Problem 7 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Rewrite the number without radicals or exponents.. $$ 8^{2 / 3} $$

Step-by-Step Solution

Answer

The simplified expression without radicals or exponents is \(8^{\frac{2}{3}} = 4\).

Step 1: Convert the exponent to a radical

Since we have a fractional exponent, we can rewrite it as a radical. Recall that the expression \(a^{\frac{m}{n}}\) can be written as \(\sqrt[n]{a^m}\). In this case, a=8, m=2, and n=3, so we have: \[ \left(8^{\frac{2}{3}}\right) =\sqrt[3]{8^2} \]

Step 2: Simplify the expression within the radical

Now, we need to simplify the expression within the radical, i.e., \(8^2\). This is simple exponentiation: \[ 8^2 = 64 \] So, our expression can now be written as: \[ \sqrt[3]{8^2} = \sqrt[3]{64} \]

Step 3: Find the cube root

The final step is to find the cube root of 64. We need to find a number, when multiplied by itself three times, gives us 64. That number is 4, as \(4 \times 4 \times 4 = 64\). So the cube root of 64 is 4. Thus, our simplified expression without radicals or exponents is: \[ 8^{\frac{2}{3}} = 4 \]

Key Concepts

Fractional Exponents

Fractional exponents might sound a bit daunting at first, but they actually give us a very convenient way to express roots and powers in one neat expression. When you see a fractional exponent, like in the problem 8\(^{2/3}\), it's helpful to break it down. Generally, the expression \(a^{\frac{m}{n}}\) can be read as finding the \(n\)-th root of \(a\) raised to the power of \(m\). This comes in handy because it combines two operations: root extraction and exponentiation.

With fractional exponents, think of the denominator as indicating the root, and the numerator as indicating the power. In our example of \(8^{2/3}\), the 3 in the denominator tells us to take the cube root, while the 2 in the numerator tells us to square the result. This method keeps things organized and straightforward when dealing with complex expressions.

Using fractional exponents is like having a multi-tool for mathematics – it simplifies dealing with complex roots and powers while maintaining a clean notation. Understanding how to manipulate these exponents is crucial to simplifying radical expressions.

Cube Root

The cube root forms the core of many problems involving radicals, particularly those involving cube powers. The cube root of a number is a value that, when multiplied by itself three times, yields the original number. In the exercise, we dealt with finding the cube root of 64. So, we were searching for some number \(x\) such that \(x \times x \times x = 64\).

Determining the cube root may seem tricky, but understanding how it works can simplify many calculations. Cube roots can often be identified through small perfect cubes, like 1, 8, 27, 64, 125, and so on. Recognizing these common cubes makes it easier to solve such problems.

For example, the cube root of 27 is 3 because \(3^3 = 27\). Similarly, as seen in the exercise, the cube root of 64 is 4, because \(4^3 = 64\). This understanding is key when breaking down expressions containing cube roots into simpler forms and integrating them with other mathematical expressions.

Exponentiation

Exponentiation is a fundamental mathematical operation that involves raising a number to a power. When you see something like \(8^2\), this means you'll be multiplying 8 by itself once more, giving us \(8 \times 8 = 64\). This process gets significant when handling expressions with radical components, such as \(8^{2/3}\).

In the context of the problem, after rewriting \(8^{2/3}\) into \(\sqrt[3]{8^2}\), you first perform the exponentiation \(8^2\) to simplify what's under the radical. Upon doing the exponentiation, the intermediate step is \(64\).

Exponentiation allows us to express and compute large numbers more easily and is essential when simplifying expressions involving fractional exponents. Mastery of this not only aids in handling textbook problems but also in comprehending more advanced mathematical theories that rely on these foundational concepts.