Michael Bowen's VC Course Pages

Answers

Math V46 Section A.6

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Introduction

Please try these problems on your own before looking at my solutions. There may be other ways to solve these; I make no claim that the methods shown are the quickest, prettiest, or most efficient.

Note that the basic properties of exponents (from section A.5) apply to all types of exponents, including positive exponents, negative exponents, fractional exponents, decimal exponents, and exponents containing variables (letters), making them very important for the solution of problems in this section as well. It would probably be a good idea to summarize the exponent properties in section A.5 on your cheat sheet, even though I did not assign any problems in that section. If you can solve a problem using positive exponents, you should be able to solve a problem containing fractional or negative exponents using exactly the same methods. Of course, additions and subtractions, if necessary, may be more difficult with fractional exponents, but the underlying principles are exactly the same.

Use the following rules to determine whether your exponential expression (your final answer) is fully simplified (you may wish to add these to your cheat sheet also):

Numeric factors or coefficients are fully reduced to lowest terms.
No variable (letter) appears more than one time anywhere in the expression. (Exception: Variables may appear more than once if they are up in the power; for example, ${x^n}{y^{2n}}$ is OK.)
All exponents are strictly positive; use the properties of negative exponents to move them around until they become positive.
The answer is not a complex fraction (which means a fraction that has smaller fractions inside it); use the methods of section A.4 to simplify complex fractions.
The answer should contain no parentheses, except possibly up in the exponent (and only if the exponent contains variables). Exception: Expressions raised to fractional powers, such as $\left( {x - 4} \right)^{{3}/{2}}$ are OK, since these cannot be "FOILed" or distributed.

Also note that a negative power applied to a parenthesized fraction may be removed by taking the reciprocal of the fraction and changing the power to positive. For example, $$\begin{equation} \label{eq:neg_power_ex_1} {{\left( {\frac{{3a}}{{2b}}} \right)^{ - 1}} \Longrightarrow {\left( {\frac{{2b}}{{3a}}} \right)^1} \Longrightarrow {\frac{{2b}}{{3a}}}}\end{equation}$$ and $$\begin{equation} \label{eq:neg_power_ex_2} {{\left( {\frac{{7x}}{{5y}}} \right)^{ - 3/2}} \Longrightarrow {\left( {\frac{{5y}}{{7x}}} \right)^{3/2}}} \end{equation}$$

Problem 31

Method 1

Using the negative-power-of-a-fraction property described in the previous paragraph, along with other exponent properties, $${\left( {\frac{{4{x^{ - 2}}}}{{{y^4}}}} \right)^{ - 1/2}} \Longrightarrow {\left( {\frac{{{y^4}}}{{4{x^{ - 2}}}}} \right)^{1/2}} \Longrightarrow \frac{{{{\left( {{y^4}} \right)}^{1/2}}}}{{{4^{1/2}}{{\left( {{x^{ - 2}}} \right)}^{1/2}}}} \Longrightarrow \frac{{{y^{\left( {4/1} \right) \cdot \left( {1/2} \right)}}}}{{\sqrt 4 \cdot {x^{\left( { - 2/1} \right) \cdot \left( {1/2} \right)}}}} \Longrightarrow \frac{{{y^2}}}{{{2^1}{x^{ - 1}}}} \Longrightarrow \frac{{{x^1}{y^2}}}{2} \Longrightarrow \boxed{\frac{{{x}{y^2}}}{2}}$$

Things to notice:

The negative-power-of-a-fraction property does not change the signs of the powers inside the parentheses; it just flips the fraction over as-is (just be sure to reverse the sign on the power outside the parentheses).
The power outside a parenthesized fraction (whether positive or negative) applies to both constants ($4$) and variables ($x$ and $y$). Many students think that constants are "immune" from exponents, but they are not; the $4$ needs a $1/2$ power just like everything else inside the parentheses. The only difference between the $x$ and the $4$ is that the $x$ is a disguised number and the $4$ is an undisguised number. So treat them the same.
Fractional powers applied to constants may (or may not) be easier to simplify if the fractional power is converted to the corresponding root.
In the third-to-last step, the power of $-1$ on the $x$ does not apply to the $2$ in front of it. The $2$ has its own (usually invisible, although I made it visible here for clarity) power of positive $1$. Therefore, when it comes time to simplify the negative exponents (which should always be the last step), the $x$ moves upstairs but the $2$ stays downstairs (only items with negative powers get moved). The $2$ is not tied to the $x$ in any way, so it is perfectly fine to separate them.
It is counterproductive to rewrite ${x^{ - 1}}$ as $\frac{1}{x}$, as this gives you an unnecessary complex fraction to simplify. Simply swap negative powers between numerator and denominator (or vice-versa) to change their signs to positive. If your answer does not contain a fraction but does contain negative powers, add a denominator of $1$ to your answer, and then swap the negative powers to the denominator and convert them to positive powers. For example, if you are near the end of a problem and the result so far is ${x^5}{y^{ - 6}}$, then place a $1$ underneath to get $\frac{{{x^5}{y^{ - 6}}}}{1}$, and swap the factor with a negative power to the denominator to get $\frac{{{x^5}}}{{{y^6}}}$.

