Chapter Four of Maths in Focus

4 Exponential and Logarithmic ­Functions TERMINOLOGY Exponential equation: Equation where the pronumeral is the index or exponent such as 3 x = 9 x Exponential function: A function in the form y = a where the variable x is a power or exponent

Logarithm: A logarithm is an index. The logarithm is the power or exponent of a number to a certain base i.e. 2 x = 8 is the same as log2 8 = x

Chapter 4 Exponential and Logarithmic Functions

Introduction This chapter introduces a new irrational number, ‘e’, that has special properties in calculus. You will learn how to differentiate and integrate the exponential function f (x) = e x. The definition and laws of logarithms are also introduced in this chapter, as well as differentiation and integration involving logarithms.

DID YOU KNOW? John Napier (1550–1617), a Scottish theologian and an amateur mathematician, was the first to invent logarithms. These ‘natural’, or ‘Naperian’, logarithms were based on ‘e’. Napier was also one of the first mathematicians to use decimals rather than fractions. He invented the notation of the decimal, using either a comma or a point. The point was used in England, but a few European countries still use a comma. Henry Briggs (1561–1630), an Englishman who was a professor at Oxford, decided that logarithms would be more useful if they were based on 10 (our decimal system). These are called common logarithms. Briggs painstakingly produced a table of logarithms correct to 14 decimal places. He also produced sine tables—to 15 decimal places—and tangent tables—to 10 decimal places. The work on logarithms was greatly appreciated by Kepler, Galileo and other astronomers at the time, since they allowed the computation of very large numbers.

Differentiation of Exponential Functions When differentiating exponential functions f (x) = a x from first principles, an interesting result can be seen. The derivative of any exponential function gives a constant which is multiplied by the original function.

EXAMPLE Sketch the derivative (gradient) function of y = 10 x.

Solution

continued

161

162

Maths In Focus Mathematics Extension 1 HSC Course

The graph of y = 10 x always has a positive gradient that is becoming steeper. So the derivative function will always be positive, becoming steeper.

The derivative function of an exponential function will always have a shape similar to the original function. We can use differentiation from first principles to find how close this derivative function is to the original function.

EXAMPLE Differentiate f (x) = 10x from first principles.

Solution f l(x) = lim

f ( x + h ) − f ( x)

h 10 x + h − 10 x = lim h "0 h 10 x (10 h − 1) = lim h "0 h 10 h − 1 x = 10 lim h "0 h h "0

You can explore limits using a graphics package on a computer or a graphical calculator.

Using the 10 x key on the calculator, and finding values of is small, gives the result:

10 h − 1 when h h

f l(x) Z 2.3026 ´ 10x or

d (10x ) Z 2.3026 ´ 10x dx

Drawing the graphs of y = 2.3026 ´ 10 x and y = 10 x together shows how close the derivative function is to the original graph. y y = 10 x y = 2.3026 × 10 x

12 10 8 6 4 2 −2

−1

1

2

x

Chapter 4 Exponential and Logarithmic Functions

163

Similar results occur for other exponential functions. In general, d x (a ) = kax where k is a constant. dx

Application If y = a x then

dy dx

You will study exponential growth and decay in Chapter 6.

= ka x = ky

This means that the rate of change of y is proportional to y itself. That is, if y is small, its rate of change is small, but if y is large, then it is changing rapidly. This is called exponential growth (or decay, if k is negative) and has many applications in areas such as population growth, radioactive decay, the cooling of objects, the spread of infectious diseases and the growth of technology.

Different exponential functions have different values of k.

EXAMPLES 1.

d x (2 ) Z 0.6931 ´ 2 x. dx y y = 0.6931 × 2x y = 2x

12 10 8 6 4 2 −3 −2 −1

2.

1

2

x

3

d x (3 ) Z 1.0986 ´ 3 x. dx y

12 10

y = 3x y = 1.0986 × 3x

8 6 4 2 −3 −2 −1

1 2 3

x

164

Maths In Focus Mathematics Extension 1 HSC Course

Notice that the derivative function of y = 3 x is very close to the original function. We can find a number close to 3 that gives exactly the same graph for the derivative function. This number is approximately 2.71828, and is called e. d x (e ) = e x dx

e is an irrational number like π.

DID YOU KNOW? A transcendental number is a number beyond ordinary numbers. Another transcendental number is π.

The number e was linked to logarithms before this useful result in calculus was known. It is a transcendental (irrational) number. This was proven by a French mathematician, Hermite, in 1873. Leonhard Euler (1707–83) gave e its symbol, and he gave an approximation of e to 23 decimal places. Currently, e is known to about 100 000 decimal places. Euler studied mathematics, theology, medicine, astronomy, physics and oriental languages. He did extensive research into mathematics and wrote more than 500 books and papers. Euler gave mathematics much of its important notation. He caused π to become standard notation and used i for the square root of –1. He first used small letters to show the sides of triangles and the corresponding capital letters for their opposite angles. Also, he introduced the symbol S for sums and f(x) notation.

ex

KEY

Use this key to find powers of e. For example, to find e2: Press SHIFT e x 2 = e 2 7.389056099 To find e: Press SHIFT e x 1 = e 1 2.718281828

EXAMPLES 1. Sketch the curve y = e x.

