Eighth Grade Mathematics: Multiplication and Division in Scientific Notation
Performing arithmetic with numbers that have dozens of zeros is cumbersome in standard notation, but scientific notation transforms multiplication and division into streamlined algebraic operations. By separating numerical coefficients from powers of ten, you can apply basic arithmetic to the coefficients while applying the Product and Quotient Rules of Exponents to the powers of ten. In this chapter, you will master multiplying and dividing numbers in scientific notation, normalizing results into proper form, and solving real-world scale problems.
Multiplying Numbers in Scientific Notation
To multiply two numbers expressed in scientific notation, use the commutative and associative properties of multiplication to regroup the expression into two distinct parts: the product of the coefficients, and the product of the powers of ten.
Multiplication Algorithm:
(a * 10^m) * (c * 10^n) = (a * c) * 10^(m + n)
Step 1: Multiply the decimal coefficients a and c. Step 2: Apply the Product Rule of Exponents to add the powers of ten: 10^m * 10^n = 10^(m + n). Step 3: Inspect the product of the coefficients. If it is greater than or equal to 10, adjust the decimal point and increase the exponent by the number of places shifted (normalization).
Example: Multiply (4.0 * 10^5) * (3.0 * 10^7)
Step 1: Regroup:
(4.0 * 3.0) * (10^5 * 10^7)
Step 2: Compute components:
12.0 * 10^(5 + 7) = 12.0 * 10^12
Step 3: Normalize (12.0 is not between 1 and 10):
12.0 = 1.2 * 10^1
(1.2 * 10^1) * 10^12 = 1.2 * 10^13
Notice that when the decimal point in the coefficient moves one place to the left, the exponent must increase by 1 to maintain numerical equivalence.
Dividing Numbers in Scientific Notation
To divide two numbers expressed in scientific notation, divide the coefficient of the numerator by the coefficient of the denominator, and apply the Quotient Rule of Exponents to subtract the exponent of the denominator from the exponent of the numerator.
Division Algorithm:
(a * 10^m) / (c * 10^n) = (a / c) * 10^(m - n) (where c != 0)
Step 1: Divide the decimal coefficient a by c. Step 2: Apply the Quotient Rule of Exponents to the powers of ten: 10^m / 10^n = 10^(m - n). Step 3: Inspect the resulting coefficient. If it is less than 1, adjust the decimal point to the right and decrease the exponent accordingly (normalization).
Example: Divide (2.4 * 10^4) / (8.0 * 10^9)
Step 1: Regroup:
(2.4 / 8.0) * (10^4 / 10^9)
Step 2: Compute components:
0.3 * 10^(4 - 9) = 0.3 * 10^(-5)
Step 3: Normalize (0.3 is less than 1):
0.3 = 3.0 * 10^(-1)
(3.0 * 10^(-1)) * 10^(-5) = 3.0 * 10^(-6)
When shifting the decimal point one place to the right to make 0.3 into 3.0, the exponent decreases by 1 (from -5 down to -6).
The Normalization Rule Summary
Adjusting non-standard scientific notation to proper form is a common stumbling block. The normalization rule can be remembered by a balanced seesaw principle: if you change the coefficient in one direction, the exponent must change in the opposite direction to keep the value balanced.
The Normalization Seesaw Principle:
- Move decimal LEFT (coefficient gets SMALLER) ---> Exponent goes UP (ADD to exponent)
- Move decimal RIGHT (coefficient gets LARGER) ---> Exponent goes DOWN (SUBTRACT from exponent)
Examples of Normalization:
45.8 * 10^6 ---> Move left 1 ---> 4.58 * 10^7
0.072 * 10^5 ---> Move right 2 ---> 7.2 * 10^3
0.0051 * 10^(-4) -> Move right 3 -> 5.1 * 10^(-7)
Chapter Exercises
Exercise 1: Multiply (6.5 * 10^7) * (4.0 * 10^(-3)). Write the final answer in proper scientific notation, showing all regrouping and intermediate steps.
Exercise 2: Divide (1.44 * 10^(-2)) / (1.2 * 10^(-7)). Express the quotient in proper scientific notation, paying careful attention to subtracting negative exponents.
Exercise 3: Light travels through a vacuum at a speed of approximately 3.0 * 10^8 meters per second. The average distance from the Sun to the Earth is approximately 1.5 * 10^11 meters. Calculate the time in seconds it takes for sunlight to reach Earth using scientific notation. Convert your final answer to minutes and seconds.
Solutions and Explanations
Solution 1: To multiply (6.5 * 10^7) * (4.0 * 10^(-3)), regroup the expression into coefficients and powers of ten: (6.5 * 4.0) * (10^7 * 10^(-3)). Multiplying the coefficients gives 6.5 * 4.0 = 26.0. Applying the Product Rule to the powers of ten gives 10^(7 + (-3)) = 10^4. This yields 26.0 * 10^4. Since 26.0 is not between 1 and 10, normalize the number by shifting the decimal point one place to the left: 26.0 = 2.6 * 10^1. Combining this with 10^4 gives 2.6 * (10^1 * 10^4) = 2.6 * 10^(1 + 4) = 2.6 * 10^5.
Solution 2: To divide (1.44 * 10^(-2)) / (1.2 * 10^(-7)), regroup into the quotient of coefficients and the quotient of powers of ten: (1.44 / 1.2) * (10^(-2) / 10^(-7)). Dividing the coefficients gives 1.44 / 1.2 = 1.2. Applying the Quotient Rule to the powers of ten requires subtracting the exponent of the denominator from the exponent of the numerator: -2 - (-7) = -2 + 7 = 5. Thus, 10^(-2) / 10^(-7) = 10^5. The combined expression is 1.2 * 10^5. Because 1.2 satisfies 1 <= 1.2 < 10, the result is already in proper scientific notation.
Solution 3: To find the travel time of sunlight, use the distance formula time = distance / speed. Substitute the given values: time = (1.5 * 10^11 meters) / (3.0 * 10^8 meters per second). Regrouping gives (1.5 / 3.0) * (10^11 / 10^8). Dividing the coefficients yields 1.5 / 3.0 = 0.5. Applying the Quotient Rule of Exponents yields 10^(11 - 8) = 10^3. This gives 0.5 * 10^3 seconds. To normalize into proper scientific notation, move the decimal point one place to the right to obtain 5.0, and decrease the exponent by 1: 5.0 * 10^2 seconds. In standard notation, 5.0 * 10^2 equals 500 seconds. To convert 500 seconds into minutes and seconds, divide 500 by 60: 500 / 60 gives 8 with a remainder of 20. Therefore, it takes approximately 5.0 * 10^2 seconds, or 8 minutes and 20 seconds, for sunlight to travel from the Sun to the Earth.