Problem 17

Question

Find the value of \(\frac{2}{5^{-2}}+6^{-2} \cdot 2^{3} \cdot 3^{2}\)

Step-by-Step Solution

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Answer
Answer: The value of the given expression is \(52\).
1Step 1: Simplify Exponents
First, we need to simplify the exponents by finding the inverse of the base. For example, for \(5^{-2}\), we need to find the value of \(\frac{1}{5^2}\): \begin{align*} 5^{-2} &= \frac{1}{5^2} \\ &= \frac{1}{25} \end{align*} Similarly, for \(6^{-2}\) and \(2^3\), we find the values: \begin{align*} 6^{-2} &= \frac{1}{6^2} \\ &= \frac{1}{36} \end{align*} \begin{align*} 2^3 &= 2 \cdot 2 \cdot 2 \\ &= 8 \end{align*} Now we have: $$\frac{2}{5^{-2}}+6^{-2} \cdot 2^{3} \cdot 3^{2} = \frac{2}{\frac{1}{25}} + \frac{1}{36} \cdot 8 \cdot 9$$
2Step 2: Perform Multiplication and Division
Now that we have the simplified exponents, we can perform the multiplication and division in the expression above. For the first term, we get: \begin{align*} \frac{2}{\frac{1}{25}} &= 2 \cdot 25 \\ &= 50 \end{align*} For the second term, we have: \begin{align*} \frac{1}{36} \cdot 8 \cdot 9 &= \frac{8 \cdot 9}{36}\\ &= \frac{72}{36} \end{align*} So, our expression now is: $$50 + \frac{72}{36}$$
3Step 3: Perform the Addition
Lastly, we need to perform the addition. Since the terms are all whole numbers, we can simply add them: \begin{align*} 50 + \frac{72}{36} &= 50 + 2 \\ &= 52 \end{align*} The value of the given expression is \(52\).

Key Concepts

Understanding Inverse ExponentsThe Role of Multiplication in ExpressionsPerforming Addition with Simplified TermsProcess of Simplification
Understanding Inverse Exponents
When dealing with exponents, one crucial concept is the inverse. An inverse of a number essentially means taking the reciprocal. In the context of exponents, such as in the expression \(5^{-2}\), we are looking for the reciprocal of \(5^2\). Here's how it works:- A negative exponent indicates taking the reciprocal of the base raised to the positive of that power.- So, \(5^{-2}\) becomes \(\frac{1}{5^2}\), which simplifies to \(\frac{1}{25}\).- Similarly, for another example, \(6^{-2}\) becomes \(\frac{1}{6^2} = \frac{1}{36}\).Understanding this concept helps in simplifying and correctly manipulating numbers with negative exponents, making calculations more manageable and less error-prone.
The Role of Multiplication in Expressions
Multiplication is a foundational operation in mathematics, and in many expressions, it plays a key role in simplifying and solving equations. Let's break it down in simpler terms. In the expression \(6^{-2} \cdot 2^3 \cdot 3^2\):
  • Simplifying each part involves calculating the exponents first: \(2^3 = 8\) and \(3^2 = 9\).
  • This gives us a multiplication problem: \(\frac{1}{36} \cdot 8 \cdot 9\).
  • By multiplying these, you reduce the equation step by step: \(8 \cdot 9 = 72\), then \(\frac{72}{36} = 2\).
Breaking down multiplication into steps helps avoid mistakes and ensures accuracy in calculations.
Performing Addition with Simplified Terms
Once the steps of dealing with exponents and multiplication are completed, the next step in evaluating an expression is addition. Addition combines simplified terms into a final result.In our expression, after all the prior steps, we have it reduced to:
  • \(50 + \frac{72}{36}\), which simplifies to \(50 + 2\).
  • Add the numbers directly, yielding \(52\).
In essence, when all factors are simplified to their numerical values, addition allows combining them into one final answer.
Process of Simplification
Simplification is a method to make complex mathematical expressions easier to understand and work with by breaking them down into more basic parts. Let's consider the expression step by step. Initially, we had \(\frac{2}{5^{-2}}+6^{-2} \cdot 2^{3}\cdot 3^{2}\). By applying inverse, multiplication, and addition, we simplified:
  • Exponents changed using inverse properties.
  • Performed multiplication to condense values.
  • Added the results for the final expression.
Simplification is essential for turning complicated expressions into more straightforward calculations, ensuring the result is accurate and easy to achieve.