Solution Use ex on your calculator to draw up a table of values: x

-3

-2

-1

0

1

2

3

y

0.05

0.1

0.4

1

2.7

7.4

20.1

Chapter 4 Exponential and Logarithmic Functions

165

2. Differentiate 5e x.

Solution d x (e ) = e x dx d d ` (5e x) = 5 (e x) dx dx = 5e x 3. Find the equation of the tangent to the curve y = 3e x at the point (0, 3).

Solution dy dx

= 3e x

At (0, 3), `

dy dx

= 3e 0 dy

=3 m=3

dx

Equation y − y1 = m (x − x1) y − 3 = 3 (x − 0 ) = 3x y = 3x + 3

continued

gives the gradient of the tangent.

166

Maths In Focus Mathematics Extension 1 HSC Course

4. Differentiate

2x + 3 . ex

Solution This is the quotient rule from Chapter 8 of the Preliminary Course book.

dy dx

= =

u lv − v lu v2 2 . e x − e x (2 x + 3)

( e x) 2 2e x − 2xe x − 3e x = e 2x − e x − 2xe x = e 2x − e x (1 + 2x) = e 2x − (1 + 2x) = ex

4.1 Exercises 1. Find, correct to 2 decimal places, the value of (a) e 1.5 (b) e − 2 (c) 2e 0.3 1 (d) 3 e (e) − 3e − 3.1 2. Sketch the curve (a) y = 2e x (b) y = e − x (c) y = − e x 3. Differentiate (a) 9e x (b) − e x (c) e x + x 2 (d) 2x 3 − 3x 2 + 5x − e x (e) (e x + 1) 3 (f) (e x + 5)7 (g) (2e x − 3) 2 (h) xe x ex (i) x

(j) x 2 e x (k) (2x + 1) e x (l)

ex 7x − 3

5x (m)   x e 4. If f (x) = x 3 + 3x − e x, find f l(1) and f m(1) in terms of e. 5. Find the exact gradient of the tangent to the curve y = e x at the point (1, e). 6. Find the exact gradient of the normal to the curve y = e x at the point where x = 5. 7. Find the gradient of the tangent to the curve y = 4e x at the point where x = 1.6, correct to 2 decimal places. 8. Find the equation of the tangent to the curve y = − e x at the point (1, –e).

Chapter 4 Exponential and Logarithmic Functions

9. Find the equation of the normal to the curve y = e x at the point where x = 3, in exact form.

167

11. Find the first and second derivatives of y = 7e x. Hence show

10. Find the stationary point on the curve y = xe x and determine its nature. Hence sketch the curve.

that

d 2y dx 2

= y.

12. If y = 2e x + 1, show that

d2y dx 2

= y − 1.

Function of a function rule Remember that the function of a function rule uses the result dy dy du = . ´ dx du dx

You studied this in Chapter 8 of the Preliminary Course book.

EXAMPLE Differentiate e x

2

+ 5x − 3

.

Solution Let u = x 2 + 5x − 3 Then y = e u dy du = 2x + 5  and  = eu dx du dy dx

dy

du ´ du dx = e u (2x + 5) =

= e x + 5x − 3 (2x + 5) = (2x + 5) e x + 5x − 3 2

Can you see a quick way to do this?

2

If y = e f (x) then

Proof Let     u = f (x) Then  y = e u dy du = f l(x) = e u  and    du dx dy dy du = ´ dx du dx = e u f l(x) = f l(x) e f (x)

dy dx

= f l(x) ef (x)

168

Maths In Focus Mathematics Extension 1 HSC Course

EXAMPLES 1. Differentiate e 5x − 2

Solution y l = f l(x) e f (x) = 5e 5x − 2 2. Differentiate x 2 e 3x. This is the product rule from Chapter 8 of the Preliminary Course book.

Solution dy dx   

= ulv + vlu = 2x . e3x + 3e3x . x2 = xe3x (2 + 3x)

3. Given y = 2e 3x + 1, show that

d2 y dx 2

= 9 (y − 1) .

Solution y = 2e 3x + 1 dy = 6e 3x dx d2 y = 18e 3x dx 2 = 9 (2e 3x) = 9 (2e 3x + 1 − 1) = 9 (y − 1)

4.2 Exercises 1. Differentiate (a) e 7x (b) e − x (c) e 6x − 2 (d) e x + 1 (e) ex + 5x + 7 (f) e 5x (g) e − 2x (h) e 10x (i) e 2x + x (j) x 2 + 2x + e 1 − x (k) (x + e 4x)5 (l) xe 2x 2

e 3x (m)   2 x (n) x 3 e 5x e 2x + 1 (o) 2x + 5

3

2. Find the second derivative of (e 2x + 1)7. 3. If f (x) = e 3x − 2, find the exact value of f l(1) and f m(0) . 4. Find the gradient of the tangent to the curve y = e 5x at the point where x = 0.

Chapter 4 Exponential and Logarithmic Functions

5. Find the equation of the tangent to the curve y = e 2x − 3x at the point (0, 1). 6. Find the exact gradient of the normal to the curve y = e 3x at the point where x = 1. 7. Find the equation of the tangent to the curve y = e x at the point (1, e). 2

8. If f (x) = 4x 3 + 3x 2 − e − 2x, find f m(−1) in terms of e.

d2y dx 2

= 16y.



d2y dx

2

−3

dy dx

+ 2y = 0, given

y = 3e 2x .

