In general, the format for using exponents is: (base)exponent where the exponent tells you how many of the base are being multiplied together.
Consider: 2 • 2 • 2 is the same as 23, since there are three 2’s being multiplied together. Likewise, 5 • 5 • 5 • 5 = 54, because there are four 5’s being multiplied together.
Exponents are also referred to as “powers”. For example, 23 can be read as “two cubed” or as “two raised to the third power”.
When we multiply negative numbers together, we must utilize parentheses to switch to exponent notation. (-4)(-4)(-4)(-4)(-4)(-4) = (-4)6
The examples below show that (-4)6 and -46 are not the same.
The missing parentheses mean that -46 will multiply six 4’s together first (by order of operations), and then take the negative of that answer.
Example: (-3)2 = (-3) × (-3) = +9 The 2 is “attached” to the parentheses, so everything inside the parentheses is squared.
-32 is not the same as (-3)2 -32 = -(32) The 2 is “attached” to the 3, but not to the negative sign. -32 = -(32) = -(3 × 3) = -(9) = -9 The expression -32 (with missing parentheses) means to multiply two 3’s together first (by order of operations), and then take the negative of that answer.
Even powers of negative numbers allow for the negative values to be arranged in pairs. This pairing guarantees that the answer will always be positive. Odd powers of negative numbers, however, always leave one factor of the negative number not paired. This one lone negative term guarantees that the answer will always be negative.
Zero Exponents
The number zero may be used as an exponent. The value of any expression raised to the zero power is 1. (Except zero raised to the zero power is undefined.)
Negative Exponents
Negative numbers as exponents have a special meaning. The rule is as follows: For example:
Exponents and Units
When working with units and exponents (or powers), remember to adjust the units appropriately. (36 ft)3 = (36 ft) • (36 ft) • (36 ft) = (36 • 36 • 36) (ft • ft • ft) = 46656 ft3
Exponents can be very useful for evaluating expressions. It is also useful to learn how to use your calculator when working with exponents
Multiplication of a decimal by 10, 100, 1000 etc.:
Method: On multiplying a decimal number by 10, 100, 1000, … the decimal point is shifted to the right by one, two, three, … places respectively. For example,
Multiplication of a decimal by a whole number:
Method : Multiply the whole number by decimal (without the decimal point). Mark the decimal point in the product from right side to have as many decimals as there are in the given decimal. For example, 12 × 3.82 First find the product of 12 and 382 (ignoring decimal) 382 × 12 Now, 3.82 × 12 = 45.84 (mark the point after two digits from right).
Multiplication of a decimal by a decimal:
Method : 1. Multiply the decimal numbers as of ordinary number (ignoring decimal points) 2. Mark the decimal point in the product after as many places (from the right) as the sum of the decimal places in the each number. For example, 82.53 × 7.4 First find the product of 8253 and 74 (ignoring decimal point) Now, 82.53 × 7.4 = 610.722 (mark the decimal point after (2 + 1 = 3) digits from right).
Multiplication of Decimal Numbers Problems with Solutions
1. Multiply : (i) 1.6 by 0.3 (ii) 8.03 by 2.9 (iii) 0.657 by 27 Solution:
2. Find the following products : (i) 23.25 × 5 (ii) 2.325 × 25 Solution: (i) 23.25 × 5 So, 23.25 × 5 = 116.25 Step 1 : Multiply the multiplicand by the multiplier without bothering about the decimal point. Step 2 : Count the number of digits in the multiplicand after decimal point. It is 2 in this case. Count two digits from the unit place in the product and put a decimal point. Therefore, 23.25 × 5 = 116.25
(ii) 2.325 × 25 So, 2.325 × 25 = 58.125 Step 1 : Multiply the multiplicand by the multiplier without bothering about the decimal point. Step 2 : The multiplicand has 3 places of decimal. Count three digits from the unit place of the product and put the decimal point. Therefore, 2.325 × 25 = 58.125
3. Multiply 6.7 × 4.25 × 12.3 Solution: (i) 6.7 × 4.25 × 12.3 = (6.7 × 4.25) × 12.3 = 28.475 × 12.3 = 350.2425 Also we can make the grouping as (ii) 6.7 × 4.25 × 12.3 = 6.7 × (4.25 × 12.3) = 6.7 × 52.275 = 350.2425 We find that (6.7 × 4.25) × 12.3 = 6.7 × (4.25 × 12.3) Hence, To find the product of three decimal fractions, we can regroup them in any order, the result is the same in both cases. Thus, multiplication of decimals is associative.
