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Chapter 0: Problem 2

Simplify the expression without using a calculating utility. $$ \begin{array}{llll}{\text { (a) } 2^{-4}} & {\text { (b) } 4^{1.5}} & {\text { (c) } 9^{-0.5}}\end{array} $$

Short Answer

Expert verified
(a) \(\frac{1}{16}\), (b) \(8\), (c) \(\frac{1}{3}\)

Step by step solution

01

Understanding Negative Exponents

In part (a), we have the expression \(2^{-4}\). Remember, a negative exponent means we take the reciprocal of the base and change the exponent to positive. That is, \(a^{-n} = \frac{1}{a^n}\).
02

Simplifying \(2^{-4}\)

By applying the negative exponent rule, \(2^{-4} = \frac{1}{2^4}\). Calculating \(2^4\) gives us 16, so \(2^{-4} = \frac{1}{16}\).
03

Interpreting Fractional Exponents

In part (b), we need to simplify \(4^{1.5}\). Fractional exponents signify roots; here, \(a^{m/n} = \sqrt[n]{a^m}\). In this case, 1.5 is \(\frac{3}{2}\).
04

Simplifying \(4^{1.5}\)

Apply the rule for fractional exponents: \(4^{1.5} = 4^{3/2} = (\sqrt{4})^3\). Since \(\sqrt{4} = 2\), this becomes \(2^3 = 8\).
05

Understanding Negative Fractional Exponents

For part (c), \(9^{-0.5}\), we combine rules for negative exponents and fractional exponents. \(a^{-m/n} = \frac{1}{a^{m/n}}\).
06

Simplifying \(9^{-0.5}\)

Convert the expression: \(9^{-0.5} = \frac{1}{9^{0.5}} = \frac{1}{\sqrt{9}}\). \(\sqrt{9} = 3\), so \(9^{-0.5} = \frac{1}{3}\).

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Key Concepts

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

Negative Exponents
Negative exponents can be a bit tricky at first, but they follow a simple rule. When a base has a negative exponent, such as in the expression \(a^{-n}\), it means you take the reciprocal of the base raised to the positive power. In simpler terms, \(a^{-n} = \frac{1}{a^n}\). For example, if we simplify \(2^{-4}\), the negative exponent tells us to flip the base and change the exponent to positive.
So, \(2^{-4} = \frac{1}{2^4}\). Calculating \(2^4\) results in \(16\), making \(2^{-4} = \frac{1}{16}\).
  • Negative sign indicates a reciprocal.
  • The exponent's absolute value decides the power.
Fractional Exponents
Fractional exponents might seem confusing, but they are just another way to represent roots. This is seen in the expression \(a^{m/n}\), which is equivalent to the \(n\)-th root of the base raised to the \(m\)-th power. It means \(a^{m/n} = \sqrt[n]{a^m}\).
In our case, for \(4^{1.5}\), the fraction \(1.5\) is seen as \(\frac{3}{2}\). Therefore, \(4^{1.5} = 4^{3/2} = (\sqrt{4})^3\). Since \(\sqrt{4}\) equals \(2\), this simplifies to \(2^3\), which is \(8\).
  • Fractional exponent signifies roots.
  • Numerator as the power and denominator as the root.
Simplifying Expressions
Simplifying expressions with both negative and fractional exponents involves applying the rules for both concepts. This may initially look daunting, but with practice, it becomes second nature. Take for instance \(9^{-0.5}\), here we combine the negative and fractional exponent rules.
Negative exponent means reciprocal, and fractional exponent signifies a root. Thus, \(9^{-0.5}\) transforms into \(\frac{1}{9^{0.5}}\) and further into \(\frac{1}{\sqrt{9}}\). Calculating \(\sqrt{9}\) gives us \(3\), hence \(9^{-0.5} = \frac{1}{3}\).
  • Combine rules of exponents efficiently.
  • Practice breaking down steps for simplicity.

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Most popular questions from this chapter

The number of hours of daylight on a given day at a given point on the Earth's surface depends on the latitude \(\lambda\) of the point, the angle \(\gamma\) through which the Earth has moved in its orbital plane during the time period from the vernal equinox (March \(21),\) and the angle of inclination \(\phi\) of the Earth's axis of rotation measured from ecliptic north \(\left(\phi \approx 23.45^{\circ}\right) .\) The number of hours of daylight \(h\) can be approximated by the formula $$ h=\left\\{\begin{array}{ll}{24,} & {D \geq 1} \\ {12+\frac{2}{15} \sin ^{-1} D,} & {|D|<1} \\ {0,} & {D \leq-1}\end{array}\right. $$ $$ \begin{array}{l}{\text { where }} \\ {\qquad D=\frac{\sin \phi \sin \gamma \tan \lambda}{\sqrt{1-\sin ^{2} \phi \sin ^{2} \gamma}}}\end{array} $$ and \(\sin ^{-1} D\) is in degree measure. Given that Fairbanks, Alaska, is located at a latitude of \(\lambda=65^{\circ} \mathrm{N}\) and also that \(\gamma=90^{\circ}\) on June 20 and \(\gamma=270^{\circ}\) on December \(20,\) approximate (a) the maximum number of daylight hours at Fairbanks to one decimal place (b) the minimum number of daylight hours at Fairbanks to one decimal place.

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