12. Show

d2 y dx 2

= b 2 y for y = ae bx .

13. Find the value of n if y = e 3x satisfies the equation d 2y dy +2 + ny = 0. 2 dx dx 14. Sketch the curve y = e x + x − 2, showing any stationary points and inflexions. 2

9. Find any stationary points on the curve y = x 2 e 2x and sketch the curve. 10. If y = e 4x + e − 4x, show that

11. Prove

169

x2 + 1 , ex showing any stationary points and inflexions.

15. Sketch the curve y =

Integration of Exponential Functions Since

d x (e ) = e x, then the reverse must be true. dx

# e x dx = e x + C To find the indefinite integral (primitive function) when the function of a function rule is involved, look at the derivative first.

EXAMPLE Differentiate e 2x + 1. Hence find Find

# 2e 2x + 1 dx.

# e 2x + 1 dx.

Solution d 2x + 1 (e ) = 2e 2x + 1 dx `

# 2e 2x + 1 dx = e 2x + 1 + C # e 2x + 1 dx = 12 # 2e 2x + 1 dx =

1 2x + 1 e +C 2

Integration is the inverse of differentiation.

170

Maths In Focus Mathematics Extension 1 HSC Course

In general

# e ax + b dx = 1a e ax + b + C Proof d ax + b (e ) = ae ax + b dx ` # ae ax + b dx = e ax + b + C # e ax + b dx = 1a # ae ax + b dx 1 = a e ax + b + C

EXAMPLES 1. Find

# (e 2x − e − x) dx.

Solution

# (e 2x − e − x) dx = 12 e 2x − (−11) e − x + C =

1 2x e + e−x + C 2

2. Find the exact area enclosed between the curve y = e 3x, the x-axis and the lines x = 0 and x = 2.

Solution 2

Area = # e 3x dx 0

2 1 = ; e 3x E 3 0 1 6 1 0 = e − e 3 3 1 6 = (e − e 0) 3 1 = (e 6 − 1) units 2 3

3. Find the volume of the solid of revolution formed when the curve y = e x is rotated about the x-axis from x = 0 to x = 2.

Solution Use index laws to simplify (e x) 2.

y = ex ` y 2 = ( e x) 2 = e 2x

Chapter 4 Exponential and Logarithmic Functions

b

V = π # y2 dx a

2

= π # e2x dx 0

2 1 = π ; e2x E 2 0 1 4 1 0 =πc e − e m 2 2 1 4 1 =πc e − m 2 2 π 4 = (e − 1) units3 2

4.3 Exercises 1. Find these indefinite integrals.

# e dx (b) # e 4x dx (c) # e − x dx (d) # e 5x dx (e) # e − 2x dx (f) # e 4x + 1 dx (g) # − 3e 5x dx (h) # e 2t dt (i) # (e 7x − 2) dx (j) # (e x − 3 + x) dx 2x

(a)

2. Evaluate in exact form. (a)

#0

1

e 5x dx

2

(b) # − e− x dx 0

(c)

#1

4

2e 3x + 4 dx

3

(d) # (3x 2 − e 2x) dx 2

(e)

#0

2

(f)

#1

2

(e 2x + 1) dx (e x − x) dx

3

(g) # (e 2x − e − x) dx 0

3. Evaluate correct to 2 decimal places. (a)

#1

3

2

e − x dx

(b) # 2e 3y dy 0

(c)

#5

6

(ex + 5 + 2x − 3) dx

1

(d) # (e 3t + 4 − t) dt 0

(e)

#1

2

(e 4x + e 2x) dx

4. Find the exact area enclosed by the curve y = 2e 2x, the x-axis and the lines x = 1 and x = 2. 5. Find the exact area bounded by the curve y = e 4x − 3, the x-axis and the lines x = 0 and x = 1. 6. Find the area enclosed by the curve y = x + e − x, the x-axis and the lines x = 0 and x = 2, correct to 2 decimal places. 7. Find the area bounded by the curve y = e 5x, the x-axis and the lines x = 0 and x = 1, correct to 3 significant figures. 8. Find the exact volume of the solid of revolution formed when the curve y = e x is rotated about the x-axis from x = 0 to x = 3. 9. Find the volume of the solid formed when the curve y = e − x + 1 is rotated about the x-axis from x = 1 to x = 2, correct to 1 decimal place.

171

172

Maths In Focus Mathematics Extension 1 HSC Course

10. Use Simpson’s rule with 3 function values to find an

#1

2

xe x dx,



approximation to



correct to 1 decimal place.

11. (a) Differentiate x 2 e x .

(b) Hence find # x(2 + x)e dx. x

12. Find # x 2e x + 1 dx using the substitution u = x 3 + 1.

14. The curve y = e x + 1 is rotated about the x-axis from x = 0 to x = 1. Find the exact volume of the solid formed. 15. Find the exact area enclosed between the curve y = e 2x and the lines y = 1 and x = 2.

3

13. Use the substitution u = x 2 to 2 evaluate # xe x dx (give exact 0 value). 2

Application

You will study these formulae in Chapter 6.

The exponential function occurs in many fields, such as science and economics. P = P0 e kt is a general formula that describes exponential growth. P = P0 e - kt is a general formula that describes exponential decay.

