Problem 101
Question
Evaluate each expression using exponential rules. Write each result in standard form. $$ \frac{8 \times 10^{-1}}{16 \times 10^{5}} $$
Step-by-Step Solution
Verified Answer
The expression evaluates to \( 5 \times 10^{-7} \) in standard form.
1Step 1: Simplify the Coefficients
The expression \( \frac{8}{16} \) simplifies to \( \frac{1}{2} \) since both numbers are divisible by 8. Therefore, the coefficient part simplifies to \( \frac{1}{2} \).
2Step 2: Apply Exponential Rules
Given \( 10^{-1} \) in the numerator and \( 10^{5} \) in the denominator, we use the law of exponents: \( \frac{a^m}{a^n} = a^{m-n} \). So, \( 10^{-1 - 5} = 10^{-6} \).
3Step 3: Combine Simplified Parts
Combine the simplified coefficient \( \frac{1}{2} \) and the result from the exponents \( 10^{-6} \) to form \( \frac{1}{2} \times 10^{-6} \).
4Step 4: Convert to Standard Form
To express \( \frac{1}{2} \times 10^{-6} \) in standard form, calculate \( \frac{1}{2} = 0.5 \). Thus, the expression in standard form is \( 0.5 \times 10^{-6} = 5 \times 10^{-7} \).
Key Concepts
Standard FormLaw of ExponentsSimplifying Fractions
Standard Form
Standard form is a way of writing numbers that makes them easier to read and compare, especially when dealing with very large or very small numbers. It is often used in scientific notation to express numbers neatly. In standard form, a number is written as the product of two parts:
In the context of your exercise, converting \( \frac{1}{2} \times 10^{-6} \) to standard form means turning 0.5 into \(5\) and adjusting the exponent from \(10^{-6}\) to \(10^{-7}\) after taking the decimal into account. Thus, it becomes \(5 \times 10^{-7}\), making it much simpler to understand and use.
- A decimal number, usually between 1 and 10.
- And a power of ten.
In the context of your exercise, converting \( \frac{1}{2} \times 10^{-6} \) to standard form means turning 0.5 into \(5\) and adjusting the exponent from \(10^{-6}\) to \(10^{-7}\) after taking the decimal into account. Thus, it becomes \(5 \times 10^{-7}\), making it much simpler to understand and use.
Law of Exponents
The law of exponents is a set of rules used to simplify expressions involving powers of the same base. One key rule is the quotient rule, which is applied when dividing two exponential terms with the same base. This law states that you subtract the exponents:
\( 10^{-1 - 5} = 10^{-6} \).
These rules simplify complex exponent problems by reducing them to more manageable expressions, letting you solve them with ease.
- \( \frac{a^m}{a^n} = a^{m-n} \)
\( 10^{-1 - 5} = 10^{-6} \).
These rules simplify complex exponent problems by reducing them to more manageable expressions, letting you solve them with ease.
Simplifying Fractions
Simplifying fractions is the process of reducing them to their simplest form. This involves dividing the numerator and denominator by their greatest common factor (GCF).
For the fraction \(\frac{8}{16}\), the GCF is 8, since 8 is the largest number that divides both 8 and 16 evenly. So, we divide both the numerator and the denominator by 8:
For the fraction \(\frac{8}{16}\), the GCF is 8, since 8 is the largest number that divides both 8 and 16 evenly. So, we divide both the numerator and the denominator by 8:
- \(\frac{8}{16} = \frac{8 \div 8}{16 \div 8} = \frac{1}{2}\)
Other exercises in this chapter
Problem 100
In your own words, describe the different methods that can be used to find the product: \((5 x+1)^{2}\).
View solution Problem 100
Simplify each expression. $$ \frac{x^{12} y^{13}}{x^{5} y^{7}} $$
View solution Problem 101
Suppose that a classmate asked you why \((2 x+1)^{2}\) is not \(\left(4 x^{2}+1\right)\). Write down your response to this classmate.
View solution Problem 101
Simplify each expression. $$ \left(5^{0}\right)^{3}+\left(y^{0}\right)^{7} $$
View solution