1
1
Algebraic techniques
An electrician, a bank worker, a plumber and so on all have tools of their trade. Without these tools, and a good working knowledge of how to use them, it would be impossible for them to perform their jobs successfully.
MathsWorld Mathematical Methods Units 3 & 4
Algebra of polynomials
Polynomials
A polynomial is an algebraic expression of the form
where n is a positive integer or zero and the coefficients an, an− 1, …, and a0 are real numbers. The term anx
n
, an≠ 0, is called the leading term of the polynomial. The degree of the polynomial is n.
The remainder and factor theorems
The remainder theorem states that if a polynomial P(x) is divided by the linear expression (x−a) then the remainder is given by P(a).
If the remainder is 0 then (x−a) is a factor of P(x). The factor theorem states that if (x−a) is a factor of P(x) then P(a) = 0 and, conversely, if P(a) = 0 then (x− a) is a factor of P(x). Dividing by (bx− c) is the same as dividing by . In this case the remainder is given by .
E x a m p l e
1
If (x− 1) is a factor of P(x) = 3x3+kx2+bx− 5 and P(−2) =−45, find the values of k and b. Hence find the remainder when P(x) is divided by (x+ 1).
S o l u t i o n
(x− 1) is a factor of P(x) so P(1) = 0. 3 +k+b− 5 = 0
k+b= 2 (1)
P(−2) =−45
−24 + 4k − 2b − 5 =−45 4k − 2b =−16
2k − b =−8 (2)
3k =−6 (1) + (2)
k =−2
Substitute k =−2 into equation (1).
−2 + b = 2 b = 4
Hence P(x) = 3x3− 2x2+ 4x − 5. The remainder is given by P(−1).
P(−1) =−3 − 2 − 4 − 5
=−14
P x( ) anx n
an–1x
n–1
… a1x a0
+ + + +
=
b x c
b
---–
⎝ ⎠
⎛ ⎞
P c
b
---⎝ ⎠ ⎛ ⎞
t i p
The solve command can be used to solve for b and k.
Define P(x) then solve P(1) = 0 and P(−2) =−45 for b and k.
CAS 10.1, 10.3
1.1
Long division of polynomials and factorisation
Long division is a useful technique, particularly for factorising polynomials by hand.E x a m p l e
2
Factorise:
a x3+ 4x2− 7x − 10 b x3− 7x + 6
S o l u t i o n
a Let P(x) = x3+ 4x2− 7x − 10.
Use the factor theorem to find a factor. P(−1) = (−1)3+ 4(−1)2− 7(−1) − 10
= 0 So (x + 1) is a factor.
Now use long division to find the other factor. x2+ 3x − 10
x + 1
)
x3+ 4x2− 7x − 10−(x3+ x2) 3x2− 7x
−(3x2+ 3x)
−10x − 10
−(−10x − 10) 0 So, P(x) = x3+ 3x2− 4x − 10
= (x + 1)(x2+ 3x − 10)
= (x + 1)(x + 5)(x − 2)
b Let P(x) = x3− 7x + 6.
Use the factor theorem to find a factor. P(1) = (1)3− 7(1) + 6
= 0
t i p
If the coefficient of the leading term of
P(x) is 1, when using the factor theorem to find a factor of P(x) (i.e. finding a value of a for which P(a) = 0), the value of a
must be a factor of the constant term. In this case you would only test the values 1, −1, 2, −2, 5, −5 and 10, −10.
GC 2.2 CAS 2.2
A table of values can be used to find factors.
A table of values for P(x) =x3+ 4x2− 7x− 10 is shown.
From this table it can be seen that P(−5) =P(−1) =P(2) = 0.
t i p
Now use long division to find the other factor. x2+ x − 6
x − 1
)
x3+ 0x2− 7x + 6−(x3− x2) x2− 7x
−(x2− x)
−6x + 6
−(−6x + 6) 0 So, P(x) = x3− 7x + 6
= (x − 1)(x2+ x − 6)
= (x − 1)(x + 3)(x − 2)
Other techniques that do not involve the use of long division can also be used to factorise cubic polynomials.
E x a m p l e
3
Factorise:
a x3+ 5x2+ 6x b x3+ 4x2− 4x − 16
S o l u t i o n a x3+ 5x2+ 6x
= x(x2+ 5x + 6) (taking out a common factor first)
= x(x + 3)(x + 2) bx3+ 4x2− 4x − 16
= x2(x + 4) − 4(x + 4) (grouping terms)
= (x + 4)(x2− 4)
= (x + 4)(x + 2)(x − 2)
Equality of polynomials
Two polynomials P1(x) and P2(x) such thatP1(x) = anxn+ an − 1xn − 1+ an − 2xn − 2+ … + a1x + a0
P2(x) = bnxn+ bn − 1xn − 1+ bn − 2xn − 2+ … + b1x + b0
are equal for all values of x if and only if an = bn, an − 1 = bn − 1, …, a1 = b1, a0 = b0.
E x a m p l e
4
Find the values of a, b and k for which 8x3+ (a − 6)x2+ (b − a)x = kx3 for all values of x. S o l u t i o n
8x3+ (a − 6)x2+ (b − a)x = kx3
8x3+ (a − 6)x2+ (b − a)x = kx3+ 0x2+ 0x
t i p
When performing long division it is recommended to keep the same powers of x in vertical columns. In some cases it may be necessary to include zero terms to make this easier.
CAS 1.9
t i p
Equating coefficients gives k = 8 (x3) a − 6 = 0 (x2)
a = 6 b − a = 0 (x)
b = a
= 6
E x a m p l e
5
Factorise 2x3− 5x2− 4x + 3 without using long division.
S o l u t i o n
Use the factor theorem to find one factor. P(−1) = 2(−1)3− 5(−1)2− 4(−1) + 3
= 0
So (x + 1) is a factor. The other factor must be quadratic (i.e. of the form ax2+ bx + c). (x + 1)(ax2+ bx + c) = 2x3− 5x2− 4x + 3
Expanding gives
ax3+ bx2+ cx + ax2+ bx + c = 2x3− 5x2− 4x + 3 ax3+ (a + b)x2+ (b + c)x + c = 2x3− 5x2− 4x + 3 Equating coefficients of like powers of x:
a = 2, c = 3 a + b =−5 2 + b =−5 b =−7
So, 2x3− 5x2− 4x + 3 = (x + 1)(2x2− 7x + 3)
= (x + 1)(2x − 1)(x − 3)
exercise
1.1
1 Find the remainder when 2x3 + 3x2 − 4x+ 1 is divided by (x+ 2).
2 Use the remainder theorem to find the remainder when 2x4 −x2 + 3x− 2 is divided by (2x− 1).
3 When x4+ ax2 +bx− 2 is divided by (x− 1) the remainder is −2, and when it is divided by (x+ 1) the remainder is −6. Find the values of a and b.
4 x3+ax2+ 4x− 5 and x2+ 4x− 8 both have the same remainder when divided by (x− 1). Find the value of a.
5 The polynomial x3+px2+qx+ 6 is exactly divisible by (x+ 2) and (x− 3). Find the values of p and q.
6 Factorise:
a x3 − 2x2 − 11x+ 12 b x3+x2− x− 1 c 3x3+ x2 − 10x
d x3 − 2x2 − 9x+ 18 e 6x3 +x2− 5x− 2 f 4x3− 4x2− 15x+ 18
3
− 2
+ − 3
+ 2
− −
t i p
a and c can be found by inspection before expanding. Then it is only necessary to determine either the
x2 or x term in the expansion of the left hand side of the equation in order to find b.
7 If x3+ 4x2− 3x+ 2 = (x+ 1)(ax2+bx+c) +d for all values of x, find the values of a, b, c
and d.
8 If x3 + 4x2 − 12x+ 14 =x3+ (mx +n)2+ 5 for all values of x, find the values of m and n.
9 If x2 =a(x+ 2)2 +b(x+ 2) +c for all values of x, find the values of a, b and c.
Solution of polynomial equations
Solving an equation means finding the value(s) of the variable that makes the equation true.
