1 2---x–1

 

 

2

1 + log

=

f x( ) = 2lnx g x( ) = –5lnx f x( ) = ln(x–e) f x( ) = ln(1–ex) f x( ) = 5–lnx h x( ) = lnx–e

f x( ) = log2 x f x( ) = log₁₀x² h x( ) = lnx g x( ) 1

---x

   ln

= h x( ) = ln(1–x²) f x( ) = log2(x²–4)

5. Sketch the graph of the following functions, clearly stating domains and labelling asymptotes.

(a) (b) (c)

(d) (e) (f)

6. Given the function , sketch the graph of

i. ii.

8. Sketch the graph of the following functions and find their range.

(a) (b)

(c) (d)

9. Sketch the graph of the following functions.

(a) (b) (c) + 1

10. Sketch the graph of the following functions, clearly stating domains and labelling asymptotes.

(a) (b)

(c) (d)

(e) (f)

11. Sketch the graph of clearly labelling its asymptote,

and intercept(s) with the axes. Hence, find .

12. Sketch the graph of (a) (b)

In document IB Maths SL Textbook (Page 154-160)