The equation ax= b is an example of a general exponential (or indicial) equation and 2x= 32 is an example of a more specific exponential equation.
To solve one of these equations it is necessary to write both sides of the equation with the same base if the unknown is an index or with the same index if the unknown is the base.
WorkEd ExamplE 12
Solve for x in each of the following.
a 2x= 32 b 3x== 1
27 c 2 × 3x= 162 d 2(1 - x)= 16
Think WriTE
a 1 Write 32 with base 2, the same as the left-hand side.
a 2x= 32 2x= 25
2 The indices are equal because the base is 2 on each side of the equation.
x = 5
b 1 Write 27 with base 3. b 3 1
271 33
x=
=
2 Write 1
33 as a number with base 3. 3x = 3-3
3 Equate the indices. x = -3
c 1 Divide both sides by 2 to leave 3x on the left-hand side.
c 2 × 3x = 162 3x = 81
2 Write 81 as a number with base 3. 3x= 34
3 Equate the indices. x = 4
d 1 Write 16 with base 2. d 21 - x = 16
21 - x = 24
2 Equate the indices. 1 - x = 4
3 Solve for x. x = -3
WorkEd ExamplE 13
Solve 5x× 252x - 3= 625 for x: a using index laws b with a CAS calculator.
Think WriTE/display
a 1 Write all numbers with the same base. 5x× 252x - 3= 625 5x× (52)2x - 3= 54
2 Simplify. 5x× 52(2x - 3)= 54
3C
3 Remove the brackets in the index. 5x× 54x - 6= 54
4 Add the indices on the left-hand side. 55x - 6= 54
5 Equate the indices. 5x-6 = 4
6 Solve the equation. 5x = 10
x = 2
b 1
2 Write the solution. Solving 5x × 25(2x - 3) = 625 for x gives x = 2.
Sometimes it is possible to use the methods for solving quadratic equations to help solve indicial equations. Remember that 22x = (2x)2.
WorkEd ExamplE 14
Solve for x in the following.
a (2x- 16)(2x+ 4) = 0 b 32x- 12 × 3x+ 27 = 0 c 4x- 2x + 3+ 16 = 0
Think WriTE
a 1 Use the Null Factor Law to solve by making each bracket equal to zero.
a (2x- 16)(2x+ 4) = 0 (2x- 16) = 0 or (2x+ 4) = 0
2 Solve each equation. 2x= 16 or 2x= -4
3 Write 16 as a number with base 2 but -4 can not be written with base 2.
2x= 24 or no real solution
4 Solve by equating the indices. x = 4
b 1 Write 32x as (3x)2. b 32x- 12 × 3x+ 27 = 0 (3x)2- 12 × 3x+ 27 = 0
2 Let 3x = a to make a simpler quadratic equation to solve.
a2 - 12a + 27 = 0, where a = 3x
3 Factorise. (a - 3)(a - 9) = 0
4 Use the Null Factor Law by making each
bracket equal to zero. a - 3 = 0, a - 9 = 0
5 Solve for a. a = 3, a = 9
6 Substitute back a = 3x. 3x = 3, 3x = 9
7 Write numbers with base 3. 3x= 31, 3x= 32
eBookplus eBookplus
Tutorial int-0530 Worked example 14
On a Calculator page, press:
• MENU b
• 3:Algebra 3
• 1:Solve 1
Complete the entry line as:
solve(5x × 25(2x - 3)= 625,x) Then press ENTER ·.
8 Equate the indices. x = 1, x = 2
c 1 Rewrite 4x as (2x)2 and 2x + 3 as 2x× 23. c 4x – 2x + 3 + 16 = 0 (2x)2- 2x× 23+ 16 = 0
2 Rewrite 23 as 8. (2x)2 - 2x × 8 + 16 = 0
3 Let 2x = a to make a simpler quadratic equation to solve.
a2 - a × 8 + 16 = 0 where a = 2x
4 Replace a × 8 with 8a because the coefficient precedes the pronumeral.
a2- 8a + 16 = 0
5 Factorise. (a - 4)(a - 4) = 0
6 Use the Null Factor Law. a - 4 = 0, a - 4 = 0
7 Solve for a. a = 4 and a = 4
8 The two factors are equal because a2- 8x + 16 = 0 is a perfect square.
9 Substitute back a = 2x. 2x= 4
10 Write 4 as a number with base 2. 2x= 22
11 Solve by equating the indices. x = 2
Remember to always make the right-hand side equal to zero when solving quadratic equations.
It is a good idea to substitute your answer back into the original equation to check the accuracy of your work.
If the base is not the same and the numbers cannot be written with the same base, then logarithms can be used. It is possible to take the logarithm of both sides of an equation provided the same base is used.
