SM SAINS SERI PUTERI, KUALA LUMPUR
INDICES AND LOGARITHMS NAME: ……………………………………………………………………………
CLASS: …………………………………………………………………………….
1
CHAPTER 5 :
5.1
INDICES AND LOGARITHMS
CONCEPTUAL MAP
INDICES AND LOGARITHMS
Indices n
a
base
Integer indices n
a) a = a x a x a … x a x a
-n
index
x
If N = a , then loga N = x
Fractional indices
___________ ___________ = ___________ = ___________
n factors b) Negative index, a
Logarithms
= _________
0
c) Zero index, a = __________
Laws of indices m
a xa
n
Equations involving indices
= _________
a) If ax = ay, then _________
am ÷ an = _________
b) If mx = nx, then ________
(am)n
= _________
2
1. INDICES AND LAW OF INDICES 1.1 Finding the value of number given in the form of :-
Type of indices (a) Integer indices
In General (i) positive indices an
Examples
=
10 4
= a × a × a × ..... × a
( −4) 3
n factors
(0.3) 5
a = base(non zero number) n = index(positive integer) (ii) zero index: a0
where a
= 1,
= =
3
3 = 4 4o = (-55)o =
≠0
0.0038o = o
− 6 = 7 3 −2 =
(ii) negative indices a
−n
=
1 a
(−5) −3 =
n
(0.4) − = 4
(0.06)-3 = (b) Fractional indices
1
1
(i) a n = n a
92 =
n = positive integer
1
125 3 =
≠0
a
m
(ii) a n = n a m =
( a) n
4
m
27 = 3
3
36 2 =
64
−
4 3
= 3
( 0.09) − 2 = Exercise : Practice 1, Practice 2 , practice 3 page 71-73
1.2 Using Laws of Indices to find values of numbers in index form that are multiplied, divided or raised to a power.
LAWS OF INDICES
a
m
×
a
n
=
a
m+ n
a
m
÷
a
n
=
a
m− n
(a m ) n = a mn n
m
m
(ab ) = a b
m
a a = n b b
n
3
Exercise 1. Simplify each of the following: (a) 53 × 5 2
3 (b) ( −2)
(c) 4 5 ÷ 4 2
3 (d) (5.32)
×( −2)
÷ (5.32)
3
−5
1
1
−
(e) 16
−5
− (f) 10 125 2
2
2
5
2
3 −3
(g)
2 × 25 2 8
−
2 3
1.3 Simplifying Algebraic Expressions Using Laws of Indices Exercise
Simplify: 1. a3 x a2
2. 35n
÷
32n
3. (54n)2
4
5. 4 5 n − 2
1
4. ( s 2 t 4 ) 2
7.
2
4 n −5
16
÷
2 n +3
1
64 2
×
8. 3 6 n −4
÷4 ×4 2n
2 n −3
× 81 ÷ 27 5n
6.
p 4 n p 5 n
−5
+3
× p
3 n −2
2 n −5
Solve the following equation: (i)
16x = 32
(ii) 25x – 2 = 1252x – 2
x 2 x +3 + 2 x +2 − 2 x + 5 in terms of y. (iii) If 2 = y, express 2
n n n n n +1 in terms of x and y. (iv) If 2 = x and 3 = y, express 8 − 3 + 27
5
n +1 Show that 3
(v)
− 3 + + 9(3 + ) 2 n
n 1
can be divided exactly by 21 for all
positive integers of n.