Logarithms ‘Logarithm’ is another name for the index or power of a number. Logarithms are related to exponential functions, and allow us to solve equations like 2 x = 5. They also allow us to change the subject of exponential equations such as y = e x to x.

Definition If y = a x, then x is called the logarithm of y to the base a.

If y = a x, then x = loga y

Logarithm keys log is used for log 10 x In is used for log e x

Chapter 4 Exponential and Logarithmic Functions

173

EXAMPLES 1. Find log 10 5.3 correct to 1 decimal place.

Solution log 10 5.3 = 0.724275869 = 0.7 correct to 1 decimal place

Use the log key.

2. Evaluate log e 80 correct to 3 significant figures.

Solution log e 80 = 4.382026634 = 4.38 correct to 3 significant figures 3. Evaluate log 3 81.

Solution Let

log3 81 = x 3 x = 81

Then i.e. ` So

(by definition)

3 =3 x=4 log3 81 = 4. x

4

4. Find the value of log 2

Solution 1 =x 4 1 2x = Then 4 1 = 2 2 = 2−2 x = −2 ` 1 So log2 = − 2. 4 Let

log2

1 . 4

Use the In key.

174

Maths In Focus Mathematics Extension 1 HSC Course

Class Investigation 1. Sketch the graph of y = log 2 x. There is no calculator key for logarithms to the base 2. Use the definition of a logarithm to change the equation into index form, and the table of values: x

y

–3

–2

–1

0

1

2

3

2. On the same set of axes, sketch the curve y = 2 x and the line y = x. What do you notice?

4.4 Exercises 1. Evaluate (a) log2 16 (b) log4 16 (c) log5 125 (d) log3 3 (e) log7 49 (f) log7 7 (g) log5 1 (h) log2 128 2. Evaluate (a) 3 log2 8 (b) log5 25 + 1 (c) 3 – log3 81 (d) 4 log3 27 (e) 2 log10 10 000 (f) 1 + log4 64 (g) 3 log4 64 + 5 (h) 2 + 4 log6 216 log3 9 (i) 2 log8 64 + 4 (j) log2 8 3. Evaluate 1 (a) log 2 2 (b) log 3 3 (c) log4 2

(d) log 5

1 25

(e) log 7 4 7 1 (f) log3 3 3 1 (g) log4 2 (h) log8 2 (i) log 6 6 6 (j) log2

2 4

4. Evaluate correct to 2 decimal places. (a) log10 1200 (b) log10 875 (c) loge 25 (d) ln 140 (e) 5 ln 8 (f) log10 350 + 4.5 log10 15 (g) 2 (h) ln 9.8 + log10 17 log10 30 (i) loge 30 (j) 4 ln 10 – 7

Chapter 4 Exponential and Logarithmic Functions

5. Write in logarithmic form. (a) 3 x = y (b) 5 x = z (c) x 2 = y (d) 2 b = a (e) b 3 = d (f) y = 8 x (g) y = 6 x (h) y = e x (i) y = a x (j) Q = e x 6. Write in index form. (a) log3 5 = x (b) loga 7 = x (c) log3 a = b (d) logx y = 9 (e) loga b = y (f) y = log 2 6 (g) y = log 3 x (h) y = log 10 9 (i) y = ln 4 (j) y = log 7 x 7. Solve for x, correct to 1 decimal place where necessary. (a) log 10 x = 6 (b) log 3 x = 5 (c) logx 343 = 3 (d) logx 64 = 6 (e) log 5

1 =x 5

(f) log x

3=

1 2

(g) ln x = 3.8 (h) 3 log 10 x − 2 = 10 3 2 1 (j) log x 4 = 3 (i) log 4 x =

8. Evaluate y given that log y 125 = 3. 9. If log 10 x = 1.65, evaluate x correct to 1 decimal place. 10. Evaluate b to 3 significant figures if log e b = 0.894. 11. Find the value of log2 1. What is the value of loga 1? 12. Evaluate log5 5. What is the value of loga a? 13. (a) Evaluate ln e without a calculator. (b) Using a calculator, evaluate (i) loge e3 (ii) loge e2 (iii) loge e5 (iv) loge e 1 (v) loge e (vi) eln 2 (vii) eln 3 (viii) eln 5 (ix) eln 7 (x) eln 1 (xi) eln e 14. Sketch the graph of y = log e x. What is its domain and range? 15. Sketch y = 10 x, y = log10 x and y = x on the same number plane. What do you notice about the relationship of the curves to the line? 16. Change the subject of y = log e x to x.

175

176

Maths In Focus Mathematics Extension 1 HSC Course

Class Discussion 1. Investigate these questions on the calculator. Can you see some patterns? (a) loge e (b) loge e2 (c) loge e3 (d) loge e4 (e) loge e5 Can you write a rule for loge ex? 2. Evaluate using a calculator. Can you write a rule to show this pattern? (a) eln 1 (b) eln 2 (c) eln 3 (d) eln 4 (e) eln 5 Can you write a rule for eln x? 3. Do these rules work if x is negative?

Logarithm laws

This corresponds to the law a m ´ a n = a m + n from Chapter 1 of the Preliminary Course book.