WorkEd ExamplE 15
Solve for x in the following. Give your answers in exact form using base 10 and correct to 3 decimal places.
a 5x= 10 b 2(x + 1)= 12
Think WriTE
a 1 Take the logarithm of both sides to base 10.
a 5x= 10
log10 (5x) = log10 (10)
2 Use loga (mp) = p loga (m) and loga (a) = 1. x log10 (5) = 1
3 Divide both sides by log10 (5). x= 1
10 5
log ( ) (exact form)
4 Use a calculator to simplify. x ≈ 1.431, correct to 3 decimal places.
b 1 Take the logarithm of both sides to base 10.
b 2(x + 1)= 12
log10 (2(x + 1)) = log10 (12)
2 Use loga (mp) = p loga (m) to simplify. (x + 1) log10 (2) = log10 (12)
3 Divide both sides by log10 (2).
( ) log ( )
log ( ) x+ =1 12 2
10 10
4 Use a calculator to simplify the right-hand
side. ∴ x =log ( )−
log ( )
10 10
12
2 1 (exact form)
5 Solve for x. x ≈ 2.585, correct to 3 decimal places.
Note: Logarithms in bases other than 10 may be used.
Inequalities are worked in exactly the same way except that there is a change of sign when dividing or multiplying both sides of the inequality by a negative number.
WorkEd ExamplE 16
Solve the following equations for x, giving your answers both in exact form and correct to 3 decimal places.
a 2x > 5 b 0.5x≤ 1.4
Think WriTE/display
1
2 Write the answers in exact form.
The calculator defaults to base e when solving exponential equations in exact form.
Note: ln (A) ⇔ loge (A). This will be discussed later in this chapter.
Note: In part b, the inequality ≤ has been changed to ≥ because loge12<0.
a Solving 2x> 5 for x gives
x e
e
>log ( ) log ( ) 5 2 .
b Solving 0.5x ≤ 1.4 for x gives
x
e
e
≥
−log
log ( ) . 7 5 2
3 a Solving 2x > 5 for x gives x > 2.322,
correct to 3 decimal places.
b Solving 0.5x ≤ 1.4 for x gives x ≥ -0.485, correct to 3 decimal places.
The equations
1. ax= y and 2x= 32 are exponential equations.
Write numbers with the same base to help simplify problems. The most commonly 2.
used bases are 2, 3 and 5.
If the base is the same, equate the indices.
3.
If the indices are the same, equate the bases.
4.
rEmEmBEr On a Calculator page, press:
• MENU b
• 3:Algebra 3
• 1:Solve 1
Complete the entry lines as:
solve(2x>5,x) solve(0.5x≤1.4,x)
Press ENTER · after each entry.
Press Ctrl / ENTER · after solving each equation to obtain a decimal approximation.
Write the answers correct to 3 decimal places.
5. Use the Null Factor Law to solve quadratic equations.
A negative number cannot be expressed in index form, for example,
6. -4 cannot be
expressed with base 2.
a
7. 2x = (ax)2
Take the logarithm of both sides of an equation or inequation using the same base.
8.
Change the sign of an inequality when multiplying or dividing by a negative 9.
number.
log
10. a (x) > 0 if x > 1 log
11. a (x) < 0 if 0 < x < 1
Exponential equations
1 WE12 Solve for x in each of the following.
a 3x= 81 b 10-x= 1000 c 1
2x=32
d 7 1
49
x= e 243x = 3
2 Solve for x in each of the following.
a 3 × 2x= 48 b 6x - 2 = 216 c 5 1
125
2x−1= d 22x - 6 = 1 3 WE13 Solve for x in each of the following.
a 3x× 3x - 1 = 243 b 5x× 52x + 1 = 625
c 2x× 4x - 1 = 16 d 3
93xx−+21 =81 4 WE14 Solve for x in the following.
a (3x- 9)(3x- 1) = 0 b 22x- 6 × 2x+ 8 = 0 c 62x- 7 × 6x+ 6 = 0 d 4x- 6 × 2x- 16 = 0 e 9x= 2 × 3x+ 3
5 Solve for x in the following.
a 25x+ 4 × 5x- 5 = 0 b 42x- 20 × 4x= -64 6 WE15 Solve for x correct to 3 decimal places.
a 2x = 5 b (0.3)x - 1 = 10 c (1.4)2 - x = 6 d 3 × 5x = 27 e 5 × 7x = 1 f 2x × 3x + 1 = 10 7 WE16 Solve for x correct to 3 decimal places.
a 3x> 5 b 22x≤ 7 c (0.2)x> 3
d 7x≥ 0.5 e (0.4)x> 0.2
8 mC The value of x for which 5 × 2x= 1255, correct to 3 decimal places, is:
A 7.971 B 897.750 C 897.749
D 7.972 E 2.059
9 mC The solution to the equation 102x = 3 × 10x + 4 is:
A log10 (-1), log10 (4) B -1, 4 C 10x + 1, 10x - 1
D 0, 0.602 E log10 (4)
ExErCisE
3C
eBookplus eBookplus Digital doc SkillSHEET 3.1 Index form
eBookplus eBookplus Digital doc SkillSHEET 3.2 Solving equations
eBookplus eBookplus Digital doc SkillSHEET 3.3 Solving indicial equations by equating the bases
eBookplus eBookplus Digital doc SkillSHEET 3.4 Solving liner inequations