2. LOGARITHMS AND LAWS OF LOGARITHMS. 2.1 Express equation in index form to logarithmic form and vice versa Definition of logarithm If a is a positive number and a
≠
⇔
` (INDEX FORM)
1 , then
Read as: Logarithm of N to the base a is equal to x.
log a N = x
(LOGARITHMIC FORM)
Notes! Since a1 = a then loga a = 1
N = Number a = base x = index
Since a0 = 1 then loga 1 = 0 We can use this relation to convert from index form to logarithm form or vice versa. EXERSICE: 1. Convert each of the following from index form to logarithmic form: INDEX FORM LOGARITHMIC FORM 2 (a) 100 = 10 (b) 64 = 43 (c) 625 = 252 (d) 10-3 = 0.001 (e) 36 = 62 (f) 0.0081 = 0.34 6
(g) 49 = 72 (h) 0.0001 = 10 -4
2. Convert each of the following from logarithmic form to index form: LOGARITHMIC FORM INDEX FORM (a) log10 1000 = 3 (b) log6 216 = 3 (c) log9 243 =
5 2
(d) log10 0.00001 = –5 (e) log2 32 = 5 (f) log3 81 = 4 (g)
log10 0.0001 = - 4 log
(h)
4 8 27
9
=
2 3
3. Convert each of the following from index form to logarithmic form: INDEX FORM
LOGARITHMIC FORM
(a) 50 = 1 (b) 320 = 1 (c) 0.140 = 1 (d)
2 5
0
=1
(e) 1080 = 1
Conclusion: loga 1 =
4. Convert each of the following from index form to logarithmic form: INDEX FORM
LOGARITHMIC FORM
7
(a) 51 = 5 (b) 321 = 32 (c) 0.141 = 0.14 (d)
2 5
1
=
2 5
(e) 1081 = 108
Conclusion: loga a =
2.2 Finding logarithm of a number
Logarithm to the base of 10 is known as the ‘common logarithm’. The value of common logarithms can be easily obtained from a scientific calculator or from the mathematical table. In common logarithm, if log10 N = x , then , antilog N = 10 x. [lg N = log 10 N] Exercise Use a scientific calculator to evaluate each of the following;
(a) log10 124 =
(b) log10 4.76 =
(c) log10 0.007826 =
(d) log 10 23 =
(e) log10 4.21.2 =
2
(f) log10 15.6 3
=
2.3 Finding logarithm of a numbers by using the Laws of Logarithms
LAWS OF LOGARITHMS
loga xy = loga x + loga y
log a
x = loga x - loga y y
loga xm = m loga x 8
Exercise:
1. Find the value of (a) log 2 4 + log 2 8
(c) log 2
(b) log 5 100 – log 5 4
1 64
2. Given that log a 2= 1.0570 and log a 3 = 1.0919 . Evaluate each of the following. (a) log a 18
(c) log a 27
(b) log a 1.5
(d) log a 4.5
(e) loga 6a
2.4 Simplifying logarithmic expressions using Laws of Logarithms. Exercise
Simplify the following logarithmic expression to the lowest form. 1.
log3 25 + log3 x – log35
3. 3 log2 x + 2 log2 y
2.
1 – log 5 xy + log5 3 x
4. 5 log5 xy – log5 x3 y2 + log5 4
9
5. log4 x + 2 log4 3 y – log4 xyz 3
7. 2 + 3log8 xy3 – log8 xy2
6.
1 2
log 5 x −
1 2
log 5 y + 2 log 5 xy
8. 4 log 6 a −
1 2
log 6 b − log 6 b c 2
2
+ log
1 6
ab 2
Given that a = log2 x and b = log2 y, express the following in terms of a and b. 1.
log2 x2 y2
2. log 2
x y 4
10
3. log 2
8 y x
5
3
2.5 Changing the Base of Logarithms 1. log a b
=
log c b log c a
2. log a b
=
1 log b a
Examples 1. Evaluate log 16 64 .
2. Evaluate log 27 3 .
3. Given that log p 4 = a and log p5 = b. Express the following in terms of a and/or b. (a) log4 p
(b) log5 p
11
(c) log4 5
(d) log4 80
4. Evaluate log2 3 x log3 4 x log4 5 x log5 8
2.6 Solving Problems Involving Logarithms
Examples 1.
Given that a = log2 x and b = log2 y, express the following in terms of a and b. (a) logy x3
(b) log xy 32
2.