Because logarithms are closely related to indices there are logarithm laws that correspond to the index laws. loga (xy) = loga x + loga y

Proof Let   x = a m and y = a n Then   m = log a x and n = log a y xy = a m ´ a n = am + n ` loga (xy) = m + n (by definition) = loga x + loga y This corresponds to the law a m ÷ a n = a m − n .

x log a b y l = log a x − log a y

Chapter 4 Exponential and Logarithmic Functions

177

Proof Let     x = a m and y = a n Then     m = loga x and n = loga y x m n y =a ÷a = am − n x ` loga b y l = m − n = loga x − loga y

(by definition)

This corresponds to the law (a m) n = a mn .

loga x n = n loga x

Proof Let     x = a m Then   m = loga x

`

x n = (a m) n = a mn n loga x = mn = n loga x

(by definition)

Examples 1. Evaluate log6 3 + log6 12.

Solution log6 3 + log6 12 = log6 (3 ´ 12) = log6 36 =2

log 6 36 = 2, since 6 2 = 36.

2. Given log5 3 = 0.68 and log5 4 = 0.86, find (a) log 5 12 (b) log5 0.75 (c) log 5 9 (d) log 5 20 continued

178

Maths In Focus Mathematics Extension 1 HSC Course

Solution (a) log5 12 = log5 (3 ´ 4) = log5 3 + log5 4 = 0.68 + 0.86 = 1.54 3 (b) log 5 0.75 = log 5 4 = log 5 3 − log 5 4 = 0.68 − 0.86 = − 0.18 (c) log 5 9 = log 5 3 2 = 2 log 5 3 = 2 ´ 0.68 = 1.36

log 5 5 = 1, since 5 1 = 5.

(d) log 5 20 = log 5 (5 ´ 4) = log 5 5 + log 5 4 = 1 + 0.86 = 1.86 3. Solve log2 12 = log2 3 + log2 x.

Solution log 2 12 = log 2 3 + log 2 x = log 2 3x So 12 = 3x 4=x

4.5 Exercises 1. Use the logarithm laws to simplify (a) loga 4 + loga y (b) loga 4 + loga 5 (c) loga 12 – loga 3 (d) loga b – loga 5 (e) 3 logx y + logx z (f) 2 logk 3 + 3 logk y (g) 5 loga x – 2 loga y (h) loga x + loga y – loga z (i) log10 a + 4 log10 b + 3 log10 c (j) 3 log3 p + log3 q – 2 log3 r

2. Given log7 2 = 0.36 and log7 5 = 0.83, find (a) log7 10 (b) log7 0.4 (c) log7 20 (d) log7 25 (e) log7 8 (f) log7 14 (g) log7 50 (h) log7 35 (i) log7 98 (j) log7 70

Chapter 4 Exponential and Logarithmic Functions

3. Use the logarithm laws to evaluate (a) log5 50 – log5 2 (b) log2 16 + log2 4 (c) log4 2 + log4 8 (d) log5 500 – log5 4 (e) log9 117 – log9 13 (f) log8 32 + log8 16 (g) 3 log2 2 + 2 log2 4 (h) 2 log4 6 – (2 log4 3 + log4 2) (i) log6 4 – 2 log6 12 (j) 2 log3 6 + log3 18 – 3 log3 2 4. If loga 3 = x and loga 5 = y, find an expression in terms of x and y for (a) loga 15 (b) loga 0.6 (c) loga 27 (d) loga 25 (e) loga 9 (f) loga 75 (g) loga 3a a (h) loga 5 (i) loga 9a 125 (j) loga a 5. If loga x = p and loga y = q, find, in terms of p and q. (a) loga xy (b) loga y3 y (c) loga x

(d) loga x2 (e) loga xy5 x2 (f) loga y (g) loga ax a (h) loga 2 y (i) loga a3y x (j) loga ay 6. If loga b = 3.4 and loga c = 4.7, evaluate c (a) loga b (b) loga bc2 (c) loga (bc)2 (d) loga abc (e) loga a2c (f) loga b7 a (g) loga c (h) loga a3 (i) loga bc4 (j) loga b4c2 7. Solve (a) log4 12 = log4 x + log4 3 (b) log3 4 = log3 y – log3 7 (c) loga 6 = loga x – 3 loga 2 (d) log2 81 = 4 log2 x (e) logx 54 = logx k + 2 logx 3

Change of base Sometimes we need to evaluate logarithms such as log2 7. We use a change of base formula.

log a x =

log b x log b a

179

180

Maths In Focus Mathematics Extension 1 HSC Course

Proof Let   y = log a x Then  x = a y Take logarithms to the base b of both sides of the equation: logb x = logb a y = y logb a logb x =y ` logb a = loga x You can use the change of base formula to find the logarithm of any number, such as log 5 2. You change it to either log 10 x or log e x, and use a calculator.

Example Find the value of log 5 2, correct to 2 decimal places.

Solution log 5 2 = You can use either log or In

log 2

log 5 Z 0.430676558 = 0.43

(by change of base)

Exponential equations You can also use the change of base formula to solve exponential equations such as 5 x = 7. You studied exponential equations such as 2 x = 8 in the Preliminary course. Exponential equations such as 2 x = 9 can be solved by taking logarithms of both sides, or by using the definition of a logarithm and the change of base formula.

Chapter 4 Exponential and Logarithmic Functions

181

Examples 1. Solve 5 x = 7 correct to 1 decimal place.