Linear equations
A linear equation in one variable is an equation involving polynomials whose degree is 1. Thus 2x + 5 = 3 − 2x is a linear equation in the variable x.
Some equations reduce to linear equations when suitably transformed. For example, the equation can be expressed as x = 7(2x − 3).
Linear equations are usually solved by isolating the variable on one side of the equation.
E x a m p l e
6
Solve for x the equation 2(2x − 3) − 5 = 3(x + 2) + 5.
S o l u t i o n
Expand the bracketed terms and simplify. 4x − 11 = 3x + 11
Isolate the variable by subtracting 3x from both sides and adding 11 to both sides. x = 22
Check that x = 22 is the correct solution.
LHS = 2(44 − 3) − 5 = 77; RHS = 3(22 + 2) + 5 = 77 = LHS
E x a m p l e
7
Solve for x.
S o l u t i o n
Clear the fractions by multiplying both sides by (2x − 1)(x + 2). Thus:
3(x + 2) = 2(2x − 1) 3x + 6 = 4x − 2
8 = x Hence x = 8.
x 2x–3
--- = 7
3 2x–1
--- 2
x+2
---=
t i p
On a TI-89, exact solutions can be found using the solve command. For example, the solution to the equation is shown below.
3 2x–1
--- 2
x+2
Quadratic equations
The standard form of a quadratic equation is ax2+ bx + c = 0, where x is the variable and a,
b and c are constants with a ≠ 0. Methods for solving quadratic equations include factorisation,
completing the square, using the quadratic formula and numerical or graphical approximation. When quadratic equations are expressed as the product of linear factors, the null factor law can be used to find solutions.
E x a m p l e
8
Solve each of the following for x.
a 3x2− 48 = 0 b 2x2= 15 − x
c x2+ 6x − 3 = 0 d 3x2+ 2x − 6 = 0
S o l u t i o n
a 3x2− 48 = 0
3(x2− 16) = 0 (common factor)
3(x − 4)(x + 4) = 0 (difference of two squares)
x = 4, −4 (using the null factor law)
b 2x2= 15 − x
2x2+ x − 15 = 0 (taking all terms to one side) (2x − 5)(x + 3) = 0 (factorising by inspection)
x = (using the null factor law)
c x2+ 6x − 3 cannot be factorised by inspection so complete the square. x2+ 6x − 3 = 0
x2+ 6x + 9 − 9 − 3 = 0 (x + 3)2− 12 = 0
= 0
So x = .
d 3x2+ 2x − 6 cannot be factorised by inspection so use the quadratic formula.
If ax2+ bx + c = 0 then .
Here, a = 3, b = 2 and c =−6.
x =
=
=
=
Null factor law
If ab = 0, then either a = 0 or b = 0 or both a and b = 0.
5 2
---,–3
x+3 2 3–
( )(x+3+2 3) 3
– +2 3,–3–2 3
t i p
To avoid errors when cancelling, taking out a common factor in the numerator may be advantageous.
=
= =
2
– ± 2 19 6
---2(–1 ± 19) 6
---1
– ± 19
---x –b ± b
2
4ac – 2a ---=
2
– ± 4 4 3– ( )( )–6 6
---2
– ± 76
6
---2
– ± 2 19
6
---1
– ± 19
Equations reducible to quadratics
Some equations are reducible to quadratics using a suitable substitution, as is shown in the next example.
E x a m p l e
9
Solve for x: .
S o l u t i o n
Let so the equation becomes a2− 5a + 6 = 0.
Now factorise and solve: (a − 3)(a − 2) = 0 a − 3 = 0 or a − 2 = 0
So, or .
x2− 3x − 1 = 0 or x2− 2x − 1 = 0 Using the quadratic formula:
x = or
x = or
CAS 10.2 GC 2.3 CAS 2.3
t i p
The solve command can be used to find exact or approximate solutions to a quadratic equation. Recall that in exact or auto mode, press to get approximate solutions.
♦ ENTER
Approximate solutions can be found using a graphics calculator.
To find the approximate solutions to 3x2+ 2x− 6 = 0, plot the graph of y= 3x2+ 2x− 6 and use the zero
command in the CALC menu to find the x-axis intercepts. The screenshot on the right shows the zero to the right
of the origin. [−10, 10] by [−10, 10]
t i p
x 1
x ---–
⎝ ⎠
⎛ ⎞2 5 x 1 x ---–
⎝ ⎠
⎛ ⎞
– +6 = 0
x 1
x ---–
⎝ ⎠
⎛ ⎞2 5 x 1 x ---–
⎝ ⎠
⎛ ⎞
– +6 = 0
a x 1
x ---– =
x 1
x
---– –3 = 0 x 1
x
---– –2 = 0
3 ± ( )–3 2–4 1( )( )–1 2
--- x 2 ± ( )–2
2
4 1( )( )–1 –
2
---=
3 ± 13
2
Equations involving higher degree polynomials
A polynomial equation has the form anxn+ an − 1xn − 1+ … + a1x + a0 = 0. Techniques for
solving linear (degree 1) and quadratic (degree 2) equations have already been considered. With cubic (degree 3) and higher degree polynomial equations it is necessary to use either factorisation or technology to find solutions.
E x a m p l e
1 0
Solve each of the following for x.
a x3+ 3x2− 5x − 15 = 0 b x5− 3x3+ 4x = 0 c x3= x − 1
S o l u t i o n
a x3+ 3x2− 5x − 15 = 0
Factorising the left-hand side by grouping gives:
x2(x + 3) − 5(x + 3) = 0 (x + 3)(x2− 5) = 0
= 0 x = b x5− 3x3+ 4x = 0
Factorising gives:
x(x4− 3x2+ 4) = 0 x(x2− 4)(x2+ 1) = 0 x(x − 2)(x + 2)(x2+ 1) = 0
x = 0, 2, −2
c x3= x − 1
x3−x+ 1 = 0
However, x3−x+ 1 cannot be factorised into rational factors, so the equation is solved using technology. A graphical approach can be used:
Sketch the graph of y=x3−x+ 1 and find any zeros or x-intercepts. From the screenshot on the right, there is exactly one solution, x=−1.325, correct to 3 decimal places.
Alternatively, find the point of intersection of the graphs of y=x3 and y=x− 1, as shown in the screenshot on the right.
t i p
The factor theorem could also be used.
x+3
( )(x– 5)(x+ 5) 3
– , 5,– 5
GC 2.3, 2.4 CAS 2.3, 2.4
[−2, 2] by [−2, 2]
[−2, 2] by [−5, 1]
On a TI-89, the solve command can be used. In this example, the technology cannot provide an exact solution, so a numerical approximation is given.
exercise
1.1
10 Solve each of the following for x.
a 2(x+ 1) = 3(x+ 2) − 14 b 9(3x+ 4) − 55x= 64
c d
e (x+ 2)2= (x− 1)2+ 6 f (x+ 1)(x− 2) − 2(x− 2) = (x+ 1)2
11 Find all the exact solution(s) to each of the following equations.
a (x+ 3)(x− 3) + 5 = 0 b 4x2 + 6x− 2 = 0 c x2 + 6x+ 7 = 0
d 5x2− 16x− 12 = 0 e (x+ 5)2= 15 − 10x f 2x4= 32x2
g x4 + 5x2 − 36 = 0 h i x3 − 2x2 − 7x− 4 = 0
j 2x3− 3x2− 11x+ 6 = 0 k x4−x3+ x− 1 = 0 l x6 = 8 − 7x3
12 Solve each of the following equations, giving answers correct to 3 decimal places.
a x2 − 2x− 5 = 0 b 2x2 = 5 − 2x c x3 − 3x2 +x− 1 = 0
d x4 − 4x2 − 2 = 0 e x3=−3 +x f x6 = 5 + 2x3
Binomial expansions
Pascal’s triangle
Pascal’s triangle is named after the French mathematician Blaise Pascal. Although Pascal was not responsible for its discovery (it was discovered by the Chinese about 500 years before his birth), he was the first mathematician to perform any extensive study of it.