Given that log2 x = m and log5 x = n, express logx 100 in terms of m and n.
12
Simplify log9 x3 – log27 x2 + log3 81
3.
4. Simplify log125 x – log25 x2 + log5 x
2 5. Simplify log s 2 5 × log5 s × log3
1 3
2.7 Solving Equation Involving Indices
An index equation is an equation that contains a variable in index form. There are three methods in solving index equations: 1. Equating the indices 2. Equating the bases 3. Taking the common logarithm on both sides of the equation.
Examples Solve the following equations Method 1: Equating the indices a. 4 x
c.
+1
(5 x 1 (5x +
b. 9 x +1
=16
−2
= 27 x −3
=125
13
Method 2: Equating the bases a. x 4
−3 b. x
= 81
−4 c. ( 2 x )
=
=
1 125
1 16
Method 3. Taking the Common Logarithms a. 3 x
=5
c. 3 x −2
b. 2 x
−1
=5
x −2
= 7 x +3
2.8 Solving Equation Involving Logarithms
A logarithmic equation is an equation that contains a variable in a logarithmic expression. There are three types of logarithmic equations: 1. Equations that can be solved by using the definition of logarithm: loga y
= x ⇔ y =a x
2. Equations that have the same bases 3. Equations that have different bases
Examples Solve the following logarithmic equations: Type 1: Equations that can be solved by using the definition of logarithm: loga y
= x ⇔ y =a x
14
a. log 2 x = 2
b. log x 16 = 2
2 = 5 x
c. log 2
Type 2: Equations that have the same bases a. log2 5 + log2 x = 4
c.
b. log3 3 x = 1 – log 3 8
log5 ( x + 2) = 2 + log 5 x
d. log3 4 + log3 (x – 5) = 1 + log 3 (2x – 1)
Type 3: Equations that have different bases a. log4 ( x + 2) = log2 3
15
b. log9 (3 x + 6) = 1 + log 3 2
2. Express each of the following in term of log a x and log a y . (a) log a
x
(b) log a xy3
2
y
y
(c) log a
a 2 x 3
(d) log a
x
3
y
3. Write each of the following expressions as single logarithm: (a) lg 3 +
1 2
(c) 2 lg x +
lg 25
1 3
lg y - 1
(e) lg 6 + 2 lg 4 – lg 8
(b) 3 lg 2 + 4 lg 4 – 2 lg 6
(d) 3 lg x –
1 2
lg y + 2
(f) log a x + log a y - 1
16
(g) 2log a x - 1 +
1 2
(h) log 3 x + 2log 3 y - 1
log a y
(i) log 2 x + log 2 y 2
(j) log b x + log b y + 1
Exercise ; Practice 9, Practice 10 , Formative exercise 5.2 , page 78 - 81
3. CHANGE OF BASE OF LOGARITHMS 3.1 Find the logarithm of a number by changing the base of the logarithm to a suitable base. 3.2 Solve problems involving the change of base and laws of logarithms.
_________ __________________________________________________________________
3.1 Finding logarithm of a number by changing the base of the logarithm to a suitable base The base of logarithms can be changed to other base by using a formula : log a b =
logc b logc a
When c = b, then loga b = =
logb b logb a 1 log b a
ACTIVITY 7: 1. Find the value of each of the following. Give your answer correct to four significant figures.
(a) log 5 9
(b) log 3 8
(c) log 30 4
(d) log 2 0.5
2. Find the value of each of the following without using calculator.. log 1 8 (a) log 3 7. log 7 2. log 2 3 = (b) log 2 6. = 6
17
(c) log
5
9.