Solution    5 x = 7 Using the definition of a logarithm, this means: log 5 7 = x log 7

=x

(using change of base formula)

log 5 1.2 = x

You can use either log or ln.

If you do not like to solve the equation this way, you can use the logarithm laws instead. Taking logs of both sides: log 5 x = log 7 x log 5 = log 7 `

x=

n

Use loga x = n loga x

log 7

log 5 = 1.2 correct to 1 decimal place

2. Solve 4 y − 3 = 9 correct to 2 decimal places.

Solution 4y − 3 = 9 Using the logarithm definition and change of base: log 4 9 = y − 3 log 9 =y−3 log 4 log 9 +3=y log 4 4.58 = y Using the logarithm laws: Taking logs of both sides: log 4 y − 3 = log 9 (y − 3) log 4 = log 9 y−3= y=

log 9 log 4 log 9

+3 log 4 = 4.58 correct to 2 decimal place

182

Maths In Focus Mathematics Extension 1 HSC Course

4.6 Exercises 1. Use the change of base formula to evaluate to 2 decimal places. (a) log4 9 (b) log6 25 (c) log9 200 (d) log2 12 (e) log3 23 (f) log8 250 (g) log5 9.5 (h) 2 log4 23.4 (i) 7 – log7 108 (j) 3 log11 340 2. By writing each equation as a logarithm and changing the base, solve the equation correct to 2 significant figures. (a) 4 x = 9 (b) 3 x = 5 (c) 7 x = 14 (d) 2 x = 15 (e) 5 x = 34 (f) 6 x = 60 (g) 2 x = 76 (h) 4 x = 50 (i) 3 x = 23 (j) 9 x = 210 3. Solve, correct to 2 decimal places. (a) 2 x = 6 (b) 5 y = 15 (c) 3 x = 20

(d) 7 m = 32 (e) 4 k = 50 (f) 3 t = 4 (g) 8 x = 11 (h) 2 p = 57 (i) 4 x = 81.3 (j) 6 n = 102.6 4. Solve, to 1 decimal place. (a) 3 x + 1 = 8 (b) 53n = 71 (c) 2 x − 3 = 12 (d) 4 2n − 1 = 7 (e) 7 5x + 2 = 11 (f) 8 3 − n = 5.7 (g) 2 x + 2 = 18.3 (h) 37xk − 3 = 32.9 (i) 9 2 = 50 (j) 6 2y + 1 = 61.3 5. Solve each equation correct to 3 significant figures. (a) e x = 200 (b) e 3t = 5 (c) 2e t = 75 (d) 45 = e x (e) 3000 = 100e n (f) 100 = 20e 3t (g) 2000 = 50e 0.15t (h) 15 000 = 2000e 0.03k (i) 3Q = Qe 0.02t (j) 0.5M = Me 0.016k

Chapter 4 Exponential and Logarithmic Functions

183

Derivative of the Logarithmic Function Drawing the derivative (gradient) function of a logarithm function gives a hyperbola.

Example Sketch the derivative function of y = log 2 x.

Solution The gradient is always positive but is decreasing.

If y = log e x then

Proof

dx

1 = x

dy

dy

dx

1 dx dx dy Given  y = log e x Then   x = e y dx = ey dy dy 1 ` = dx e y 1 =x    

dy

=

=

1 dx

is a special result that

dy can be proved by differentiating from first principles.

184

Maths In Focus Mathematics Extension 1 HSC Course

Function of a function rule

If y = log e f (x), then

Proof Let     u = f (x) Then     y = log e u dy 1 = ` du u du = f l(x) Also dx dy dy du . = dx du dx 1 = u . f l(x) 1 . = f l(x) f (x)

Examples 1. Differentiate log e (x 2 − 3x + 1) .

Solution d 2x − 3 [loge (x 2 − 3x + 1)] = 2 dx x − 3x + 1 2. Differentiate log e

x+1 . 3x − 4

Solution x+1 3x − 4 = loge (x + 1) − loge (3x − 4)

Let y = loge dy dx

3 1 − x + 1 3x − 4 1 (3x − 4) − 3 (x + 1) = (x + 1) (3x − 4) 3x − 4 − 3x − 3 = (x + 1) (3x − 4) −7 = (x + 1) (3x − 4) =

dy dx

= f l(x) .

f l(x) 1 = f ( x) f ( x )

Chapter 4 Exponential and Logarithmic Functions

185

3. Find the gradient of the normal to the curve y = loge (x 3 − 5) at the point where x = 2.

Solution dy

3x 2 dx x 3 − 5 When   x = 2, dy 3 (2) 2 =     dx 2 3 − 5 m1 = 4    

=

The normal is perpendicular to the tangent i.e. m1 m2 = − 1 4m 2 = − 1 1 ` m2 = − 4 4. Differentiate y = log 2 x.

Solution y = log 2 x log e x = log e 2 1 = ´ log e x log e 2 dy 1 1 = ´ dx log e 2 x 1 = x log e 2 5. Find the derivative of 2x.

Solution 2 = e ln 2 ` 2 x = (e ln 2) x = e x ln 2 dy = ln 2e x ln 2 dx = ln 2 ´ 2 x = 2 x ln 2

This result comes from the Preliminary Course.