CAS 10.2
continued
5x+6 4
--- 100–4x
3
---–6
= 6
x
---+4 15
x ---–3
=
16 x 3 x ---+
⎝ ⎠ ⎛ ⎞2
– = 0
Constructing Pascal’s triangle
.At the top we place the number 1. This is referred to as row zero.
.At each end of every subsequent row, place the number 1.
.Each other element is found by adding the two numbers diagonally above it in the
Rows 0 to 6 of Pascal’s triangle are shown below.
For example, in row 4, the second element 4 is obtained by adding the numbers 1 and 3 diagonally above it. Similarly, the third element 6 is found by adding the numbers 3 and 3.
Pascal’s triangle and binomial expansions
Consider the expansion of (a + b)n for n = 0, 1, 2, 3, 4 and 5. Since the bracketed expression consists of two terms, it is said to be binomial and hence its expansion is known as a binomial expansion.
(a + b)0= 1 (a + b)1= a + b
(a + b)2= a2+ 2ab + b2
(a + b)3= a3+ 3a2b + 3ab2+ b3
(a + b)4= a4+ 4a3b + 6a2b2+ 4ab3+ b4
(a + b)5= a5+ 5a4b + 10a3b2+ 10a2b3 + 5ab4+ b5
It is important to note the following features of each expansion. .There are n + 1 terms in each expansion.
.In each term the sum of the powers of a and b is n.
.In successive terms, the powers of a decrease from n to 0 and the powers of b increase from 0 to n.
.The coefficients of the terms in each expansion relate to the rows of Pascal’s triangle as shown below.
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
(a + b)0 1
(a + b)1 1 1
(a + b)2 1 2 1
(a + b)3 1 3 3 1
(a + b)4 1 4 6 4 1
(a + b)5 1 5 10 10 5 1
E x a m p l e
1 1
Use Pascal’s triangle to expand:
a (x − 3)4 b (2x − 3)6 c
S o l u t i o n
a To expand (x − 3)4 we obtain the coefficients from the 4th row of Pascal’s triangle.
So the expansion of (a + b)4 is (a + b)4= a4+ 4a3b + 6a2b2+ 4ab3+ b4. Here a = x, b = −3.
(x − 3)4= x4+ 4x3(−3) + 6x2(−3)2+ 4x(−3)3+ (−3)4
= x4− 12x3+ 54x2− 108x + 81
b To expand (2x − 3)6 we obtain the coefficients from the 6th row of Pascal’s triangle.
So the expansion of (a + b)6 is
(a + b)6= a6+ 6a5b + 15a4b2+ 20a3b3+ 15a2b4+ 6ab5+ b6 Here a = 2x, b = −3.
(2x − 3)6= (2x)6+ 6(2x)5(−3) + 15(2x)4(−3)2+ 20(2x)3(−3)3+ 15(2x)2(−3)4+ 6(2x)(−3)5+ (−3)6
= 64x6− 576x5+ 2160x4− 4320x3+ 4860x2− 2916x + 729
c To expand we obtain the coefficients from the 3rd row of Pascal’s triangle.
So the expansion of (a + b)3 is (a + b)3= a3+ 3a2b + 3ab2+ b3
Here a = 2x, b = .
=
=
1 4 6 4 1
1 6 15 20 15 6 1
1 3 3 1
2x 1
x ---–
⎝ ⎠
⎛ ⎞3
2x 1
x ---–
⎝ ⎠
⎛ ⎞3
1 x
---2x 1
x ---–
⎝ ⎠
⎛ ⎞3 ( )2x 3
3 2x( )2 1 x ---–
⎝ ⎠
⎛ ⎞ 3 2x( ) 1 x ---–
⎝ ⎠ ⎛ ⎞2 1
x ---–
⎝ ⎠ ⎛ ⎞3
+ + +
8x3–12x 6
x
--- 1
x3 ---– +
CAS 10.4
t i p
Expansion of (
a
+
b
)
nRecall from Year 11 Combinatorics that . The elements in Pascal’s triangle can be expressed as combinations, as shown below.
For example, . This agrees with the values found previously.
Using combinations, the expansion for a high power of (a + b)n can be found without having to write out all the previous rows of Pascal’s triangle.
n r ⎝ ⎠ ⎛ ⎞ nC
r
n!
n–r
( )!r! ---= = 0 0 ⎝ ⎠ ⎛ ⎞ 1 0 ⎝ ⎠ ⎛ ⎞ 1
1 ⎝ ⎠ ⎛ ⎞ 2 0 ⎝ ⎠
⎛ ⎞ 2
1 ⎝ ⎠ ⎛ ⎞ 2
2 ⎝ ⎠ ⎛ ⎞ 3 0 ⎝ ⎠ ⎛ ⎞ 3
1 ⎝ ⎠ ⎛ ⎞ 3
2 ⎝ ⎠ ⎛ ⎞ 3
3 ⎝ ⎠ ⎛ ⎞ 4 0 ⎝ ⎠ ⎛ ⎞ 4
1 ⎝ ⎠
⎛ ⎞ 4
2 ⎝ ⎠ ⎛ ⎞ 4
3 ⎝ ⎠
⎛ ⎞ 4
4 ⎝ ⎠ ⎛ ⎞ 5 0 ⎝ ⎠ ⎛ ⎞ 5
1 ⎝ ⎠ ⎛ ⎞ 5
2 ⎝ ⎠ ⎛ ⎞ 5
3 ⎝ ⎠ ⎛ ⎞ 5
4 ⎝ ⎠ ⎛ ⎞ 5
5 ⎝ ⎠ ⎛ ⎞ 4 1 ⎝ ⎠ ⎛ ⎞ 4!
3!1!
--- 4 4
2 ⎝ ⎠ ⎛ ⎞
, 2---4!2!! 6
= = = =
Binomial expansion
In general:
(a + b)n=nC0anb0+nC1an− 1b1+ … +nCran−rbr+ … +nCn −1a1bn− 1+nCna0bn = an+nC1an− 1b + … +nCran−rbr+ … +nCn −1abn− 1+ bn
where n ∈ N.
The (r + 1)th term, or general term, is given by nCran−rbr, r = 0, 1, 2, …, n.
GC 1.8 CAS 1.8
t i p
To calculate a conbination with the TI-83/84 or TI-89 calculator, access the combination symbol nCr from the MATH:PRB menu or the CATALOG. For example, here is the calculation of .9C6
E x a m p l e
1 2
Expand (3x + y)5 using (a + b)n= an+n
C1an− 1b +nC2an− 2b2+ … + bn
S o l u t i o n (3x + y)5= (3x)5+5
C1(3x)4y +5C2(3x)3y2+5C3(3x)2y3+5C4(3x)y4+ y5 = 243x5+ 5(81x4)y + 10(27x3)y2+ 10(9x2)y3+ 5(3x)y4+ y5
= 243x5+ 405x4y + 270x3y2+ 90x2y3+ 15xy4+ y5
exercise
1.1
13 State the number of terms in the expanded form of:
a (x+ 4)7 b (2x− 5)10 c (1 − 4z)14
14 Use Pascal’s triangle to expand:
a (x+ 3)5 b (x− 2)4 c (x− 1)5 d (2x + 1)6
e (3x− 2)5 f (1 − 3x)4 g (2t+ 3y)4 h 15 Use (a+b)n= an+nC1an − 1b +nC2an − 2b2+ … +bn to expand:
a (2y −x)6 b (x2− y)4 c (x2+x)3 d (x2 −y2)5
e f g h
Find the 4th term in the expansion of each of the following.
a (x− 2)8 b (2x− 3t)4
continued
2x 1 x ---–
⎝ ⎠
⎛ ⎞6
x
2
---+2
⎝ ⎠ ⎛ ⎞4
x 1 x ---+
⎝ ⎠ ⎛ ⎞5
2
x2
---–x
⎝ ⎠ ⎛ ⎞7
2x–1
( )6
1.1The gr
eatest coef
ficient
CD
SAC analysis task
In this analysis task, you will be asked to determine a general expression for the greatest coefficient in the expansion of (1 + x)n. Note that the binomial coefficients are all positive.