Log 3 125 =
(d) log 4 5. log 5 3. log 3 7. log 7 64 =
Exercise : Practice 12, page 82
3.2 Solving Problems involving the change of base and laws of logarithms. ACTIVITY 8 1. Given that log 2 3 = 1.585 and log 2 5 = 2.322.Find the value of each of the following. (a) log3 30 (b) log 3 0.6
2. Given that log 2 a = b. Without using the calculator, express the following in terms of b: (a) loga 16 (b) log 4 2a2
(c) log 16 a
(d) log a 32a3
3. If log 3 x = r and log 3 y = s , express each of the following in terms of r and s . (a) log y x2 (b) log xy 81 2
18
4. If log 2 x = m and log 2 y = n , express each of the following in terms of m and n. 2 (a) log 4 x y
2 (b) log x y xy 2
Exercise : Practice 13 , Practice 14 , Formative exercise 5.3, page 82 – 84 .
4. EQUATION INVOLVING INDICES AND LOGARITHMS 4.1 Solving equations involving indices METHOD: 1. Comparison of indices or base (i) If the base are the same , when a x = a y , then x = y (ii) If the index are the same , when a x = b x , then a = b
2. Using common logarithm ACTIVITY 9: 1. Solve the following equations: (a) 3 x = 81
(b) 16 x = 8
(c) 8 x+1 = 4
(d) 9 x+1 = 3
(e) 9 x. 3 x-1 = 81
(f) 2 x + 3 - 42x = 0
(g) x 4
= 81
(h)
1 125
= x −3
2. Find the value of x and y in each of the following simultaneous equations.
19
x
(a) 3 .9
y −1
= 243 and
2
3 x y
4
= 32
x
= 32 y and
x
(b) 2(4 )
27 9
y
= 81
3. By using common logarithm(log 10), solve the following equations and give your answer correct to two decimal places. (a) 2 x = 3
(b) 7 − x
(c) 42 x+1 = 7
(d) 153 x −4
(e) 6 5−2 x
(f) 2 .3 = 18
=4
(g) 2 2 x.5 x +1
=7
x
=8
= 10
x
(h) 2 x .3 x
= 9 x−4
4. By using replacement method, solve each of the following equations: (a) 2 x+ 4
(c) 2 2 x
− 2 x +3 = 1
− 7.2 x +1 − 32 = 0
(b) 3 x +1 + 3 x + 2
(d) 32 x+ 2
=4
+ 3 x+ 2 − 4 = 0
20
(f) 6 2 x +1 − 7(6 x ) +1 = 0
(e) 2(4 2 x ) −17(4 x ) + 8 = 0
Exercise : Practice 15 , page 85
4.2 Solving Equations involving logarithms METHOD: 1. For two logarithms of the same base, if log a m = log a n , then m = n . 2. Convert to index form, if log a m = n , then m = a n. ACTIVITY 10: 1. Solve the following equations.
(a) lg x = lg 3 + 2 lg 2 –
3 4
lg 16
(b) lg 4 + 2 lg x = 2
(c) lg (4 x – 3) = lg ( x + 1 ) + lg 3
(d) lg (10 x + 5) – lg ( x+ 4 ) = lg 2
2. Solve the following equations : (a) lg 25 + lg x – lg ( x – 1) = 2
(b) 2 lg 3 + lg (2 x) = lg (3 x + 1)
(c) lg x + lg (2 x – 1) = 1
(d) 2 lg 5 + lg (2 x - 1) – lg ( x + 2) = 1
21
(e) log2 x2 = 3 + log2 ( x + 6)
(f)log 2 (1+ x) + log2 (5– x) – log2 ( x –2) = 3
3. Solve the following equations; (a) log x 8 = 5 − log x 4
(b) log 4 ( x + 6) = log 2 3
(c) 3 + log x 4 = 5 log x 2
(d) log 5 (3 x − 5) = 2 log 5 6 − log 5 4
4. Solve each of the following equations(change the base and used the replacement method) (a) 2 log x 3 – log3 x + 1 = 0
(b) log5 x – 2 log x 5 = 1
22
(c) log2 x – 3 = 2 log x 4
Exercise : Practice 16 , Formative Exercise 5.4, Chapter Review Exercise 1-10, page 86-88.