186

Maths In Focus Mathematics Extension 1 HSC Course

4.7 Exercises 1. Differentiate (a) x + log e x (b) 1 − log e 3x (c) ln (3x + 1) (d) loge (x 2 − 4) (e) ln (5x 3 + 3x − 9) (f) loge (5x + 1) + x 2 (g) 3x 2 + 5x − 5 + ln 4x (h) log e (8x − 9) + 2 (i) log e (2x + 4) (3x − 1) 4x + 1 (j) log e 2x − 7 (k) (1 + loge x) 5 (l) (ln x − x) 9 (m) (loge x) 4 (n) (x 2 + loge x) 6 (o) x log e x

6. Find the gradient of the normal to the curve y = log e (x 4 + x) at the point (1, log e 2). 7. Find the exact equation of the normal to the curve y = log e x at the point where x = 5. 8. Find the equation of the tangent to the curve y = log e (5x + 4) at the point where x = 3. 9. Find the point of inflexion on the curve y = x loge x − x 2 . 10. Find the stationary point on the ln x curve y = x and determine its nature.

(r) x 3 log e (x + 1)

11. Sketch, showing any stationary points and inflexions. (a) y = x − log e x (b) y = (log e x − 1) 3 (c) y = x ln x

(s) log e (log e x)

12. Find the derivative of log3 (2x + 5).

log e x x (q) (2x + 1) log e x (p)

ln x (t) x−2 (u)

2x

e loge x

(v) e x ln x (w) 5 (loge x) 2 2. If f (x) = log e 2 − x , find f l(1). 3. Find the derivative of log 10 x. 4. Find the equation of the tangent to the curve y = log e x at the point (2, log e 2). 5. Find the equation of the tangent to the curve y = log e (x − 1) at the point where x = 2.

13. Differentiate (a) 3 x (b) 10 x (c) 2 3x − 4 14. Find the equation of the tangent to the curve y = 4 x + 1 at the point (0, 4). 15. Find the equation of the normal to the curve y = log 3 x at the point where x = 3.

Chapter 4 Exponential and Logarithmic Functions

187

Integration and the Logarithmic Function 1 # dx x = # x dx = log e x + C

#

f l(x) f (x)

Integration is the inverse of differentiation.

dx = log e f (x) + C

Examples 1 1. Find the area enclosed between the hyperbola y = x , the x-axis and the lines x = 1 and x = 2, giving the exact value.

Solution

21 A = # x dx 1 = 7 loge x A 21 = loge 2 − loge 1 = loge 2

So area is log e 2 units2. 2. Find

#

x2 dx. x3 + 7

Solution

#

3x 2 x2 1 dx = # 3 dx 3 x +7 x +7 1 = loge (x 3 + 7) + C 3 3

continued

188

Maths In Focus Mathematics Extension 1 HSC Course

3. Find

#

x+1 dx. x2 + x + 4

Solution

#

2 (x + 1) x+1 1 dx = # 2 dx 2 x + 2x + 4 x + 2x + 4 2x + 2 1 = # 2 dx 2 x + 2x + 4 1 = loge (x 2 + 2x + 4) + C 2 2

4.8 Exercises 1. Find the indefinite integral (primitive function) of 2 (a) 2x + 5 4x (b) 2 2x + 1 5x 4 (c) 5 x −2 1 (d) 2x 2 (e) x 5 (f) 3x 2x − 3 (g) 2 x − 3x x (h) 2 x +2 3x (i) 2 x +7 x+1 (j) 2 x + 2x − 5 2. Find (a)

# 4x4− 1 dx

dx x+3 x2 dx (c) # 3 2x − 7 x5 dx (d) # 6 2x + 5 x+3 dx (e) # 2 x + 6x + 2 (b) #

3. Evaluate correct to 1 decimal place. 2 dx 2x + 5 5 dx (b) # 2 x + 1 7 x2 dx (c) # 3 1 x + 2 3 4x + 1 dx (d) # 0 2x 2 + x + 1 4 x − 1 dx (e) # 3 x 2 − 2x

(a)

#1

3

4. Find the exact area between the 1 curve y = x , the x-axis and the lines x = 2 and x = 3. 5. Find the exact area bounded by 1 , the x-axis and the curve y = x−1 the lines x = 4 and x = 7. 6. Find the exact area between the 1 curve y = x , the x-axis and the lines y = x and x = 2 in the first quadrant. 7. Find the area bounded by the x curve y = 2 , the x-axis and x +1 the lines x = 2 and x = 4, correct to 2 decimal places.

Chapter 4 Exponential and Logarithmic Functions

8. Find the exact volume of the solid formed when the curve 1 y= is rotated about the x-axis x from x = 1 to x = 3.

12. (a) Show that 3x + 3 1 2 . = + x2 − 9 x + 3 x − 3 3x + 3 (b) Hence find # 2 dx. x −9

9. Find the volume of the solid formed when the curve 2 y= is rotated about the 2x − 1 x-axis from x = 1 to x = 5, giving an exact answer.

13. (a) Show that

10. Find the area between the curve y = ln x, the y-axis and the lines y = 2 and y = 4, correct to 3 significant figures. 11. Find the exact volume of the solid formed when the curve y = log e x is rotated about the y-axis from y = 1 to y = 3.



x−6 5 =1− . x−1 x−1 x−6 (b) Hence find # dx. x−1

14. Find the indefinite integral (primitive function) of 3 2x − 1. 15. Find, correct to 2 decimal places, the area enclosed by the curve y = log 2 x, the x-axis and the lines x = 1 and x = 3 by using Simpson’s rule with 3 function values.