Part 1 Gathering the tools for analysis
Let Tr + 1 be the (r + 1)th term and Tr be the rth term in the expansion of (1 + x)n for 1 ≤ r ≤ n.
a Write down general expressions for Tr + 1 and Tr.
b Find a general simplified expression for the ratio of the coefficients of Tr + 1 and Tr.
c Under what condition relating n and r must the coefficient of Tr + 1 always be greater
than or equal to the coefficient of Tr?
Part 2
By using Pascal’s triangle and the results from the previous part, determine a general expression for the greatest coefficient in the expansion of (1 + x)n. Explain your reasoning carefully and illustrate your findings with some specific examples.
analysis task 1—
1.2Appr
oximations using binomial expansions
CD
SAC analysis task
Expansions of binomial expressions are not limited to positive integer powers. The expansion of (1 + x)−1 can be used to approximate reciprocals of numbers close to 1; the expansion of can be used to approximate square roots of numbers close to 1.
Part 1
(1 + x)–1= 1 − x + x2− x3+ …, where the nth term is (−x)n − 1
a Write down the next four terms in the series.
b i What is the value of x in this expansion to approximate ?
ii Use a calculator to evaluate and write down the result correct to 4 decimal places.
iiiUse the binomial expansion above with 4 terms to approximate . To how many decimal places is this accurate?
ivHow many terms are required to approximate correct to 4 decimal places?
Part 2
where the nth term is .
a Simplify this expression.
b What are the next three terms in this series in simplest form?
c Write down the value of x if this expansion is used to approximate:
i ii iii iv
v vi vii viii
d Complete the following table.
i In each case, how many terms are required for 4 decimal place accuracy?
ii As each extra term is added the approximation approaches the actual value. This is referred to as convergence. How does the value of x affect the rate of convergence?
x
Value from calculator or Excel
Approximation using
1 term 2 terms 3 terms 4 terms 5 terms 6 terms 7 terms
1+x ( )1 2⁄
1 1.3 ---1 1.3 ---1 1.3 ---1 1.3
---1+x
( )1 2⁄ 1 1
2
---x 1
2
---⎝ ⎠ ⎛ ⎞ 1
2
---–
⎝ ⎠ ⎛ ⎞x2
2
--- 1
2
---⎝ ⎠ ⎛ ⎞ 1
2
---–
⎝ ⎠ ⎛ ⎞ 3
2
---–
⎝ ⎠ ⎛ ⎞x3
6 --- …, + + + + = 1 2 ---⎝ ⎠ ⎛ ⎞ 1
2
---–
⎝ ⎠ ⎛ ⎞ 3
2
---–
⎝ ⎠ ⎛ ⎞ 5
2
---–
⎝ ⎠
⎛ ⎞ … xn–1
n–1
( )!
---,n>1
(n− 1) terms
1.4 0.6 1.3 0.7
1.2 0.8 1.1 0.9
1+x
1.4 0.6 1.3 0.7 1.2 0.8 1.1 0.9
analysis task 2—
approximations using binomial expansions
SAC
Systems of simultaneous
linear equations
A set of two or more equations in the same variables is referred to as a system of equations. In this section we will consider systems of linear equations.
Systems of linear equations in two unknowns
Recall that solving simultaneous linear equations in two unknowns algebraically involves obtaining a single linear equation in one variable by eliminating the other variable.
E x a m p l e
1
Solve these simultaneous linear equations using an algebraic method: y = 2x − 3 and 3x + 2y = 8
S o l u t i o n
y = 2x − 3 (1)
3x + 2y = 8 (2)
Substitute 2x − 3 for y in equation (2). 3x + 2(2x − 3) = 8
3x + 4x − 6 = 8 7x = 14
x = 2
Substitute x = 2 in equation (1). y = 2(2) − 3
y = 1
So the solution is x = 2 and y = 1.
E x a m p l e
2
Solve these simultaneous linear equations: x + 5y =−13 and 2x − y = 7
S o l u t i o n
These equations can be solved algebraically.
x + 5y =−13 (1)
2x − y = 7 (2)
x + 5y =−13 (1)
10x − 5y = 35 (3) = (2) × 5 11x = 22 (1) + (3)
x = 2
Substitute x = 2 in equation (2). 4 − y = 7
y =−3
So the solution is x = 2 and y =−3.
t i p
It is a good idea to check your answer. Substitute x= 2 and y= 1 into the left-hand side of equation (2). LHS = 3(2) + 2(1) = 8 = RHS.
1. 2
Simultaneous linear equations can also be solved graphically by finding the point of intersection of the straight lines corresponding to each equation.
They can also be solved using matrix methods on a graphics calculator. The command rref uses elimination methods to solve the equations. To use this method, first write each equation in the form ax + by = c. Then create a matrix whose elements are the coefficients of x and y and the constant terms in the equations.
x + 5y =−13 2x − y = 7
To solve the equations, apply the command rref (found in the MATRIX editor or in the
CATALOG) to the matrix above. The matrix is entered by entering each row enclosed in square brackets and each entry in the row separated by commas. An extra set of square brackets is required around the rows. The result when rref is applied is shown in the TI-83/84 screenshot below left and the TI-89 screenshot below right.
The rref command has transformed the original pair of equations into: 1x + 0y = 2
0x + 1y = −3
So the solution is x = 2, y = −3. GC 2.4
CAS 2.4
[−10, 10] by [−10, 10]
1 5–13 2–1 7 ⇒
GC 8.2 CAS 8.2
GC 8.1 CAS 8.1,
10.2
t i p
The matrices can be stored first and then used. On the TI-83/84 a matrix name, like [A], from the list of matrix names must be used. On the TI-89 any alpha location can be used. Note that on the TI-89, an alternative approach to solving simultaneous equations is to use the solve
command as shown in the third screenshot below.
Parallel and coincident lines
The graphs of the two equations are straight lines. If the graphs are of parallel lines, which do not intersect, there will be no solutions to the simultaneous equations. If the graphs are coincident, that is, the same straight line, there will be an infinite number of solutions to the simultaneous equations.
E x a m p l e
3
Solve the simultaneous equations 3x − 2y = 3 and 6x − 4y = k if:
a k = 6 b k = 0
S o l u t i o n a If k = 6, then:
3x − 2y = 3 (1) 6x − 4y = 6 (2)
6x − 4y = 6 (3) = (1) × 2 6x − 4y = 6 (2)
The graphs of these two equations are the same straight line, that is, the two lines are coincident. Any point on the first line will also lie on the second. There is an infinite number of solutions to the simultaneous equations consisting of the points lying on the line. Let y = t. Then:
6x − 4t = 6 6x = 6 + 4t
x =
So, for any real value of t, and y = t satisfy the equations.
b If k = 0, then: 3x − 2y = 3 (1) 6x − 4y = 0 (2)
6x − 4y = 6 (3) = (1) × 2 6x − 4y = 0 (2)
Subtracting these equations gives 0 = 6. This is impossible, so these simultaneous equations have no solution. This is because the graphs of two original equations are parallel lines that never intersect.
1 2
3
---t
+
x 1 2
3
---t
+ =
y
x
2
0
–2
–4 4
2 –2
–4 4
6x – 4y = 0
exercise
1.2
1 Solve each of the following pairs of simultaneous equations by hand then check your answers using technology.
a 3x+ 2y= 4 b = 4 c y= 14 − 4x
x− 3y= 5
= 10 3x+ 2y= 8
d = 3 e = 11 f x= 3 −y
5x+ 3y= 17 = 22 = 8
2 Find the value(s) of k for which the simultaneous equations have:
i no solution ii an infinite number of solutions
a x+ y= 3 b x−y= k c 4x+ 9y=k
2x+ 2y=k −3x+ 3y= 6 0.8x+ 1.8y= 2
3 A straight line has equation y=mx +c. Use simultaneous equations to find the equation of the straight line that passes through the points (−2, 11) and (3, −4).
4 A parabola that passes through the points (−1, 11) and (2, 5) has equation y=x2+bx+c. Use simultaneous equations to find the values of b and c.
