3 B The Logarithmic Function Base e

We have seen that e =.. 2·7183 is the most natural base to use for exponential functions in calculus. This numbere is similarly the most natural base to use for the calculus of logarithmic functions. This section introduces log_ex in preparation for the calculus of the logarithmic function developed in the rest of the chapter.

The Logarithmic Function: Because e is the most natural base to use in calculus, the functiony = log_ex is called the logarithmic function to distinguish it from other logarithmic functions like log₂x and log10x that have other bases.

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THE LOGARITHMIC FUNCTION: The logarithmic function with base e, y = log_ex,

is called the logarithmic function to distinguish it from all other logarithmic functionsy = log_ax.

Below are tables of values and the graphs of the mutually inverse functionsy = e^x andy = log_ex. These are two of the most important graphs in the course.

x 1

e² 1

e 1 e e²

log_ex −2 −1 0 1 2

x y

1 2

1 2

e e

x −2 −1 0 1 2

e^x 1

e² 1

e 1 e e²

We know already that the tangent toy = e^xat itsy-intercept (0, 1) has gradient 1.

When this graph is reflected in the liney = x, this tangent is reflected to a tangent toy = log_ex at its x-intercept (1, 0). Both these tangents must have gradient 1.

Properties of the Graph ofy = log_ex: The graph of y = log_ex alone is drawn below on graph paper. The tangent has been drawn at the x-intercept (1, 0) to show that the gradient of the curve there is exactly 1.

x y

0 1 2 3

−1 1

1e

e y x= − 1

y= log_ex

The graph of y = log_ex and its properties must be thoroughly known. Its properties correspond to the properties listed earlier of its inverse functiony = e^x.

• The domain is x > 0.

The range is all values ofy.

• The y-axis x = 0 is a vertical asymptote to the curve.

• As x → 0⁺,y → −∞.

Asx → ∞, y → ∞.

• The curve has gradient 1 at its x-intercept (1, 0).

• The curve is always concave down.

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The Notation for the Logarithmic Function — log_ex, log x and ln x: In calculus, log_ex is the only logarithmic function that matters. It is often written simply as logx and from now on, if no base is given, basee will be understood.

The function log_ex is also written as ln x, the ‘n’ standing for ‘natural’ loga-rithms. The ‘n’ also stands for ‘Napierian’ logarithms, in honour of the Scottish mathematician John Napier (1550–1617), who first developed tables of logarithms base e for calculations (first published in 1614).

Note: Be careful of the quite different convention on calculators, where log means log₁₀x. The function labelled ln is used to find logarithms base e.

Notice that the function e^x is usually located on the same button as ln because the two functionsy = e^x and y = log_ex are inverses of each other.

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THE NOTATION FOR THE LOGARITHMIC FUNCTION: In this course, log_ex = log x = ln x

all mean the same thing, that is, the logarithm base e of x.

ON CALCULATORS, HOWEVER:

ln is used to approximate log_ex. It is the inverse function of e^x . log is used to approximate log₁₀x. It is the inverse function of 10^x . WORKEDEXERCISE:

(a) Use your calculator to find, correct to four significant figures:

(i) log_e10 (ii) log_{e 10}¹ (iii) log_e100

(b) How are the answers to parts (ii) and (iii) related to the answer to part (i)?

SOLUTION:

(a) Using the function labelled ln on the calculator,

(i) log_e10 =.. 2·303 (ii) log_{e 10}¹ =.. −2·303 (iii) log_e100 =.. 4·605 (b) Using the log laws,

log_{e 10}¹ =− log_e10 and log_e10² = 2 log_e10, and these relationships are clear from the approximations above.

Combining the Logarithmic and Exponential Functions: As with any base, if the loga-rithmic and exponential functions basee are applied successively to any number, the result is the original number.

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THE LOGARITHMIC AND EXPONENTIAL FUNCTIONS ARE MUTUALLY INVERSE: log_ee^x=x and e^logex =x.

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Again, these identities follow immediately from the definition of logarithms, but here is further explanation if it is needed.

First, log_ee^x =x log_ee, by the log laws,

=x × 1

=x.