Method 2

It is not necessary to change the negative power outside the parenthesized fraction to a positive power; we could simply work with it as is. $${\left( {\frac{{4{x^{ - 2}}}}{{{y^4}}}} \right)^{ - 1/2}} \Longrightarrow \frac{{{4^{ - 1/2}}{{\left( {{x^{ - 2}}} \right)}^{ - 1/2}}}}{{{{\left( {{y^4}} \right)}^{ - 1/2}}}} \Longrightarrow \frac{{{4^{ - 1/2}}{x^{\left( { - 2/1} \right) \cdot \left( { - 1/2} \right)}}}}{{{y^{\left( {4/1} \right) \cdot \left( { - 1/2} \right)}}}} \Longrightarrow \frac{{{4^{ - 1/2}}{x^1}}}{{{y^{ - 2}}}} \Longrightarrow \frac{{{x^1}{y^2}}}{{{4^{1/2}}}} \Longrightarrow \frac{{{x^1}{y^2}}}{{\sqrt 4 }} \Longrightarrow \boxed{\frac{{x{y^2}}}{2}}$$

Things to notice:

The final answer does not depend on the method used, provided that the final answer meets all the criteria for complete simplification as spelled out in the Introduction above.
The $x$ had a positive power so it did not move when we adjusted factors having negative powers. The $4$ and the $y$ had negative powers, so we moved them to the opposite ends of the fraction to make their powers positive. There is still no difference between the treatment of constants ($4$) and of variables ($y$).
Never convert negative fractional powers into roots. Move them up or down the fraction to make them positive fractional powers, and then convert to a root if desired.

Problem 33

Note that the $12$ in the denominator of this problem is not affected by the ${1/4}$ power on the $x$. So be sure to treat it as ${12^1}$, not as ${12^{1/4}}$. Start by clearing out the parentheses, in accordance with order of operations: $$\frac{{{{\left( {8x} \right)}^{ - 1/3}}}}{{12{x^{1/4}}}} \Longrightarrow \frac{{{8^{ - 1/3}}{x^{ - 1/3}}}}{{{{12}^1}{x^{1/4}}}}$$

We cannot consider this to be fully simplifed for two reasons: (1) the variable $x$ appears twice, and (2) there are negative exponents in the expression. It is most efficient to deal with the multiple occurrences of the $x$ first. Using the exponent property $\frac{{{x^m}}}{{{x^n}}} = {x^{m - n}}$, we may write the expression as $$\frac{{{8^{ - 1/3}}{x^{\left( { - 1/3} \right) - \left( {1/4} \right)}}}}{{12}} \Longrightarrow \frac{{{8^{ - 1/3}}{x^{\left( { - 4/12} \right) - \left( {3/12} \right)}}}}{{12}} \Longrightarrow \frac{{{8^{ - 1/3}}{x^{ - 7/12}}}}{{12}}$$ where we use an LCD to subtract the fractional exponents. Then we remove the negative exponents by moving the corresponding factors from numerator to denominator while changing the signs of the exponents: $$\frac{1}{{{8^{1/3}} \cdot {x^{7/12}} \cdot 12}}$$ Note that all the factors in the numerator had negative powers, so they all moved to the denominator. It is important to realize that if this happens, the numerator does not disappear; instead, it is replaced by a $1$.

An additional bit of simplification can be achieved by noticing that the fractional power on the $8$ is really a cube root. After we work this through, it then becomes possible to multiply the constants together to achieve the final result: $$\frac{1}{{\sqrt[3]{8} \cdot {x^{7/12}} \cdot 12}} \Longrightarrow \frac{1}{{2 \cdot {x^{7/12}} \cdot 12}} \Longrightarrow \boxed{\frac{1}{{24{x^{7/12}}}}}$$

Problem 83

This was requested by quite a few students. The goal is to simplify the expression by using a sophisticated version of the GCF factoring method to convert the expression from two terms into a distributive expression with two factors. The GCF we find will form the front end of the distributive. As in all GCF problems, the distributive we obtain must, when multiplied out, yield the original unfactored expression.