Simultaneous linear equations in more than two
unknowns
Situations often arise in which there are three or more linear equations with corresponding numbers of unknowns. Such systems of equations can be solved by hand, eliminating one variable at a time until a single variable remains and solving the resulting linear equation. The values of the other variables can then be found by substitution. Systems of equations are more easily solved using technology.
E x a m p l e
4
Solve the following system of equations, using:
a algebra (by hand) b technology.
x − 2y + 3z = 3 x + 4y + 2z = 4 2x − 2y + z = 4
S o l u t i o n
a x − 2y + 3z = 3 (1) x + 4y + 2z = 4 (2) 2x − 2y + z = 4 (3)
2x − 4y + 6z = 6 (3) = (1) × 2 x + 4y + 2z = 4 (2)
4x − 4y + 2z = 8 (4) = (3) × 2
3 2
---x–y
1 2
---x 3
4
---y +
x y
2
---+ x
3
--- y
2
---+
x y
3
---–
1 3
---x 2
3
---y –
Now y can be eliminated. 3x + 8z = 10 (5) = (3) + (2) 5x + 4z = 12 (6) = (4) + (2) 3x + 8z = 10 (5)
10x + 8z = 24 (7) = (6) × 2 7x = 14
x = 2
Substitute x = 2 in equation (6) to find z. 10 + 4z = 12
4z = 2
z =
Substitute x = 2 and in equation (2) to find y. 2 + 4y + 1 = 4
4y = 1
y =
The solution is x = 2, and .
b First create a matrix whose elements are the coefficients of the variables and the constant terms in the equations.
Now use the rref command.
This has transformed the equations into 1x + 0y + 0z = 2
0x + 1y + 0z =
0x + 0y + 1z =
Read off the solution: x = 2, and .
1 2
---z 1
2 ---=
1 4
---y 1
4
---= z 1
2 ---=
1 –2 3 3
1 4 2 4
2 –2 1 4
GC 8.2 CAS 8.2
1 4 ---1 2
---t i p
The matrix can be stored first, then rref applied to the stored matrix.
y 1
4
---= z 1
Matrix representation of systems of linear equations
The system of equations in example 4 can be written in the form of a matrix equation as follows.x − 2y + 3z = 3
x + 4y + 2z = 4 ⇒
2x − 2y + z = 4
A X B
The matrix form on the right is written as AX = B, where A is the matrix formed by the coefficients of the variables, X is a column matrix containing the variables and B is a column matrix formed by the constants on the right hand side of the equations.
To recover the original equations from the matrix equation, start by multiplying the first row of A by the column matrix X and equate the result to the first entry in B:
(1)(x) + (−2)(y) + (3)(z) = 3, i.e. x − 2y + 3z = 3.
The remaining equations can be recovered in the same way.
When written in this form the equations can be solved using the command simult, as shown in the following screenshots. The command is most easily accessed via the CATALOG.
The role of parameters
A system of linear equations in which there are fewer equations than unknowns can have infinitely many solutions. The solutions are then given in terms of a parameter, another variable. Then substitution of a particular value for the parameter gives one particular solution for the other variables.
In some cases, a parametric solution occurs for an obvious geometrical reason, as in example 3 part a, where the two equations represent coincident lines.
E x a m p l e
5
Solve the system of equations: x + 4z = 13
2x − 2y − 7z =−19 4x − 2y + z = 7 S o l u t i o n
First create a matrix whose elements are the coefficients of x and y and the constant terms in the equations.
CAS
only
1 –2 3
1 4 2
2 –2 1 x y z
3 4 4
=
CAS 8.2
1 0 4 13
2 –2 –7 –19
Now use the rref command, as shown on the right.
The resulting matrix shows x + 4z = 13 and ,
so there is an infinite set of solutions.
This set of solutions can be written in terms of a parameter. Let z = t.
x + 4t = 13 and =
x = 13 − 4t
y =
The solutions are x = 13 − 4t, , z = t, where t ∈ R.
If t = 0, the solution is x = 13, , z = 0. If t = 1, then x = 9, y = 15, z = 1.
Similarly, further specific solutions can be found by assigning t any real number value.
Inconsistent systems
Sometimes there are no possible solutions to a system of linear equations. In such cases, the system is said to be inconsistent.
E x a m p l e
6
Solve the system of equations: 3x − y − 4z = 11
x + 4y + 3z = 10 2x + y − z =−1
S o l u t i o n
First create a matrix whose elements are the coefficients of x and y and the constant terms in the equations.
Now use the rref command.
The last row of this matrix shows 0x + 0y + 0z = 1. This cannot be true, so the system of equations has no solution. Alternatively, using the solve command returns the value false, which again means that the system is inconsistent.
y 15
2
---z
+ 45
2 ---= CAS 8.2
y 15
2
---t
+ 45
2 ---15 2
---(3–t)
y 15
2
---(3–t)
=
y 45
2 ---=
CAS 10.3
t i p
Using the solve command gives x=−(4@1 − 13), , z= @1.
The symbol @1 stands for a general constant or parameter. It is more usual to use a letter such as t, as in example 5.
y –15(@1–3)
2
---=
3 –1 –4 11
1 4 3 10
exercise
1.2
5 Solve the following systems of linear simultaneous equations by hand. Check your answers using technology.
a x+y +z= 3 b x+ 2y−z= 2 c 2x+y +z= 10
4x+ 2y +z=6 2x−y+ 2z= 5 x+ 5y − 3z=−7 9x+ 3y +z= 13 x+ y−z= 1 x+ 2y + 3z= 23
d 5x+z= 2 e x− 2y + 2z= 8 f x+ 3y + 2z= 1
y +z=5 3x− 2y+ z=8 2x−y− z= 6
−x+ y=1 2x+y− z=1 3x+y − 4z=−1
6 A parabola with equation y=ax2+bx+c passes through the points (−1, −10), (2, −1) and (3, −6). Find the values of a, b and c.
7 For each of the following:
i determine whether the system has a unique solution or an infinite number of solutions or no solution
ii find the solutions to those systems of simultaneous equations for which solutions exist.
a x+ 3y+ 4z= 2 b x+ 4y + 3z= 10 c x−y +z= 6 2x+y +z=5 2x+y− z=−1 x− 4y + 5z= 7
x− y− 2z=4 3x−y − 4z= 11 x+ 2y − 3z= 5
d 3x+ 2y− 5z= 4 e 2x− 3y+ z=−1 f 3x− 3y− z= 1 5x+ 3y− 8z= 6 x+ 2y− z=0 2x− 3y+ z= 1 4x+ 2y− 6z= 5 3x −y + 2z= 3 x− 2z= 0
CAS 8.2
Warning
Interpreting error messages
The alternative matrix method of solving systems of equations using the command
simult only works if there is a single (unique) solution to the simultaneous system.
The screenshots below show what happens when this method is applied in examples 5 and 6.
The error message means that there is a problem with the matrix stored as a. So this method is unable to deal with examples in which there are infinitely many solutions, or there are inconsistent equations. In such cases the rref or solve commands are more informative.
continued
8 Express the following systems of equations as matrix equations and use technology to solve them.
a x− 2y− 5z −w= 0 b x− 3y+ 6w = 1 3x− 5y− z− 3w= 1 x− 5y +z−w = 1 2x− 8y− 2z− 10w=−4 y +z+w = 1 2x− 5y + 2z− 5w=−1 x+y+ z= 0
9 A curve passes through the points (−1, −9), (1, −5) (2, −3) and (3, 7). If the curve has equation y =ax3+bx2+ cx+ d, find the values of a, b, c and d.
10 Determine the values of n for which the following system of equations has no solution. 2x+ (n+ 3)y− (n+ 4)z= 1
2y+ (n+ 4)z= 2
x − 2y +z= 1
11 Five numbers add to zero. The first is equal to the sum of the second and the fourth. The third is equal to the sum of the fourth and the fifth. The sum of the first two numbers is 2 more than the fifth number. The fifth number is three times the sum of the third and fourth numbers. Find the five numbers.