4. Find (i) 10.05 × 1.05 (ii) 100.01 × 1.1 Solution: (i) First multiply 1005 by 105 Sum of decimal places in the given decimal = (2 + 2) = 4 So, product will contain 4 places of decimals from the right side. 10.05 × 1.05 = 10.5525 (ii) 100.01 × 1.1 First multiply 10001 by 11. Sum of decimal places in the given decimals = (2 + 1) = 3 So, product will contain 3 places of decimals from the right side. 100.01 × 1.1 = 110.011
Division of Decimal Numbers
Dividing a decimal by 10, 100, 1000 etc.:
Method: On dividing a number by 10, 100, 1000, … the digits of the number and quotient are same but the decimal point in the quotient shifts to left by one, two, three, … places. For example, 3.27 ÷ 10 = 0.327 3.27 ÷ 100 = 0.0327 3.27 ÷ 1000 = 0.00327
Dividing a decimal by a whole number:
Method: (i) Divide the dividend considering it as a whole number. (ii) When the division of whole-number part of the dividend is complete, mark the decimal point in the quotient and proceed with the division as in case of whole number. For example,
Dividing a decimal by a decimal:
Method: (i) Convert the divisor into a whole number by multiplying it by 10, 100, 1000, … etc, depending upon the number of decimal places in it. Also we multiply the dividend by the same multiplier. (ii) Divide the new dividend by the whole number obtained above. For example, 22.08 ÷ 1.5
Division of Decimal Numbers Problems with Solutions
1. Find 15.225 ÷ 0.35 Solution: Thus, we note that 15.225 ÷ 0.35 = 1522.5 ÷ 0.35 = 1522.5 ÷ 35 Thus if the decimal point is moved to two places towards right in the divisor then the decimal point is also moved to the right in dividend by same number of places.
2. Find 50.76 ÷ 9.4 Solution: Hence, 50.76 ÷ 9.4 = 507.6 ÷ 94 Thus, we note that we can make the divisor as a whole number by shifting the decimal point to right by as many places as the number of the decimal places in the divisor. This way, the divisor is changed into a whole number.
3. Divide (i) 15.225 by 0.35 (ii) 50.76 ÷ 9.4 Solution:
4. Find the quotient of 0.06688 ÷ 0.038 Solution: Make the divisor a whole number by shifting the decimal point in dividend to the right by three places,
5. Find 0.024 ÷ 0.6 Solution:
5. Find 64 ÷ 0.08 Solution: [Shift the decimal points two places to the right in both the numbers]
Dividing of a whole number by a decimal: For example,
Law-2: If a and b are non-zero integers and m is a positive integer, then
am × bm = (a × b)m
Eg : 53 × 33 = (5 × 5 × 5) × (3 × 3 × 3) = (5 × 3) × (5 × 3) × (5 × 3) = 15 × 15 × 15 = (15)3 So, 53 × 33 = (5 × 3)3 = (15)3 Here, we find that 15 is the product of bases 5 and 3. Also, if a and b are non-zero integers, then a5 × b5 = (a × a × a × a × a) × (b × b × b × b × b) = (a × b) (a × b) (a × b) (a × b) (a × b) = (ab)5
Law-3: If a is a non-zero integer and m and n are two whole numbers such that m > n, then When an exponential form is divided by another exponential form whose bases are same, then the resultant is an exponential form with same base but the exponent is the difference of the exponent of the divisor from the exponent of the dividend.
Law-4: Division of exponential forms with the same exponents and different base: If a and b are any two non-zero integers, have same exponent m then for am ÷ bm, we write Law-5: If ‘a’ be any non-zero integer and m and n any two positive integers then
Law-6: Law of zero Exponent: We know that By using Law-3 of exponents, we have 26 ÷ 26 = 26–6 = 20 Thus, 20 = 1 In general am ÷ am = am – m = a0 and also Hence, a0 = 1 Any non-zero integer raised to the power 0 always results into 1.
Use Of Exponents In Expressing Large Numbers
We know that 100 = 10 × 10 = 102, 1000 = 10 × 10 × 10 = 103, 10000 = 10 × 10 × 10 × 10 = 104 We can write a number followed by large number of zeroes in powers of 10. For example, we can write the speed of light in vacuum = 300,000,000 m/s = 3 × 1,00,000,000 m/s = 3 × 108 m/s = 30 × 107 m/s = 300 × 106 m/s Similarly, the age of universe = 8,000,000,000 years (app.) = 8 × 109 years (app.) We can also express the age of universe as 80 × 108 years or 800 × 107 years, etc. But generally the number which preceded the power of 10 should be less than 10. Such a notation is called standard or scientific notation. So 8 × 109 years is the standard form of the age of the universe. Similarly, the standard form of the speed of light is 3 × 108 m/s.
