Problem 52 - Algebra for College Students

Question

Simplify each expression. Express final results without using zero or negative integers as exponents. \(\left(a^{3} b^{-3} c^{-2}\right)^{-5}\)

Step-by-Step Solution

Answer

The expression simplifies to \(\frac{b^{15} c^{10}}{a^{15}}\).

Step 1: Apply the Power of a Power

First, apply the power of a power rule which states \((x^m)^n = x^{m \cdot n}\). In this expression, we need to multiply the exponents inside the brackets by the exponent outside, which is \-5\: \((a^3 b^{-3} c^{-2})^{-5}\). This results in \((a^{3 \cdot -5} b^{-3 \cdot -5} c^{-2 \cdot -5})\).

Step 2: Simplify the Exponents

Now calculate the new exponents using the results from Step 1: - For \(a\): \(3 \times -5 = -15\), so it becomes \(a^{-15}\)- For \(b\): \(-3 \times -5 = 15\), so it becomes \(b^{15}\)- For \(c\): \(-2 \times -5 = 10\), so it becomes \(c^{10}\).

Step 3: Express with Positive Exponents

Convert any negative exponents into positive exponents by using the reciprocal: - \(a^{-15}\) becomes \(\frac{1}{a^{15}}\).So, the expression is transformed from \(a^{-15} b^{15} c^{10}\) into \frac{b^{15} c^{10}}{a^{15}}\.

Key Concepts

Power of a Power Rule

When simplifying expressions that involve an exponent raised to another exponent, we use the Power of a Power Rule. This important rule tells us that for any base \(x\) raised to an exponent \(m\), and then again raised to another exponent \(n\), we multiply the exponents together. Mathematically, this is represented as:

• \((x^m)^n = x^{m \cdot n}\)

Let's look at this rule with an example from the exercise. The expression \((a^3 b^{-3} c^{-2})^{-5}\) applies the Power of a Power Rule. By multiplying the inside exponents \((3, -3, -2)\) by the outside exponent \(-5\), we fully utilize this rule:

• For \(a\), the calculation is: \(3 \times -5 = -15\)
• For \(b\), it’s: \(-3 \times -5 = 15\)
• And for \(c\), it’s: \(-2 \times -5 = 10\)

This step is crucial because it prepares us to resolve any negative exponents afterwards.

Negative Exponents

Negative exponents may look intimidating, but they have a straightforward rule: they represent the reciprocal of the base raised to the corresponding positive exponent. Essentially, when you have a negative exponent, you "flip" the base to the denominator. Here's how this looks:

• \(x^{-m} = \frac{1}{x^m}\)

In the exercise, once we calculated the new exponents, say \(a^{-15} b^{15} c^{10}\), the only negative exponent we had was \(a^{-15}\). By applying the negative exponent rule, \(a^{-15}\) is converted into \(\frac{1}{a^{15}}\). This transformation easily helps us express the whole expression without negative exponents, making it clearer and easier to understand.

Positive Exponents

With positive exponents, our task is much simpler. The exponent denotes how many times the base is multiplied by itself. For any base \(x\) and positive exponent \(m\), it simply means multiplying \(x\) by itself \(m\) times:

• \(x^m = x \times x \times \ldots \text{(m times)}\)

In our example, after converting the negative exponents, we are left with \(b^{15}\) and \(c^{10}\). These are already in their cleanest form, representing the base \(b\) multiplied by itself 15 times, and base \(c\) multiplied by itself 10 times. Positive exponents like these are straightforward to handle, so they usually conclude our simplification process.

Reciprocal of Exponents

The concept of reciprocals is key when dealing with negative exponents but can also be applied in general mathematics operations. Reciprocals involve swapping the numerator and denominator of a fraction. For instance, the reciprocal of \(x\) is \(\frac{1}{x}\).

• If \(a^m\) becomes \(a^{-m}\), its reciprocal is \(\frac{1}{a^m}\).

In simplifying expressions, especially when turning negative exponents into positive ones, understanding reciprocals is fundamental. It helps in maintaining the expression within a simplified, easily readable form. In our example, the transformation of \(a^{-15}\) into \(\frac{1}{a^{15}}\) is a direct application of the reciprocal concept, and this is how it ensures that all exponents remain positive, enhancing clarity.