Number
sense with the spence
In 1976, a Los Angeles secretary officially married a 50-pound rock. The ceremony was attended by more than 20 people, at least 1 of whom, we can
Algebra of exponential and
logarithmic functions
Definition of a logarithm
Recall that if a > 0 and ax= y then logay = x. This means that x is the power to which a is raised in order to obtain y. For example, log39 = 2, since 32= 9.
Laws of logarithms
The laws of logarithms can be deduced from the index laws.
Logarithm of a product (first log law)
loga(xy) = logax + logay, where x, y > 0.
(In words, the log of a product is equal to the sum of the logs.) Let m = logax and n = logay.
x = am and y = an (definition of a logarithm)
xy = am× an
= am + n (1st index law)
loga(xy) = m + n (logarithm form)
= logax + logay
Logarithm of a quotient (second log law)
loga = logax − logay, where x, y > 0.
(In words, the log of a quotient is equal to the difference of the logs.) Let m = logax and n = logay.
x = am and y = an (definition of a logarithm)
=
= am − n (2nd index law)
loga = m − n (logarithm form)
= logax − logay
Logarithm of a power (third log law)
loga(xn) = nlogax, where x > 0.(In words, the log of a power is equal to the product of the power and the log.) Let m = logax.
x = am (definition of a logarithm)
xn = (am)n
= amn (3rd index law)
loga(xn) = mn (logarithm form)
x y
---⎝ ⎠ ⎛ ⎞
x y
--- a
m
an
---x y
---⎝ ⎠ ⎛ ⎞
1.3
1. 3
Two special cases of the third law are important. If n = 0, loga(x
0
) = 0, i.e. loga1 = 0. Of course, this also follows from the fact that a 0
= 1. If n =−1, loga(x−1) =−logax, i.e. loga = −logax.
Changing the base of logarithms
When a decimal approximation to a logarithm is required, the change of base rule is useful, together with a calculator that will evaluate logarithms with base 10.
Suppose we need an approximation for logax. Let y = logax, so that x = ay.
Take logarithms (base b) of each side: logbx = logbay
= y logba = (logax)(logba) Therefore logax = .
In the special case where b = 10, we have:
.
E x a m p l e
1
Correct to 3 decimal places, log35 = 1.465 and log32 = 0.631. Use these to find an
approximate value for the following.
a log310 b log340 c d log345
S o l u t i o n
a log310 = log3(5 × 2) b log340 = log3(8 × 5) = log35 + log32 = log38 + log35
= 1.465 + 0.631 = log323+ log35
= 2.096 = 3 log32 + log35
= 3 × 0.631 + 1.465
= 3.358
c = log34 − log35 d log345 = log3(9 × 5)
= log322− log35 = log39 + log35
= 2 log32 − log35 = 2 + 1.465
= 2 × 0.631 − 1.465 = 3.465
=−0.203
1 x
---Comparison of index and logarithm laws
Index form Log form
.First law: am× an= am +n loga(xy) = logax + logay
.Second law: = am −n loga = logax − logay
.Third law: (am)n= amn loga(xn) = n logax
.Special value: a0= 1 loga1 = 0
am an
--- x
y
--⎝ ⎠ ⎛ ⎞
logbx
logba
---logax
log10x
log10a ---=
log345
---E x a m p l e
2
Evaluate
S o l u t i o n
=
= = 3
E x a m p l e
3
Express a in terms of b if:
a log10a = 2 log10b + 1 b log4a = 0.5b + 3
S o l u t i o n
a log10a = 2 log10b + 1 b log4a = 0.5b + 3
= log10b2+ log1010 a = 40.5b+ 3
= log1010b2 = 40.5b× 43
a = 10b2 = (40.5)b× 64
= 64 × 2b
E x a m p l e
4
Find the value of log1113, correct to 3 decimal places.
S o l u t i o n
Use the change of base rule to change the logarithm base 11 to base 10.
logax =
log1113 =
= 1.070 (correct to 3 decimal places) log58
log52
---log58
log52
--- log52
3
log52
---3 log52
log52
---log10x
log10a
---log1013
log1011
---GC 1.1 CAS 1.1
Let the calculator do all the work by entering log(13)/log(11). This gives 1.069667. If you find log13 and log11 and round each to 3 decimal places before dividing, the calculator gives 1.069164, leading to an answer of 1.069 which is not correct to 3 decimal places. On a TI-89, log (base 10) is available from the CATALOG.
t i p
exercise
1.3
1 Evaluate each of the following without the use of a calculator.
a log464 b log93 c
d e f log318 − log32
g log69 + log624 h 2 log26 − log29 i
2 Express each of the following as a single logarithm.
a log5x− 3 log5(x+ 1) b 2 log103 + 5 log10x
c 2 log3(x − 1) − log34 − log3x d
e f
3 If logax= 0.3 and logay= 0.2, find:
a loga(xy) b c d 2 logaxy
3
4 If log52 = 0.43 and log53 = 0.68 evaluate each of the following.
a log54 b log515 c log560
d log524 e log540 f
5 Find an expression for a in terms of b, if:
a log10a= 2 − b b log10a= 2b− 1
c log10a= 2 log10b− 3 d
6 If x= log2a and y= log2b, express in terms of x and y.
7 Use the change of base rule and a calculator to evaluate the following, correct to 3 decimal places.
a log47 b log59 c log210
Solving exponential equations and inequations
Consider the exponential equation 3x= 81. The solution can be obtained as follows:3x= 81 = 34 x = 4
It is not always possible to equate indices as in the example above. Sometimes it is necessary to use logarithms.
log3
1 27
---log581
log53
--- log364
log316
---log1014 log10
7 5
---–
log2x 2
y 2 log2xy 2
log2
x y2 ---+ –
log3x 2 1
3
---log3 x 6
y3
( )–4 log3x
+ 1
2
---log10x+log10 x
loga
x y2
--- loga
x y
---log5
10 3
---log10 a 2
1 2
---log10 b
– =
log2
4a3b2 b
---⎝ ⎠
E x a m p l e
5
Solve for x, giving answers correct to 3 decimal places.
a 2x− 3= 15 b 5−0.4x= 1.7 c 3x+ 1= 5x d 2x + 3 = 2x
S o l u t i o n a 2x− 3= 15
log102x− 3= log1015 (taking the log of both sides)
(x − 3) log102 = log1015
x − 3 =
x =
= 6.907 (correct to 3 decimal places)
b 5−0.4x= 1.7 log105−0.4x= log101.7 −0.4xlog105 = log101.7
x =
=−0.824 (correct to 3 decimal places)
c 3x+ 1= 5x log103x+ 1= log105x
(x + 1) log103 = x log105
log103 = x log105 − x log103
x =
= 2.151 (correct to 3 decimal places)
d 2x + 3 = 2x
We need to use technology to find a solution. Sketch the graphs of y = 2x + 3 and y = 2x; the x-coordinates at the points of intersection of the two graphs are the solutions to this equation.
With a TI-89, the solve command can also be used to find solutions.
Hence, correct to 3 decimal places, the solutions are x =−1.296 and x = 3.247.
log1015
log102
---log1015
log102
---+3
log101.7
0.4 log105
–
---log103
log105 log– 103
---GC 2.4 CAS 2.4,
10.2
[−5, 10] by [−5, 15] [−5, 10] by [−5, 15]
E x a m p l e
6
Find, correct to 3 decimal places:
a {x: 3x+ 1< 4} b {x: 4(0.5)x≥ 1}
S o l u t i o n a {x: 3x+ 1< 4}
3x+ 1< 4 log10 3x+ 1< log10 4
(x + 1) log10 3 < log10 4
x log10 3 < log10 4 − log10 3
x <
x < 0.262 The solution is {x:x < 0.262}.
GC 2.4 CAS 2.4
Solutions to any exponential equation can be obtained graphically. In example 3 part a, the solution can be found as the point of intersection between the graphs of y= 2x− 3 and y= 15.
[−1, 10] by [0, 20]
t i p
t i p
The solve command can be used to find the exact or approximate solutions to exponential equations, for example, to solve 3x= 8 for x.
