Problem 16 - Basic Engineering Mathematics

Question

Simplify \(\frac{p^{1 / 2} q^{2} r^{2 / 3}}{p^{1 / 4} q^{1 / 2} r^{1 / 6}}\) and evaluate when \(p=16, q=9\) and \(r=4\), taking positive roots only.

Step-by-Step Solution

Answer

The simplified expression is \(p^{1/4} q^{3/2} r^{1/2}\) and evaluates to 108 when \(p = 16\), \(q = 9\), \(r = 4\).

Step 1: Apply the Quotient Rule for Exponents

The quotient rule for exponents states \(\frac{a^m}{a^n} = a^{m-n}\). Apply this rule to each variable in the expression \(\frac{p^{1 / 2} q^{2} r^{2 / 3}}{p^{1 / 4} q^{1 / 2} r^{1 / 6}}\), resulting in \(p^{1 / 2 - 1 / 4} q^{2 - 1 / 2} r^{2 / 3 - 1 / 6}\).

Step 2: Calculate the Exponent Differences

Compute the differences for each variable's exponent:- For \(p\): \(1/2 - 1/4 = 1/4\)- For \(q\): \(2 - 1/2 = 3/2\)- For \(r\): \(2/3 - 1/6 = 1/2\)Thus, the expression simplifies to \(p^{1/4} q^{3/2} r^{1/2}\).

Step 3: Substitute Given Values

Now substitute \(p = 16\), \(q = 9\), and \(r = 4\) into the simplified expression \(p^{1/4} q^{3/2} r^{1/2}\) resulting in \(16^{1/4} \times 9^{3/2} \times 4^{1/2}\).

Step 4: Evaluate Each Term

Calculate each term separately:- \(16^{1/4} = 2\) (since \(\sqrt[4]{16} = 2\))- \(9^{3/2} = 27\) (since \((\sqrt{9})^3 = 3^3 = 27\))- \(4^{1/2} = 2\) (since \(\sqrt{4} = 2\))This results in \(2 \times 27 \times 2\).

Step 5: Calculate the Final Result

Multiply the evaluated values: \(2 \times 27 \times 2 = 108\). Thus, the final value of the expression with the given substitutions is 108.

Key Concepts

Quotient Rule for Exponents

The quotient rule for exponents is a fundamental principle used to simplify expressions involving powers with the same base. When dividing like bases, you can subtract the exponent of the divisor from the exponent of the dividend. This is given by the formula:

• \(\frac{a^m}{a^n} = a^{m-n}\)

In the original exercise, this rule was applied to simplify the expression \(\frac{p^{1 / 2} q^{2} r^{2 / 3}}{p^{1 / 4} q^{1 / 2} r^{1 / 6}}\). By using the quotient rule, each variable in the numerator and denominator is individually simplified. For example:

• For \(p\): Subtracting exponents gives \(p^{1/2 - 1/4}\).
• For \(q\): Subtracting exponents results in \(q^{2 - 1/2}\).
• For \(r\): Subtracting exponents generates \(r^{2/3 - 1/6}\).

This systematic approach allows for a clear and organized method of reducing the expression to its simplest form.

Simplifying Expressions

Simplifying expressions involves reducing complex algebraic expressions to their simplest form. This is achieved through the application of various algebraic rules and properties, such as the quotient rule for exponents.In our exercise, after applying the quotient rule, we simplify each part of the expression by solving the exponent differences:

• For \(p\): \(1/2 - 1/4 = 1/4\)
• For \(q\): \(2 - 1/2 = 3/2\)
• For \(r\): \(2/3 - 1/6 = 1/2\)

These calculations are straightforward as each part of the algebraic fraction is treated separately. This ensures clarity and prevents potential errors from complex computations. Ultimately, simplifying leads to the expression \(p^{1/4} q^{3/2} r^{1/2}\), making it easier to substitute given values in subsequent steps.

Evaluating Expressions

After simplification, the next step is evaluating the expression by substituting the provided values. This involves calculating the numerical result of the algebraic expression when specific numbers are plugged in for the variables.In the given exercise, the expression \(p^{1/4} q^{3/2} r^{1/2}\) is evaluated using these substitutions: \(p = 16\), \(q = 9\), and \(r = 4\). Each component is computed individually:

• For \(p^{1/4}\): Calculate \(16^{1/4}\), which equals 2, as the fourth root of 16 is 2.
• For \(q^{3/2}\): Calculate \(9^{3/2}\). First, find \(\sqrt{9} = 3\); then \(3^3 = 27\).
• For \(r^{1/2}\): Calculate \(4^{1/2}\), which is 2, as the square root of 4 is 2.

By multiplying these results together, \(2 \times 27 \times 2\), the final solution of 108 is obtained. Evaluating ensures we understand not only the steps of simplification but also how to reach a numerical solution.