Problem 10 - Calculus with Concepts in Calculus

Question

Approximate the number. $$ (16.05)^{1 / 4}(7.95)^{2 / 3} $$

Step-by-Step Solution

Answer

The approximate value of the expression is 8.

Step 1: Simplify the expression

First, we will break down the expression \[ (16.05)^{1/4}(7.95)^{2/3} \]into simpler calculations. Let's approximate the roots separately: - The 4th root of 16.05 is close to the 4th root of 16,- The 2/3 power of 7.95 is close to the 2/3 power of 8.

Step 2: Approximate the 4th root of 16

The 4th root of 16 can be calculated as:\[\sqrt[4]{16} = (16)^{1/4} = (2^4)^{1/4} = 2.\]Therefore, the approximation for \( (16.05)^{1/4} \) is about 2.

Step 3: Approximate the 2/3 power of 8

The 2/3 power of 8 can be calculated as:\[(8)^{2/3} = (2^3)^{2/3} = 2^2 = 4.\]Therefore, the approximation for \( (7.95)^{2/3} \) is about 4.

Step 4: Calculate the product

Now that we have the approximations:\( (16.05)^{1/4} \approx 2 \) and \( (7.95)^{2/3} \approx 4 \),we multiply them together:\[2 \times 4 = 8.\]Thus, the approximate value of the expression is 8.

Key Concepts

Fourth Root

In mathematics, the fourth root of a number is a value that, when multiplied by itself four times, gives the original number. This root is represented using fractional exponents as \(x^{1/4}\). For instance, let's consider the number 16: finding its fourth root means determining what number raised to the power of four equals 16. As shown in the solution, we have: \[\sqrt[4]{16} = (16)^{1/4} = (2^4)^{1/4} = 2\]This process hinges on understanding powers of numbers. Since \(2^4 = 16\), the fourth root of 16 is simply 2. The same logic is applied when approximating the fourth root of numbers slightly different from perfect powers, like 16.05. It's reasonable to approximate \(16.05^{1/4}\) as close to 2 because 16.05 is very close to 16.

• Used to simplify calculations involving higher powers.
• Particularly handy when dealing with even powers.
• Useful for approximating radical expressions.

Fractional Exponents

Fractional exponents extend the concept of integer exponents to non-integer powers. With fractional exponents, the numerator represents the power to which the base is raised, and the denominator represents the root that is being taken. For example, in \(x^{2/3}\), the number 2 signifies squaring the base, and 3 indicates taking the cube root. Let's illustrate this using the number 8:\[(8)^{2/3} = (2^3)^{2/3} = 2^2 = 4\]Here, \(8 = 2^3\), and taking it to the power of \(2/3\) involves:- Cubing the base first (since 8 is the cube of 2).- Raising to the power of 2 (square the result of cube rooting).Thus the \(2/3\) power of 8 is 4. When the number involved, such as 7.95, is close to 8, approximating with such powers becomes practical.

• Simplifies complex root and power operations.
• Allows for easy switch between roots and powers.
• Key in understanding logarithmic functions.

Multiplication of Exponents

Understanding the principle of multiplying exponents is crucial when handling expressions with multiple terms raised to power. When multiplying two exponents with the same base, we add the powers:\[a^m \times a^n = a^{m+n}\]In the context of simplifying expressions involving roots and fractional powers, it's about seamlessly managing different higher roots and powers:1. Calculate each part individually.2. Apply the basic rules of exponentiation.Back to our problem, after determining \(16.05^{1/4}\) and \(7.95^{2/3}\), we have:- Approximated values: 2 and 4.- Multiplying these results gives: \(2 \times 4 = 8\).This general rule simplifies working with complex calculations, particularly when variables or larger expressions are at play.

• Simplifies working with expressions with powers.
• Essential for polynomial and radical functions.
• Reduces computational complexity.