Any negative powers remaining in the numerator are then rendered positive by moving them down to the denominator, and the expression inside the back end of the distributive is simplified using order-of-operations techniques such as FOIL, distributing, and/or combining like terms. The result is typically a more compact expression, typically a single, relatively simple rational expression (fraction) whose numerator resembles a distributive or a FOIL.

This is likely the most sophisticated form of factoring most of you will encounter in your mathematical careers. Let us see how this works.

The original problem is $$\begin{equation} \label{eq:orig_prob_83}{ - \frac{1}{2}\left( {x - 2} \right){\left( {x + 3} \right)^{ - 3/2}} + {\left( {x + 3} \right)^{ - 1/2}}} \end{equation}$$ When seeking a GCF, we need to find a piece (factor) in the first term $ - \frac{1}{2}\left( {x - 2} \right){\left( {x + 3} \right)^{ - 3/2}}$ and a piece (factor) in the second term ${\left( {x + 3} \right)^{ - 1/2}}$ which match (except for their exponents, which are typically different). A visual inspection should readily make clear that each term contains $\left( {x + 3} \right)$ raised to various powers. This is not yet our full GCF, but it is the GCF's foundational building block.

To complete construction of the GCF, we must determine the proper exponent to apply to this $\left( {x + 3} \right)$. As discussed in lecture, the correct exponent to use is always the smaller of the two exponents appearing on the $\left( {x + 3} \right)$ in each term. In this case, the candidate exponents are ${ - 3/2}$ (first term) and ${ - 1/2}$ (second term). The smaller of these two is ${ - 3/2}$ (by "smaller" we mean "farther to the left on the real number line"), so this is the exponent we apply to the common element $\left( {x + 3} \right)$ to obtain the true GCF, which is $$\begin{equation} \label{eq:true_gcf} {{\text{GCF}} = {\left( {x + 3} \right)^{ - 3/2}}} \end{equation}$$

From this result, it immediately follows that the distributive format we are seeking will have the structure $$\begin{equation} \label{eq:distrib_structure} {\underbrace {{{\left( {x + 3} \right)}^{ - 3/2}}}_{{\text{“front end” (GCF)}}} \cdot \underbrace {\left( {{\text{rest of distributive}}} \right)}_{{\text{“back end”}}} \Longrightarrow {\left( {x + 3} \right)^{ - 3/2}} \cdot \left[ {\left( {{\text{first-term leftovers}}} \right) + \left( {{\text{second-term leftovers}}} \right)} \right]} \end{equation}$$ Apparently, the next task will be to determine the "leftovers".

(Aside)

In an "easy" GCF factoring problem such as $$\begin{equation} \label{eq:easy_problem} {3{x^4} + 7{x^9} \Longrightarrow {x^4} \cdot \left( {3 + 7{x^5}} \right)} \end{equation}$$ the "leftovers" terms $3$ and $7{x^5}$ in the parentheses on the right are obtained (whether we consciouly realize it or not) by dividing each of the original terms in the problem by the GCF which composes the front end of the distributive. In other words, the $3$ comes from $\displaystyle \frac{{3{x^4}}}{{{x^4}}}$, and the $7{x^5}$ comes from $\displaystyle \frac{{7{x^9}}}{{{x^4}}}$. Applying the exponent property $\displaystyle \frac{{{b^m}}}{{{b^n}}} = {b^{m - n}}$ allows us to rewrite these fractions without denominators. With this change in notation, we may now say that the $3$ comes from $3{x^{4 - 4}}$, and the $7{x^5}$ comes from $7{x^{9 - 4}}$. This is not the final form of the answer, but it gives us a starting point for determining the "leftovers" terms' values. The common pattern in both terms is that, in each exponent, the subtraction is of the form $$\begin{equation} \label{eq:subtraction_pattern} {\left( {{\text{original power of common base}}} \right) - \left( {{\text{power of GCF}}} \right)} \end{equation}$$

(Continuation of solution)