PRACTICE MAKES PERFECT
1. Evaluate each of the following without using calculator or mathematical table. (a) 2 log4 2 –
1
log4 9 + log4 12
[2]
(b) log10 3 – log10 2 + 2 log 10 4 – log10 6 + 2 log 10 5
[2]
2
2. Given that log 7 4 = 0.712 and log 7 5 = 0.827, find the value of: (a) log7 20 (b) log7 0.8
[1.539] [- 0.115]
35
(c) log7 64 (d) log7 100
[- 0.309] [2.366] 2
1
1
3
3. Find the value of 8 3 × 16 − 2 ÷ 4 2 [ ] 8 n+1 n+3 n+5 4. Prove that 2 + 2 + 2 can be divisible by 42 for all positive integers of n. 5. By replacing u = 3 x, solve the equation 2.3 2 x
− 3x = 1 .
[ x = 0]
6. Given that log 3 2 = 0.63, without using calculator or mathematical table, solve the equation 3 7. If 2 x −2.3 x +2
2 x −5
=
3 x
+2
.
2
[ x = 6.37]
= 6 2 x , prove that
6 x
1
= 2 . Hence, find the value of x correct to three 4
significant figures.
[ x = 0.452]
8. Solve the following equations: (a) log2 (5 – x ) + log2 ( 1 + x) = 3 + log 2 ( x – 2) (b) 5 log x 2 + 2 log x 4 – 3 log x 2 = 3 9. Prove that log x 4 =
2 log 2 x
[ x = 3] [ x = 4]
.Hence, solve the equation log x 4 + log2 x + 3 = 0
23
[ x
=
1 2
,x
1
=
4
]
10. Given that lg 2 = p, express the followings in term of p : (a)
lg
[
2
1 2
p
(b) lg 40
[1 + 2 p]
(c) log8 4
[
=
5 x y 11. Find the value of x and y if 4 .16
12. Solve the equation
625 x +2
=
1 16
and 6 x .362 y = 1.
[ x
2 3
]
=−4, y = 1 ] 9
9
1 5 x .25 x −1
Answer:- x
=−
Answer:- x
=1
13. Solve the equation 2 x.8 x = 4 5 x−3
14.
]
2 5
Show that 3 x+2 + 3 x−1 − 5(3 x ) is divisible by 13. Answer : 13(3
x − 1
15.
Solve the equation 64 2 x −3
+ 2 = 34 Answer:-
16.
x
Solve the equation 3 2
)
2 x +1
23 12
= 10 Answer: 0.6477
17.
Solve the equation log 4 [log 2 ( 2 x − 3) ] = log 9 3 Answer:3.5
18.
Solve the equation log( x − 2) − log(4 x −1) = log
1
x
Answer:- 5.2361,0 .7639 19.
Solve the equation log 2x - 4 log x16 = 0 Answer: 16 ,
20.
Solve the equation 2(7
x −1
) =5
= 3.7232
Solve 3log2 x = 81 Answer : x
22.
16
x
Answer: x 21.
1
Solve the equation 3
2 x+1
= 16
x+0.5
- 2 (3
)-3=0 Answer: 0.5
23.
Solve the equation 10 2 x+1 - 7 (10 x) = 26 Answer:0.3010
24.
Given that log3 5 = m and log9 2 = n , express log 3 50 in terms of m and n
24
Answer:- 2m + 2n 25.
Given that log x 2 = k and log x 7 = h , express
log
x
3.5 x
in terms of k and h
Answer:- h − k + 26.
Given that 2 log 2 ( x + y ) = 3 + log 2 x + log 2 y , show that ( x
27.
If log2a + log2b = 4, show that log 4ab = 2 and that log 8ab = 4/3. If log 2a + log2b = 4, show that log 4ab = 2 and that log 8ab = 4/3.
2
+ y = 6 xy )
1 2
2
25
