Solving Indicial Equations PDF

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242

M at hs Q u es t 1 1 M at he ma ti ca l M et ho ds

Indicial equations -1-

We can solve equations of the form

x 3 = 2 as follows: 1

Take the cube of both sides:

 x -3-  

3

= 23

The left-hand side becomes x, so x = 8. However, when the unknown (or variable) is not a base number but is an index number, a different approach is required.

Method 1: Exact solutions without a calculator To attempt to solve index equations exactly, express both sides of the equation to the same base and equate the powers. If a m = a n, then m = n.

WORKED Example 8 Find the value of x in each of the following equations. a 3x = 81 b 4x − 1 = 256 c 63x − 1 = 362x − 3 THINK

WRITE

a

a

b

c

3x = 81

1

Write the equation.

2

Express both sides to the same base.

3

Equate the powers.

1

Write the equation.

2

Express both sides to the same base.

3

Equate the powers.

4

Solve the linear equation for x by adding one to both sides.

1

Write the equation.

2

Express both sides to the same base.

63x − 1 = (62)2x − 3

3

Remove the brackets by multiplying the powers.

63x − 1 = 64x − 6

4

Equate the powers.

5

Subtract 3x from both sides to make x the subject.

6

Add 6 to both sides to solve the equation.

3x = 34 Therefore, x = 4. b

4x − 1 = 256 4x − 1 = 44 Therefore, x − 1 = 4. x=5

c

63x − 1 = 362x − 3

Therefore, 3x − 1 = 4x − 6 −1 = x − 6 x=5

More complicated equations can be solved using the same technique.

Chap te r 5 E xp on en tial a n d log ar it hm ic f u nction s

243

WORKED Example 9 Solve for n in the following equation. 23n × 16n + 1 = 32 THINK 1 2

3

4

5 6

WRITE

Write the equation. Express both sides using the same base, 2. Remove the brackets by multiplying the powers. Multiply the terms on the left-hand side by adding the powers. Equate the powers. Solve the linear equation for n.

23n × 16n + 1 = 32 23n × (24)n + 1 = 25 23n × 24n + 4 = 25 27n + 4 = 25 Therefore, 7n + 4 = 5 7n = 1 n = 1--7

In some cases indicial equations can be expressed in a quadratic form and solved using the Null Factor Law. Look for numbers in index form similar to a2x and ax appearing in different terms.

WORKED Example 10 Solve for x if 52x − 4(5x) − 5 = 0. THINK 1 2

3 4 5 6 7 8 9

WRITE

Write the equation. Rewrite the equation in quadratic form. Note that 52x = (5x)2. Substitute y for 5x. Rewrite the equation in terms of y. Factorise the left-hand side. Solve for y using the Null Factor Law. Substitute 5x for y. Equate the powers. State the solution(s).

()52x − 4(5x) − 5 = 0 (5x)2 − 4(5x) − 5 = 0 Let y = 5x Therefore, y2 − 4y − 5 = 0. (y − 5)(y + 1) = 0 Therefore, y = 5 or y = −1 5x = 5 or 5x = −1 ⇒ 5x = 51 and 5x = −1 has no solution. ⇒x=1

Note that in step 8, the possible solution 5x = −1 was rejected because there is no value of x for which it will be satisfied. Recall that exponential functions such as 5x are always positive.

Method 2: Using a calculator and trial and error Indicial equations which cannot have both sides expressed to the same base number do not generally have exact, rational solutions. A trial and error method using a calculator can find solutions to a desired degree of accuracy.

Kuta Software - Infinite Algebra 2

Exponential Equations Not Requiring Logarithms Solve each equation. 2x+3

=1

2) 5

3 − 2x

=5

1−2x

= 243

4) 3

2a

=3

−a

3x−2

=1

6) 4

2p

=4

8) 2

2x + 2

1) 4

3) 3

5) 4

−2 a

=6

3m

⋅6

7) 6

9) 6

2 − 3a

−m

=6

−2m

10)

2x 2

x

x

11) 10 −3 ⋅ 10 =

1 10

x

−x

−2 p − 1

=2

=2

3x

−2 x

x x x 12) 3−2 + 1 ⋅ 3−2 − 3 = 3−

x

x

13) 4 −2 ⋅ 4 = 64

x 15) 2 ⋅

1 = 32 32

x x 17) 64 ⋅ 16−3 = 163 − 2

19) 81 ⋅ 9−2

21)

() 1 6

r

b −2

= 27

3x + 2

⋅ 2163x =

r

23) 16 ⋅ 64 3 − 3 = 64

14) 6

−2 x

⋅6

−x

16) 2

−3p

⋅2

2p

18)

81

1 216

=2

2p

3n + 2

=3

4

243

−n

−3 x

⋅ 9 = 27

20) 9

1 216

=

x

22) 243k + 2 ⋅ 9 2k − 1 = 9

24) 16

2p − 3

⋅4

−2 p

=2

4

Kuta Software - Infinite Algebra 2

Exponential Equations Not Requiring Logarithms Solve each equation. 2x+3

1) 4

2) 5

=1

{ } 1−2x

3) 3

= 243

4) 3

{−2}

3 x− 2

6) 4

{} −2 a

3m

=3

2p

−a

=4 −

=6

2 − 3a

8) 2

{2}

9) 6

2a

−2 p − 1

{ }

2 3

7) 6

−x

{0}

=1

5) 4

=5

{3}

3 2

−

3 − 2x

1 4

2x + 2

=2

3x

{2}

⋅6

−m

=6

−2m

10)

{0}

2x 2

x

=2

−2 x

{0}

x

x

11) 10 −3 ⋅ 10 =

{} 1 2

1 10

x x x 12) 3−2 + 1 ⋅ 3−2 − 3 = 3−

{ } −

2 3

x

x

13) 4 −2 ⋅ 4 = 64

x

x

14) 6−2 ⋅ 6 − =

{−3}

1 216

{1}

1 = 32 32

x 15) 2 ⋅

16) 2

−3p

⋅2

2p

=2

2p

{0}

{10}

x x 17) 64 ⋅ 16−3 = 163 − 2

18)

81

3n + 2

243

{ } 7 12

b −2

4 17

x

21)

{ }

3 4

−

3x + 2

() { } 1 6

−

⋅ 2163x =

r

23) 16 ⋅ 64 3 − 3 = 64

{} 6 7

1 216

3 4

22) 243k + 2 ⋅ 9 2k − 1 = 9

{ } −

1 6

r

x

20) 9−3 ⋅ 9 = 27

= 27

{ } −

4

{ } −

19) 81 ⋅ 9−2

=3

−n

24) 162

2 3

p−3

{4}

⋅4

−2 p

=2

4...