Problem 20
Question
Simplify. $$6^{7} \cdot 6^{-10} \cdot 6^{2}$$
Step-by-Step Solution
Verified Answer
\(\frac{1}{6}\)
1Step 1: Apply the Product of Powers Law
Combine the terms using the Product of Powers law.
\(6^{7} \cdot 6^{-10} \cdot 6^{2} = 6^{7 + (-10) + 2}\)
2Step 2: Perform the Addition
Add the exponents together.
\begin{align*}
6^{7 + (-10) + 2} &= 6^{(7-10) + 2}\\
&= 6^{(-3) + 2}
\end{align*}
3Step 3: Continue Adding the Exponents
Continue adding the exponents.
\begin{align*}
6^{(-3) + 2} &= 6^{(-1)}
\end{align*}
Now, we have simplified the expression to \(6^{-1}\). This expression is as simple as it gets, but we can also express this as a fraction.
4Step 4: Express as a Fraction
Express the result as a fraction using the property \(a^{-n} = \frac{1}{a^n}\).
\begin{align*}
6^{-1} = \frac{1}{6^{1}}
\end{align*}
As a final answer, we have: \(\frac{1}{6}\)
Key Concepts
Product of PowersNegative ExponentsSimplifying Expressions
Product of Powers
In mathematics, when we have the same base and are multiplying, we use the **Product of Powers** rule. This rule makes it easier to simplify expressions involving exponents. You simply add the exponents together when multiplying terms that share the same base. For example, with numbers like \(a^m \cdot a^n\), this becomes \(a^{m+n}\). Each term here has base \(a\).
Let's apply that to our example: \(6^7 \cdot 6^{-10} \cdot 6^2\). All terms here have the base \(6\). Therefore, we add the exponents: \(7 + (-10) + 2\). This simplifies to \(6^{-1}\). By understanding the Product of Powers rule, you've turned a cumbersome multiplication expression into something more manageable!
Let's apply that to our example: \(6^7 \cdot 6^{-10} \cdot 6^2\). All terms here have the base \(6\). Therefore, we add the exponents: \(7 + (-10) + 2\). This simplifies to \(6^{-1}\). By understanding the Product of Powers rule, you've turned a cumbersome multiplication expression into something more manageable!
Negative Exponents
Negative exponents might seem tricky, but they're quite simple once you understand the concept. A **Negative Exponent** indicates a reciprocal. Let's break that down. The expression \(a^{-n}\) equals \(\frac{1}{a^n}\). Essentially, a negative exponent "flips" the base to the denominator.
In our solution, we simplified the expression down to \(6^{-1}\). Using the rule for negative exponents, this converts to \(\frac{1}{6^1}\), or more straightforwardly, just \(\frac{1}{6}\). Recognizing and handling negative exponents will help you rewrite expressions in a more simplified, cleaner form.
In our solution, we simplified the expression down to \(6^{-1}\). Using the rule for negative exponents, this converts to \(\frac{1}{6^1}\), or more straightforwardly, just \(\frac{1}{6}\). Recognizing and handling negative exponents will help you rewrite expressions in a more simplified, cleaner form.
Simplifying Expressions
Simplifying expressions is all about making life easier! It involves reducing complex expressions into simpler, equivalent forms. The steps are methodical, making everything more straightforward. Let’s consider how this applies.
- First, use the Product of Powers rule to combine exponents. For instance, in \(6^7 \cdot 6^{-10} \cdot 6^2\), we simplify by combining the exponents (7, -10, 2) to get \(6^{-1}\).
- Next, approach any negative exponents. By recognizing \(a^{-n}\) as \(\frac{1}{a^n}\), you simplify the negative exponent to a fraction, translating \(6^{-1}\) into \(\frac{1}{6}\).
Other exercises in this chapter
Problem 20
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Solve each quadratic inequality. Graph the solution set and write the solution in interval notation. $$2 a^{2} \geq 7 a$$
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Identify the center of each ellipse and graph the equation. $$\frac{(x-2)^{2}}{16}+\frac{y^{2}}{25}=1$$
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