Problem 52 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Simplify the expression, writing your answer using positive exponents only. $$ \frac{(u v)^{-1}}{u^{-1}+v^{-1}} $$

Step-by-Step Solution

Answer

The simplified expression with only positive exponents is: \(\frac{1}{u+v}\).

Step 1: Rewrite expressions using positive exponents

The first thing we need to do is rewrite the given expression using positive exponents. We can do this by knowing that \(a^{-n} = \frac{1}{a^n}\) and apply it to every negative exponent in the expression: \[ \frac{(u v)^{-1}}{u^{-1}+v^{-1}} = \frac{\frac{1}{(uv)^1}}{\frac{1}{u^1}+\frac{1}{v^1}} \]

Step 2: Find a common denominator for the denominator of the given expression

To simplify the given expression further, let's find a common denominator for the denominator part. In this case, the common denominator will be the product of \(u^1\) and \(v^1\), which is \(uv\). Now let's rewrite the denominator using the common denominator: \[ \frac{\frac{1}{(uv)^1}}{\frac{1}{u^1}+\frac{1}{v^1}} = \frac{\frac{1}{(uv)^1}}{\frac{v^1+u^1}{uv}} \]

Step 3: Simplify the expression by dividing fractions

Now we have a fraction divided by another fraction. To solve this, we can use the rule \(\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}\), so our expression becomes: \[ \frac{\frac{1}{(uv)^1}}{\frac{v^1+u^1}{uv}} = \frac{1}{(uv)^1} \times \frac{uv}{v^1+u^1} \]

Step 4: Cancel terms within the expression

Observe that we can actually cancel the term \((uv)^1\) from the multiplication: \[ \frac{1}{(uv)^1} \times \frac{uv}{v^1+u^1} = \frac{1}{1} \times \frac{1}{v^1+u^1} \] This simplifies our expression to just: \[ \frac{1}{v^1+u^1} \] And since the exponents are both 1, we can remove them. Our final answer is: \[ \frac{1}{u+v} \]

Key Concepts

Exponents and Powers

Exponents are a way of expressing repeated multiplication of the same number. When dealing with expressions like \(a^n\), the "\(n\)" is the exponent, and "\(a\)" is the base. In this expression, the base \(a\) is multiplied by itself \(n\) times. For example, \(3^4\) means \(3 \times 3 \times 3 \times 3\), which results in 81.
Using powers makes it easier to handle large numbers. An important rule is that any number to the power of zero is one, written as \(a^0 = 1\). This holds true for any non-zero base "a". Also, the power of a power rule states \((a^m)^n = a^{m\times n}\). These rules help simplify complex expressions.

Fraction Simplification

Simplifying fractions involves reducing the expression to its simplest form. This means shrinking the fraction so the numerator and denominator are the smallest values possible. Let's consider the fraction reduction process::

• Identify a common factor in both the numerator and the denominator.
• Divide both by the common factor to reduce the fraction.

When simplifying fractions that include powers or more complex expressions, the process is similar. You first find a common factor, like an algebraic term, between expressions in the numerator and the denominator and simplify where possible. This step is critical for making mathematics simpler and more manageable without changing the value of the expression.

Negative Exponents

Negative exponents are special and represent the reciprocal of the base raised to the opposite positive exponent. For example, \(a^{-n} = \frac{1}{a^n}\). This suggests moving the base \(a\) from the numerator to the denominator when the exponent is negative, flipping the expression.
Understanding negative exponents is crucial for simplifying expressions like the original exercise, where \((uv)^{-1}\) represents the reciprocal of \(uv\), or \(\frac{1}{uv}\). When you encounter a negative exponent, rewriting the expression using positive exponents is often the first step.
The rules for exponents, such as multiplying or dividing powers with the same base, also apply when dealing with negative exponents. For instance, \(a^{-m} \times a^n = a^{n-m}\), allows for the simplification of expressions with both positive and negative exponents. Learning these rules makes working with any algebraic expression easier and more intuitive.