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Chapter 5: Problem 98

Use Property 3 of Table 5.4 and Property 7 of Table 5.5 to prove Property 8 of Table 5.5

Short Answer

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Based on the step-by-step solution provided, answer the following question: **Question:** Prove Property 8 of Table 5.5, \(a^{-m} = \frac{1}{a^m}\), using Property 3 of Table 5.4, \((a^m)^n = a^{mn}\), and Property 7 of Table 5.5, \(a^0 = 1\). **Answer:** We began by applying Property 3 with \(n=0\), which gave us \((a^m)^0 = a^{m(0)}\), which simplified to \((a^m)^0 = a^0\). Next, we applied Property 7, which gave us \((a^m)^0 = 1\). We rewrote the fraction \(1/a^m\) as a product with an exponent: \(1 \cdot a^{-m} = \frac{1}{a^m}\). Then, we used Property 3 again to identify that \((a^m)^{-1} = a^{-m}\) and substituted this into the previous equation: \(1 \cdot (a^m)^{-1} = \frac{1}{a^m}\). This allowed us to prove Property 8 of Table 5.5: \(a^{-m} = \frac{1}{a^m}\).

Step by step solution

01

Start with Property 3 of Table 5.4

Begin this proof by writing down Property 3 of Table 5.4: \((a^m)^n = a^{mn}\).
02

Apply Property 7 of Table 5.5

Next, we look at Property 7 of Table 5.5: \(a^0 = 1\). Apply this property on the right-hand side of Property 3 by choosing \(n=0\): \((a^m)^0 = a^{m(0)}\).
03

Simplify the exponents

We know that \(m(0)=0\), so \((a^m)^0 = a^0\).
04

Apply Property 7 of Table 5.5 again

Now, we can use Property 7 of Table 5.5 again: \(a^0 = 1\). So, \((a^m)^0 = 1\).
05

Show \(a^{-m}\) is the same as a fraction

We want to show: \(a^{-m} = \frac{1}{a^m}\). Let's rewrite this fraction as a product with an exponent: \(1 \cdot a^{-m} = \frac{1}{a^m}\).
06

Apply Property 3 of Table 5.4, solving for \(-m\)

Using Property 3 of Table 5.4, we see that \((a^m)^{-1} = a^{-m}\). Substitute this into the equation from Step 5: \(1 \cdot (a^m)^{-1} = \frac{1}{a^m}\).
07

Prove Property 8 of Table 5.5

We have shown that \((a^m)^0 = \)1\( and \)(a^m)^{-1} = \frac{1}{a^m}\(. Combining these two statements, we can now prove Property 8 of Table 5.5: \)a^{-m} = \frac{1}{a^m}$.
08

Conclusion

Property 8 of Table 5.5, \(a^{-m} = \frac{1}{a^m}\), has been proved using Property 3 of Table 5.4, \((a^m)^n = a^{mn}\), and Property 7 of Table 5.5, \(a^0 = 1\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponent Laws
Exponent laws are fundamental principles that help us manipulate expressions with exponents efficiently. By understanding and applying these laws, we can simplify complex expressions and solve equations more easily.
Some of the most important exponent laws include:
  • Product of Powers: When multiplying like bases, you add the exponents: \(a^m \cdot a^n = a^{m+n}\).
  • Power of a Power: When raising an exponent to another power, you multiply the exponents: \((a^m)^n = a^{mn}\).
  • Power of a Product: When raising a product to a power, apply the exponent to each factor: \((ab)^n = a^n b^n\).
  • Zero Power: Any non-zero number raised to the power of zero equals one: \(a^0 = 1\).
Using these laws correctly helps in simplifying expressions and solving problems, just as seen in the step-by-step solution above.
Properties of Exponents
The properties of exponents allow us to streamline expressions by following specific rules. They are essential for resolving tasks involving powers and require a solid understanding to apply correctly.
Here are some critical properties to remember:
  • Power of Zero: As demonstrated, any non-zero number raised to the power of zero equals one. This property is crucial, as it simplifies expressions dramatically: \(a^0 = 1\).
  • Power of One: Any number raised to the power of one is identical to the number itself: \(a^1 = a\).
  • Negative Exponents: Negative exponents indicate reciprocals. When an exponent is negative, the base is on the opposite side of a fraction bar: \(a^{-m} = \frac{1}{a^m}\).
Understanding and using these properties properly allows us to transform and solve expressions effectively. This step-by-step breakdown illustrates how multiplying and simplifying powers can confirm these principles.
Negative Exponents
Negative exponents can be a bit confusing at first, but they have a straightforward meaning. They indicate the reciprocal of a number raised to a positive exponent.
For example, the expression \(a^{-m}\) can be rewritten as \(\frac{1}{a^m}\). This transformation is particularly useful because it helps in simplifying expressions that involve division of powers.
  • Reciprocal Understanding: A negative exponent tells you to "flip" the base to the denominator: \(a^{-m} = \frac{1}{a^m}\).
  • Applying Exponent Laws: As shown in Property 3 of Table 5.4, \((a^m)^{-1} = a^{-m}\), which substantiates that \(a^{-m}\) is \(\frac{1}{a^m}\).
  • Practical Example: Consider \(2^{-3}\), which is equivalent to \(\frac{1}{2^3} = \frac{1}{8}\).
Understanding negative exponents is essential for mastering exponents. Once you grasp their meaning, you'll be ready to tackle any exponent-related problem with confidence.

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Most popular questions from this chapter

Area versus net area Graph the following functions. Then use geometry (not Riemann sums) to find the area and the net area of the region described. The region between the graph of \(y=-3 x\) and the \(x\) -axis, for \(-2 \leq x \leq 2\)

A midpoint Riemann sum Approximate the area of the region bounded by the graph of \(f(t)=\cos \frac{t}{2}\) and the \(t\) -axis on \([0, \pi]\) with \(n=4\) subintervals. Use the midpoint of each subinterval to determine the height of each rectangle (check your book to see figure).

Definite integrals Use geometry (not Riemann sums) to evaluate the following definite integrals. Sketch a graph of the integrand. show the region in question, and interpret your result. $$\int_{0}^{4} f(x) d x, \text { where } f(x)=\left\\{\begin{array}{ll} 5 & \text { if } x \leq 2 \\ 3 x-1 & \text { if } x>2 \end{array}\right.$$

Average value of the derivative Suppose \(f^{\prime}\) is a continuous function for all real numbers. Show that the average value of the derivative on an interval \([a, b]\) is \(\bar{f}^{\prime}=\frac{f(b)-f(a)}{b-a},\) Interpret this result in terms of secant lines.

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