Secondly, e^logex =x can be proven by taking logarithms of both sides:

log_eLHS = log_ee^logex

= log_ex, by the previous identity,

= log_eRHS.

WORKEDEXERCISE:

Use the functions labelled ln and e^x on your calculator to demonstrate that:

(a) log_ee¹⁰ = 10 (b) e^loge10 = 10 SOLUTION:

(a) log_ee¹⁰ = log_e22 026·46 . . .

=.. 10

(b) e^loge10 =e²·302 585 ...

=.. 10

Differentiating the Logarithmic Function: The logarithmic function y = log_ex can be differentiated quickly using the known derivative of its inverse functione^x.

Let y = log_ex.

Then x = e^y, by the definition of logarithms.

Differentiating, dx

dy =e^y, since the exponential function is its own derivative,

=x, since e^y =x, and inverting, dy

dx = 1 x.

Hence the derivative of the logarithmic function is the reciprocal function.

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THE DERIVATIVE OF THE LOGARITHMIC FUNCTION IS THE RECIPROCAL FUNCTION: d

dxlog_ex = 1 x

The following worked exercise uses this derivative to confirm that the tangent at thex-intercept has gradient 1. This was established earlier in the section using the graphical argument that the graphs of y = e^x and y = log_ex are mutual reflections iny = x.

WORKEDEXERCISE:

Find the gradient of the tangent toy = log_ex at its x-intercept.

SOLUTION:

The function is y = log_ex.

Differentiating, dy

dx = 1 x. The graph crosses thex-axis at A(1, 0).

Hence, substitutingx = 1 into the formula for the derivative,

x y

1 2

1

−1

e y= log_ex gradient at x-intercept = 1.

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Transformations of the Logarithmic Graph: The usual methods of transforming graphs can be applied to y = log_ex. When the graph is shifted sideways, the vertical asymptote at x = 0 will also be shifted.

A small table of approximate values can be a very useful check, particulary when a sequence of transformations is involved. Remember thaty = log_ex has domain x > 0 and that the vertical asymptote is at x = 0.

WORKEDEXERCISE:

Use transformations of the graph ofy = log_ex, and a table of values, to generate a sketch of each function. State the domain and show the x-intercept and the vertical asymptote.

(a) y = log_e(−x) (b) y = log_ex − 2 (c) y = log_e(x + 3) SOLUTION:

(a) Graphy = log_e(−x) by reflecting y = log_ex in the y-axis.

x −e −1 −¹_e

y 1 0 −1

Domain: x < 0 x-intercept: (−1, 0)

x y

− e

1e

1

−1

−1

Asymptote: x = 0 (the y-axis)

(b) Graphy = log_ex − 2 by shifting y = log_ex down 2 units.

x ¹_e 1 e e²

y −3 −2 −1 0

Domain: x > 0 x-intercept: (e², 0)

x y

e² e

−1−2 1

Asymptote: x = 0 (the y-axis)

(c) Graphy = log_e(x + 3) by shifting y = log_ex left 3 units.

x ¹_e − 3 −2 e − 3

y −1 0 1

Domain: x > −3 x-intercept: (−2, 0)

x log_e3 y

−3 + e

−3

−2

1

Asymptote: x = −3

Exercise 3B

Note: Remember that logx and ln x both mean log_ex (except on the calculator, where log means log₁₀x and ln means log_ex).

1. Use your calculator to approximate the following, correct to four decimal places where necessary. Read the note above and remember to use the ln key on the calculator.

(a) log_e1 (b) log_e2

(c) ln 3 (d) ln 8

(e) log¹₂ (f) log¹₃

(g) ln¹₈ (h) ln₁₀¹

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2. Use the functions labelled ln and e^x on your calculator to demonstrate that:

(a) log_ee² = 2 (b) log_ee³ = 3

(c) log_ee¹= 1 (d) log_ee⁻² =−2

(e) log_ee⁻³ =−3 (f) log_ee⁻¹ =−1 3. Use the functions labelled ln and e^x on your calculator to demonstrate that:

(a) e^loge2= 2 (b) e^loge3= 3

(c) e^loge1 = 1 (d) e^loge10 = 10

(e) e^{loge 12} = ¹₂ (f) e^{loge 1 01} = ₁₀¹

x y

0 1 2 3

−1 1

e1

e 4.