Applying the principle developed in the preceding paragraph to our problem, we may proceed with determining the "leftovers" terms in the back end of the distributive; Equation $\eqref{eq:distrib_structure}$ becomes $$\begin{equation} \label{eq:leftovers_start} {{\left( {x + 3} \right)^{ - 3/2}} \cdot \left[ { - \frac{1}{2}\left( {x - 2} \right){{\left( {x + 3} \right)}^{\left( { - 3/2} \right) - \left( { - 3/2} \right)}} + {{\left( {x + 3} \right)}^{\left( { - 1/2} \right) - \left( { - 3/2} \right)}}} \right]} \end{equation}$$

The first simplifying step is to consolidate (subtract) the exponents in the back end of the distributive: $$\begin{equation} \label{eq:combine_powers} {{\left( {x + 3} \right)^{ - 3/2}} \cdot \left( { - \frac{1}{2}\left( {x - 2} \right){{\left( {x + 3} \right)}^0} + {{\left( {x + 3} \right)}^{2/2}}} \right) \Longrightarrow {\left( {x + 3} \right)^{ - 3/2}} \cdot \left( { - \frac{1}{2}\left( {x - 2} \right)\left( 1 \right) + {{\left( {x + 3} \right)}^1}} \right)} \end{equation}$$ The next simplifiction is to remove unnecessary ones: $$\begin{equation} \label{eq:remove_ones} {{\left( {x + 3} \right)^{ - 3/2}} \cdot \left( { - \frac{1}{2}\left( {x - 2} \right) + \left( {x + 3} \right)} \right)} \end{equation}$$ After that, we remove the small parentheses in the back end of the distributive by distributing the $\displaystyle -\frac{1}{2}$ in front of the first one and the invisible positive one (the plus sign) in front of the second one, and combine the like terms which result: $$\begin{equation} \label{eq:comb_like_terms} {{\left( {x + 3} \right)^{ - 3/2}} \cdot \left( { - \frac{1}{2}x + 1 + x + 3} \right) \Longrightarrow {\left( {x + 3} \right)^{ - 3/2}} \cdot \left( {\frac{1}{2}x + 4} \right)} \end{equation}$$

This is not a bad answer; the issue with it is that the instructions for the problem requested the final answer to be written without using negative exponents, and there is a negative exponent on the front end of the distributive. To remove this, we insert a denominator of $1$ below the last result in Equation $\eqref{eq:comb_like_terms}$, and then slide the factors having negative exponents down into this new denominator to change the exponents to positive. The result is $$\begin{equation} \label{eq:pos_powers} {\frac{{{{\left( {x + 3} \right)}^{ - 3/2}} \cdot \left( {\frac{1}{2}x + 4} \right)}}{1} \Longrightarrow \frac{{\left( {\frac{1}{2}x + 4} \right)}}{{{{\left( {x + 3} \right)}^{3/2}}}}}\end{equation}$$

We are very close to the end; however, a review of the rules for full simplification given in the Introduction above alerts us to the fact that the $\frac{1}{2}$ in the numerator needs to be fixed because it is a small fraction inside of a larger fraction, and therefore makes the entire expression (technically, at least) a complex fraction. Fortunately, this is easy to correct; we simply multiply the entire fraction by $\displaystyle \frac{2}{2}$ (I picked $2$ because it is the LCD of all the "small" denominators), and distribute the upper $2$ through the numerator to remove the $\frac{1}{2}$. This leads (at long last) to the final result \begin{align} \frac{2}{2} \cdot \frac{{\left( {\frac{1}{2}x + 4} \right)}}{{{{\left( {x + 3} \right)}^{3/2}}}} & \Longrightarrow \frac{{2 \cdot \left( {\frac{1}{2}x + 4} \right)}}{{2 \cdot {{\left( {x + 3} \right)}^{3/2}}}} \nonumber \\ {} & \Longrightarrow \frac{{2 \cdot \left( {\frac{1}{2}x} \right) + 2 \cdot \left( 4 \right)}}{{2{{\left( {x + 3} \right)}^{3/2}}}} \nonumber \\ {} & \Longrightarrow \frac{{\frac{2}{1} \cdot \left( {\frac{1}{2}x} \right) + 2 \cdot \left( 4 \right)}}{{2{{\left( {x + 3} \right)}^{3/2}}}} \nonumber \\ \label{eq:final_prob_83} {} & \Longrightarrow \boxed{\frac{{x + 8}}{{2{{\left( {x + 3} \right)}^{3/2}}}}} \end{align} As a reminder, you should never distribute the denominator, as the distributive property fails (i.e., is illegal to apply) whenever there is a visible exponent (in this case, ${3/2}$) attached to the parentheses.