The exact answer is given as , as the calculator uses a new base, e, as the default base for logarithms. We will explore the base e later in this chapter. The change of base rule can be used to convert answers from the calculator to a different base:
x= = = log38
x 3 ln 2
ln 3
---=
3 ln 2 ln 3
---ln 8 ln 3
---CAS 10.2
b {x: 4(0.5)x≥ 1} 4(0.5)x≥ 1
(0.5)x≥
We could take the log of both sides and solve as in part a, taking care to note that log100.5 is negative. However, in
this case it is easier to use indices. Since
, the inequality becomes
2x≤ 4 (taking the reciprocal of both sides) 2x≤ 22
x ≤ 2
The solution is {x:x ≤ 2}. 1
4
---0.5 1
2
---= 1
2
---⎝ ⎠ ⎛ ⎞x 1
4
---≥
log10 4 log– 10 3
log10 3
---E x a m p l e
7
Solve for x:
a 52x− 4 × 5x− 5 = 0 b 2x= 10 × 2−x+ 9
S o l u t i o n
a 52x− 4 × 5x− 5 = 0 (5x)2− 4 × 5x− 5 = 0 Factorising gives:
(5x+ 1)(5x− 5) = 0 Using the null factor law:
5x=−1 (which is impossible as 5x is positive for all x), or 5x= 5
x = 1 b 2x= 10 × 2−x+ 9
Multiply both sides by 2x:
(2x)2= 10 + 9 × 2x (2x)2− 9 × 2x− 10 = 0
Factorising gives: (2x+ 1)(2x− 10) = 0
2x=−1 (which is impossible), or 2x= 10
x = log210
This is the exact solution. If an approximate solution is required then the change of base rule can be used.
x = log210
=
= 3.322 (correct to 3 decimal places)
Solving logarithmic equations
The definition of a logarithm and the laws of logarithms are used in solving logarithmic equations.
E x a m p l e
8
Solve for x.
a log2(x + 3) = 5 b log2(x + 3) − log2(x − 2) = 3 c log3 x + log3(x − 8) = 2
S o l u t i o n a log2(x + 3) = 5
x + 3 = 25 x + 3 = 32 x = 29 log1010
log102
---Check:
LHS = log2 (29 + 3) = log2 32 = 5
b log2(x + 3) − log2(x − 2) = 3
= 3
= 23 x + 3 = 8(x − 2) x + 3 = 8x − 16
19 = 7x
x =
Check: LHS =
=
=
= log28 = 3
= RHS
exercise
1.3
8 Solve for x.a 5x = 25 b c 16x = 4
d 22x− 5= 1 e f
9 Solve for x.
a 42x + 1= 22x − 3 b c
d e f
10 Solve for x, giving answers correct to 3 decimal places.
a 2x = 10 b c 5x= 0.5
d 100.2x= 35 e 2 × 3−0.01x= 0.32 f 3 × 10x − 1= 125
11 Solve for x, giving answers correct to 3 decimal places.
a 2x =x+ 1 b 3 − 2x= 5−0.2x c x2 −x= 3−x
12 Working correct to 3 decimal places, find
a {x: 2x> 5} b {x: 3−x < 0.4} c {x: 52x≥ 22}
d e {x: 2x + 4> 0.2} f
c log3x + log3(x − 8) = 2
log3(x(x − 8)) = 2
x(x − 8) = 32 x2− 8x − 9 = 0 (x − 9)(x + 1) = 0
x = 9, −1 Check:
x = 9:
LHS = log39 + log3(9 − 8) = 2 + log31
= 2
= RHS x =−1:
LHS = log3(−1) + log3(−9)
It is not possible to evaluate the logarithm of a negative number. So, x = 9 is the only solution.
log2 x 3
+ x–2
---⎝ ⎠
⎛ ⎞
x+3 x–2
---19 7
---log2⎝⎛19---7 +3⎠⎞–log2⎝⎛19---7 –2⎠⎞
log240
7
--- log25
7 ---– log2 40 7 ---5 7
---t i p
When solving logarithmic equations it is necessary to always check solutions. Be sure to check the original equation.
continued
3x 1 9
---=
49c = 7 32x 1 2
---=
5x 1 25x+2
---= 3x2–5x–6 = 1 6
( )x–3
365–x
= 7x2 = ( )7x 2 27x+1 = 3x2–1 1
3
---⎝ ⎠ ⎛ ⎞x
18
=
x: 1 2
---⎝ ⎠ ⎛ ⎞x≤0.3
⎩ ⎭
⎨ ⎬
⎧ ⎫
13 Solve for x.
a 22x− 6 × 2x+ 8 = 0 b 32x− 12 × 3x+ 27 = 0
c 52x− 30 × 5x+ 125 = 0 d 22x− 15 × 2x− 16 = 0
e 2x + 4 × 2−x− 5 = 0 f 3x− 8 = 9 × 3−x
g 22x+ 1− 17 × 2x+ 8 = 0 h 21 +x= 4 − 21 −x
14 Solve for x.
a log2(x+ 1) = 3 b logx125 = 3
c 2 log3x= 4 d log3x 2
= 4
e log3(x+ 2) − log3(x+ 1) = 4 f log2(x+ 1) − log2x= log2(x+ 2)
g log105x= 2 log10x h logx(x
2
+ 6x) = 3
i log2x 2
= (log2x) 2
j log2(x+ 2) + log2(x− 2) = 1
k (log3x) 2
− 18 log3x+ 81 = 0 l log10x− 3 log104 = log10(x− 1)
The base
e
Consider the binomial expansion of , where n is a natural number.
= = = = = As So,
The irrational number e is defined as:
e=
=
= 2.71828 …
In many applications of exponential and logarithmic functions, the base number e is used. It is referred to as the natural base. The function y = ex, x ∈ R, is referred to as the natural exponential function and y = loge x, x > 0 is referred to as the natural logarithmic function. loge x is also denoted by ln x. Natural logarithms can be evaluated using the function on
1 1 n
---+
⎝ ⎠
⎛ ⎞n
1 1 n
---+
⎝ ⎠
⎛ ⎞n
1 nC1
1 n
---⎝ ⎠ ⎛ ⎞ nC
2
1 n
---⎝ ⎠ ⎛ ⎞2
C n 3 1 n ---⎝ ⎠ ⎛ ⎞3
…
+ + + +
1 n!
n–1
( )!
--- 1
n
---⎝ ⎠
⎛ ⎞ n!
2!(n–2)!
--- 1
n
---⎝ ⎠
⎛ ⎞2 n !
3!(n–3)!
--- 1
n
---⎝ ⎠ ⎛ ⎞3
…
+ + + +
1 n n( –1)!
n–1
( )!
--- 1
n
---⎝ ⎠
⎛ ⎞ n n( –1)(n–2)!
2!(n–2)!
--- 1
n
---⎝ ⎠
⎛ ⎞2 n n 1 –
( )(n–2)(n–3)!
3!(n–3)!
--- 1
n
---⎝ ⎠ ⎛ ⎞3
…
+ + + +
1 1 n n( –1) 2!
--- 1
n
---⎝ ⎠
⎛ ⎞2 n n 1 –
( )(n–2) 3!
--- 1
n
---⎝ ⎠ ⎛ ⎞3
…
+ + + +
1 1 1
2!
--- 1 1
n
---–
⎝ ⎠
⎛ ⎞ 1
3!
--- 1 1
n
---–
⎝ ⎠
⎛ ⎞ 1 2
n ---– ⎝ ⎠ ⎛ ⎞ … + + + +
n→∞ 1
n
---→0 2
n
---→0 …
, , , 1 1 n ---+ ⎝ ⎠
⎛ ⎞n
n→∞
lim 1 1 1
2! --- 1 3! --- … + + + + = 1 1 n ---+ ⎝ ⎠
⎛ ⎞n nlim→∞
1 1 1
E x a m p l e
9
Evaluate each of the following correct to 4 decimal places.
a 2e3 b c 2 loge5 d
S o l u t i o n
a 2e3= 40.1711 b = e−1 c 2 loge5 = 3.2189 d = 0.3466
= 0.3679
The techniques for solving equations, already covered in this chapter, are also valid for equations involving the natural base.