(a) Photocopy the graph of y = log_ex above, and on it draw the tangent at (1, 0), ex-tending the tangent across to they-axis.

(b) Measure the gradient of this tangent and confirm that it is equal to the reciprocal of thex-coordinate at the point of contact.

(c) Copy and complete the table of values to the right by measuring the gradienty^ of each tangent.

x 1

e 1

2 1 2 e

gradienty^ 1 (d) What do you notice about the y-intercepts of the x

tangents?

5. (a) Photocopy the graph ofy = log_ex in the previous question, and on it draw the tangent aty = 1, extending the tangent across to the y-axis.

(b) Measure the gradient of this tangent and confirm that it is equal to the reciprocal of thex-coordinate at the point of contact.

(c) Repeat for the tangents aty = 0, −1, ¹₂ and −¹₂.

(d) What do you notice about they-intercepts of the tangents?

D E V E L O P M E N T

Note: Remember that logx and ln x both mean log_ex (except on the calculator, where log means log₁₀x and ln means log_ex).

6. Sketch the graph of y = log_ex and use your knowledge of transformations to graph the following functions. Note that in each case the y-axis is a vertical asymptote and the domain isx > 0.

(a) y = log_ex + 1 (b) y = log_ex + 2 (c) y = ln x − 1 (d) y = ln x − 2

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7. (a) Copy and complete the following tables of values for the functions y = log_ex and y = − log_ex, giving your answers correct to two decimal places.

x ¹₄ ¹₂ 1 2 4

log_ex

x ¹₄ ¹₂ 1 2 4

− log_ex

(b) Sketch both graphs on the same number plane, and draw the tangent to each at the x-intercept.

(c) Find the gradients of the two tangents, and hence explain why they are perpendicular.

8. Use the log laws to simplify:

(a) e log_ee (b) ¹_e log_e ¹_e (c) 3 log_ee²

(d) log_e√ e

(e) e log_ee³− e log_ee (f) log_ee + log_e ¹_e

(g) log_ee^e (h) log_e(log_ee^e)

(i) log_e(log_e(log_ee^e)) 9. Express as a single logarithm:

(a) log_e3 + log_e2 (b) log_e100− log_e25

(c) log_e2− log_e3 + log_e6 (d) log_e54− log_e10 + log_e5 10. (a) What is the x-coordinate of the point on the curve y = log x where y = 0?

(b) Use the result d

dx logx = 1

x to find the gradient of the tangent at this point.

(c) Hence write down the equation of the tangent, and find itsy-intercept.

(d) Repeat the above steps for the points where y = −1, 1 and 2.

(e) Compare the values of the y-intercepts with those found in question 5.

11. (a) Sketch a graph ofy = log x and hence write down its domain.

(b) Write down y^ and hence explain why the graph always has positive gradient.

(c) Find y^ and hence explain why the graph is always concave down.

12. Use the graph of y = log x and your knowledge of transformations to graph the following functions. Show the vertical asymptote and state the domain in each case.

(a) y = log(x − 1) (b) y = log(x − 3)

(c) y = log(x + 1) (d) y = log(x + 2)

(e) y = − log x (f) y = log(−x)

C H A L L E N G E

13. Sketch the graph of y = − log_ex and use your knowledge of transformations to graph the following functions. Note that in each case they-axis is a vertical asymptote and the domain is x > 0.

(a) y = − log_ex + 1 (b) y = − log_ex + 2 (c) y = − log x − 1 (d) y = − log x − 2 14. The function log(1 + x) may be approximated using the power series

log(1 +x) = x −x² 2 +x³

3 −x⁴

4 +· · · , for 0 ≤ x ≤ 1.

Use this power series to approximate each of the following, correct to two decimal places.

Then compare your answers with those given by your calculator.

(a) log 1¹₂ (b) log⁵₄ (c) log¹₂ (d) log¹₃

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