Eg: Write the following numbers in standard form : (i) 4340000 (ii) 173000 (iii) 140000 Solution: (i) It is clear that 4340000 = 434 × 10000 Also, 4340000 = 4.34 × 106 ∵ 434 = 4.34 × 100 = 4.34 × 102 (ii) Also, 173000 = 1.73 × 105 (iii) Also, 140000 = 1.4 × 105
Interest: Interest is the amount paid in lieu of using some money which is not owned by us.
The amount of money deposited, lent or borrowed is called principal (P).
The additional money given at the end of a period for using the principal is called interest.
The total money we receive or pay is called the amount due at that time. Thus the sum of principal and interest is called amount. i.e. amount = principal + interest
The time for which the money is kept in the bank or for which the loan has been borrowed is called the time period.
To find the simple interest on a certain amount of money we need to know three quantities. (i) Amount deposited or borrowed is called principal (P) (ii) Rate of interest (R) (iii) Time period (T)
Note: If the rate of interest is given per annum then the time period must be expressed in terms of year. For Example Time period
Simple Interest Problems with Solutions
1. Find the simple interest when; Principal = ₹ 600, Rate = 2% per annum and Time = 20 months. Solution: We have, P = Principal = ₹ 600, R = Rate percent per annum = 2% And T = Time = 20 months = year Therefore, simple interest (S.I.) Thus S.I. = ₹ 20.
2. Find the principal when Simple Interest = ₹ 72, Rate = 3% per annum and Time = 3 months. Solution:
3. Find the rate when Principal = ₹ 700, Simple Interest = ₹ 168 and Time = 16 months. Solution:
4. Find the time when principal = ₹ 640, Rate =12½ % per annum and Simple Interest = ₹ 40. Solution:
5. Neeraj borrowed a sum of money at 10½ % per annum from a bank. If he paid ₹ 1863.75 as interest for 2½ years, find the sum. Solution: Hence, the required sum = ₹ 7100
6. A sum of money becomes \(\frac { 7 }{ 4 }\) of itself in 6 years at a certain rate of interest. Find the rate of interest. Solution:
7. If Meena gives an interest of ₹ 45 for one year at 9% rate p.a., what is the sum she has borrowed? Solution:
8. What rate gives ₹ 280 as interest on a sum of ₹ 56,000 in 2 years ? Solution:
9. Find the amount to be paid at the end of 3 years in each case : (i) Principal = ₹ 1200 at 12% p.a. (ii) Principal = ₹ 7500 at 5% p.a. Solution:
The repeated addition of numbers can be written in short form (product form). Examples: Also, we can write the repeated multiplication of numbers in a short form known as exponential form. For example, when 5 is multiplied by itself for two times, we write the product 5 × 5 in exponential form as 52 which is read as 5 raised to the power two.
Similarly, if we multiply 5 by itself for 6 times, the product 5 × 5 × 5 × 5 × 5 × 5 is written in exponential form as 56 which is read as 5 raised to the power 6. In 56, the number 5 is called the base of 56 and 6 is called the exponent of the base.
In general, we write, An exponential number as ba, where b is the base and a is the exponent. The notation of writing the multiplication of a number by itself several times is called the exponential notation or power notation.
Thus, in general we find that : If ‘a’ is a rational number then ‘n’ times the product of ‘a’ by itself is given as a × a × a × a ….. , n times and is denoted by an, where ‘a’ is called the base and n is called the exponent of an.
1. Write the following statements as repeated multiplication and complete the table:
S.No.
Statements
Repeated Multiplication
Short form
(i)
3 multiplied by 3 for 6 times
3 × 3 × 3 × 3 × 3 × 3 = 729
36
(ii)
2 multiplied by 2 for 3 times
2 × 2 × 2
23
(iii)
1 multiplied by 1 for 7 times
1 × 1 × 1 × 1 × 1 × 1 × 1
17
2. Write the base and exponent of following numbers. And also write in expanded form:
S.No.
Numbers
Base
Exponent
Expanded Form
Value
(i)
34
3
4
3 × 3 × 3 × 3
81
(ii)
25
2
5
2 × 2 × 2 × 2 × 2
32
(iii)
33
3
3
3 × 3 × 3
27
(iv)
22
2
2
2 × 2
4
(v)
17
1
7
1 × 1 × 1 × 1 × 1 × 1 × 1
1
Exponents of Negative Integers
When the exponent of a negative integer is odd, the resultant is a negative number, and when the power of a negative number is even, the resultant is a positive number.When the exponent of a negative integer is odd, the resultant is a negative number, and when the power of a negative number is even, the resultant is a positive number. or (a negative integer) an odd number = a negative integer. (a negative integer) an even number = a positive integer.
Examples:
Ex.1 Express 144 in the powers of prime factors. Solution: 144 = 16 × 9 = 2 × 2 × 2 × 2 × 3 × 3 Here 2 is multiplied four times and 3 is multiplied 2 times to get 144. ∴ 144 = 24 × 32