E x a m p l e
1 0
Solve for x:
a 3e2x= 7 b e2x− 5ex+ 4 = 0
S o l u t i o n a 3e2x= 7
e2x=
2x =
x =
The solutions above are exact solutions. A decimal approximation in each case can also be obtained, if necessary. Thus in part a the answer, correct to 3 decimal places, is x = 0.424 as given in the screenshot opposite.
E x a m p l e
1 1
Solve for x.
a b 3 ln (2x) = 6
S o l u t i o n
a logex =
x =
=
1 e
--- ln 2
GC 1.1 CAS 1.1
1 e
--- ln 2
b e2x− 5ex+ 4 = 0 (ex)2− 5ex+ 4 = 0 (ex− 1)(ex− 4) = 0 ex= 1 or ex= 4 x = ln 1 or x = ln 4 x = 0 or ln 4 7
3 ---loge7
3 ---1 2
--- ln 7
3
---logex 1
2 ---=
t i p
Always give an exact answer unless a decimal approximation is required or necessary.
b 3 ln (2x) = 6 ln (2x) = 2 2x = e2
x = 1 2
---e2
1 2 ---e1 2⁄
exercise
1.3
15 Solve for x in each of the following giving
i exact answers. ii answers correct to 4 decimal places.
a ex= 5 b e2x= 3 c 2e−x = 8
d ex+ 1= 2 e f 30e−0.2x + 5 = 65
g h e2x− 9ex− 10 = 0 i ex= 6e−x+ 5
16 Solve for x in each of the following, giving exact answers.
a ln x= 2 b 3 loge4x= 9 c ln(2x− 3) = 0
d ln(x2) = 4 e ln(3x− 2) = ln(x+ 1) f loge(15 −x
2
) = loge(x
2
−x)
g (ln x)(ln x− 3) = 0 h i ln(x+ 4) − ln(x+ 1) = lnx
17 Let a and b be positive real numbers.
a If ax+ 1= bx− 1, express x in terms of a and b using natural logarithms.
b Hence show that the solution to 2x + 1= 10x − 1 can be written in the form , where p, q and r are positive integers with p> 1.
continued
2
ex–1
--- = 5 3
e2–x --- = 9
loge
1
x --- = 3
1 plogeq loger
---+
Number
sense with the spence
77
The famous mathematician Carl Friedrich Gauss was born in 1777 and died in early 1855 at the tender age of 77. When he was a young boy in a school classroom with students far older than himself, the students were asked by their teacher to find the sum of the first 100 counting numbers. The teacher expected the students to work on this problem for an extended period of time, but within a matter of moments Gauss correctly proclaimed the answer to be 5050. How did he solve the problem so quickly?
Algebra of circular functions
The unit circle, symmetry properties and exact values
The unit circle is a circle of radius 1 unit with centre at the origin, O. Let P(x, y) be a point on the unit circle such that OP makes an angle θ with the positive direction of the x-axis. A tangent is drawn to this circle through the point A(1, 0). When the line OP is produced it meets this tangent at Q.In general, for any point P on the unit circle, we define the functions cosine and sine by:
cosine θ= cosθ= x and sine θ= sinθ= y where x is the x-coordinate of P and y is the y-coordinate of P.
We define the tangent function by: tangent θ= tanθ
= AQ, the y-coordinate of Q
tanθ=
=
= , cosθ≠ 0
The angle θ is positive if the angle of rotation from the positive direction of the x-axis is in an anticlockwise direction and negative if the angle of rotation from the positive direction of the x-axis is in a clockwise direction.
The four quadrants in the Cartesian plane are numbered according to the system shown on the right.
Recall the following properties of symmetry of the trigonometric functions.
Radians Degrees
Quadrant 2 sin (π − θ) = sinθ cos (π−θ) = −cosθ tan (π − θ) = −tanθ
sin (180 − θ)° = sinθ° cos (180 − θ)° = −cosθ° tan (180 − θ)° = −tanθ° Quadrant 3 sin (π+θ) =−sinθ
cos (π + θ) = −cosθ tan (π+θ) = tanθ
sin (180 +θ)° =−sinθ° cos (180 + θ)° = −cosθ° tan (180 +θ)° = tanθ° Quadrant 4 sin (2π−θ) =−sinθ
cos (2π − θ) = cosθ tan (2π−θ) =−tanθ
sin (360 −θ)° =−sinθ° cos (360 − θ)° = cosθ° sin (360 −θ)° =−tanθ°
x y P
M
A (1, 0)
Q
O
θ
x
y
AQ 1
---y x
---sinθ cosθ
---Quadrant 1 Quadrant 2
Quadrant 4 Quadrant 3
x
y
1. 4
And for negative angles: sin (−θ) = −sinθ cos (−θ) = cosθ tan (−θ) = −tanθ
The signs of the circular functions can be remembered using the following diagrams.
This is sometimes known as the CAST rule.
The exact values of sin, cos and tan for come from the triangles drawn below.
Other exact values are obtained directly from the unit circle.
θ 0 π 2π
sinθ 0 1 0 −1 0
cosθ 1 0 −1 0 1
tanθ 0 1 undefined 0 undefined 0
SIN is positive
ALL functions are positive
TAN is positive
COS is positive
x
y
x
y
S
A
T
C
π 6
--- π
4
--- π
3
---, ---,
π
4
π
4 2 1
1
π
6
π
3 2
1 3
(1, 0) 0, 2 (0, 1)
(–1, 0)
(0, –1)
O x
y
2 3 2
π 6
--- π
4
--- π
3
--- π
2
--- 3π
2
---1 2
--- 1
2
--- 3
2
---3 2
--- 1
2
--- 1
2
---1 3
--- 3
E x a m p l e
1
Find the exact value of:
a sin 120° b c
S o l u t i o n
a sin 120° = sin (180° − 60°) b = c =
= sin 60° = =
= = =
=−1
Solving equations involving circular functions
Recall that the steps for solving a trigonometric equation in the form f(x) = d, where f is one of sine, cosine or tangent, can be summarised as follows:
.Identify the quadrants in which the solutions lie. .Find the reference angle in the first quadrant. .Find other solutions using symmetry properties.
E x a m p l e
2
Solve for x: ,
S o l u t i o n
Identify the quadrants in which the solutions lie. The value of cosx is positive so there is a solution in the first quadrant and a solution in the fourth quadrant (CAST rule).
Find a reference angle in the first quadrant:
cos 30° = .
Solution in the first quadrant: x = 30
Find the other solution using the properties of symmetry.
Solution in the fourth quadrant is: x = 360 − 30
= 330 So x = 30, 330.
cos11π 6
--- tan 5π
4 ---–
⎝ ⎠
⎛ ⎞
cos11π 6
--- cos 2π π
6 ---–
⎝ ⎠
⎛ ⎞ tan 5π
4 ---–
⎝ ⎠
⎛ ⎞ –tan5π 4
---cosπ 6
--- tan π π
4 ---+
⎝ ⎠
⎛ ⎞
– 3
2
--- 3
2
--- tanπ
4 ---–
CAS 1.6
t i p
The TI-89can be used to find exact values of sinθ, cosθ and tanθ in DEGREE or RADIAN mode. With the mode set to RADIAN, we can still find the values for angles in degrees by using the ° symbol.
x°
cos 3
2
---= 0≤ ≤x 360
S
T A
C
---E x a m p l e
3
Solve for x: sinx = 0.32, 0 ≤ x ≤ 2π for x. Give your answers correct to 4 decimal places.
S o l u t i o n
Identify the quadrants in which the solutions lie.
The value of sin x is positive, so there is a solution in the first quadrant and a solution in the second quadrant. Find a reference angle in the first quadrant.
Using a calculator, the reference angle is 0.3257. Solution in the first quadrant: x = 0.3257.
Find the other solution using the properties of symmetry.
Solution in the second quadrant is: x =π− 0.3257
= 2.8159 So x = 0.3257, 2.8159.
E x a m p l e
4
Solve for θ: , 0 ≤θ≤ 2π
S o l u t i o n sinθ=
<