189

190

Maths In Focus Mathematics Extension 1 HSC Course

Test Yourself 4 1. Evaluate to 3 significant figures. (a) e 2 − 1 (b) log 10 95 (c) log e 26 (d) log 4 7 (e) log 4 3 (f) ln 50 (g) e + 3 5e 3 (h) ln 4 (i) e ln 6 (j) e ln 2

7. Find the volume of the solid formed if the area bounded by y = e 3x, the x-axis and the lines x = 1 and x = 2 is rotated about the x-axis.

2. Differentiate (a) e 5x (b) 2e 1 − x (c) log e 4x (d) ln (4x + 5) (e) xe x ln x (f) x (g) (e x + 1)10

9. Find the area enclosed between the curve y = ln x, the y-axis and the lines y = 1 and y = 3.

3. Find the indefinite integral (primitive function) of (a) e 4x x (b) 2 x −9 (c) e − x 1 (d) x+4 4. Find the equation of the tangent to the curve y = 2 + e 3x at the point where x = 0. 5. Find the exact gradient of the normal to the curve y = x − e − x at the point where x = 2. 6. Find the exact area bounded by the curve y = e 2x, the x-axis and the lines x = 2 and x = 5.

8. If log 7 2 = 0.36 and log 7 3 = 0.56, find the value of (a) log 7 6 (b) log 7 8 (c) log 7 1.5 (d) log 7 14 (e) log 7 3.5

10. (a) Use Simpson’s rule with 3 function values to find the area bounded by the curve y = ln x, the x-axis and the lines x = 2 and x = 4. (b) Change the subject of y = ln x to x. (c) Hence find the exact area in part (a). 11. Solve (a) 3 x = 8 (b) 2 3x − 4 = 3 (c) logx 81 = 4 (d) log6 x = 2 (e) 12 = 10e 0.01t 12. Evaluate (a)

#0

1

3e 2x dx

dx 3x − 2 2 2x 3 − x 2 + 5x + 3 dx (c) # x 1 (b) #

4

1

13. Find the equation of the tangent to the curve y = e x at the point (4, e 4). 14. Evaluate log 9 8 to 1 decimal place.

Chapter 4 Exponential and Logarithmic Functions

15. (a)  Find the area bounded by the curve y = e x, the x-axis and the lines x = 1 and x = 2. (b) This area is rotated about the x-axis. Find the volume of the solid of revolution formed. 16. Simplify (a) 5 loga x + 3 loga y (b) 2 log x k − log x 3 + log x p

18. Find the stationary points on the curve y = x 3 e x and determine their nature. 19. Use the trapezoidal rule with 4 strips to find the area bounded by the curve y = ln (x 2 − 1), the x-axis and the lines x = 3 and x = 5. 20. Evaluate to 2 significant figures (a) log 10 4.5 (b) ln 3.7

17. Find the equation of the normal to the curve y = ln x at the point (2, ln 2).

Challenge Exercise 4 1. Differentiate

loge x e 2x + x

.

2. Find # xe x dx, using the substitution u = x 2. 2

3. Find the exact gradient of the tangent to the curve y = e x + log x at the point where x = 1. e

4. If log b 2 = 0.6 and log b 3 = 1.2, find (a) log b 6b (b) log b 8 (c) log b 1.5b 2 5. Differentiate (e 4x + loge x) 9. 6. Find the shaded area, correct to 2 decimal places.

7. Find the derivative of loge

1 . 2x − 3

8. Use Simpson’s rule with 5 function values to find the volume of the solid formed when the curve y = e x is rotated about the y-axis from y = 3 to y = 5, correct to 2 significant figures. 9. Differentiate 5 x. 10. Find

# x2ex − 1 dx. 3

11. Show that

d 2 (x loge x) = x (1 + 2 loge x). dx

Hence evaluate

#1

3

2x (1 + 2 loge x) dx,

giving an exact answer.

12. Find

# 3 x dx.

13. (a)  Find the point of intersection of the curves y = log e x and y = log 10 x. (b) Find the exact equations of the tangents to the two curves at this point of intersection. (c) Find the exact length of the interval XY where X and Y are the y-intercepts of the tangents.

191

192

Maths In Focus Mathematics Extension 1 HSC Course

14. Use Simpson’s rule with 3 function values to find the area enclosed by the curve y = e 2x, the y-axis and the line y = 3, correct to 3 significant figures. x loge x . 15. Find the derivative of   ex 16. Use the substitution u = 3x 2 + 1 to find

# xe

3x 2 + 1

dx. −x

17. If y = e + e , show x

d2y dx 2

= y.

18. Use the substitution u = x − 2 to 3

evaluate

#0

2

x2 ex

3

−2

dx.

19. Prove

d2y 2

−4

dx 5x y = 3e − 2 .

dy dx

− 5y − 10 = 0, given

20. Find the equation of the curve that has f m(x) = 12e 2x and a stationary point at (0, 3). 21. Sketch y = loge (x − x 2). 22. A curve has

dy dx

2

= xe x and passes through



the point c 0,

1 m . Use the substitution 2



u = x 2 to find the equation of the curve.