OBJECTIVES

When you finish this section, you should be able to:

1. Factor and simplify algebraic expressions (p. A-1)
2. Complete the square (p. A-2)
3. Solve equations (p. A-3)
4. Solve inequalities (p. A-5)
5. Work with exponents (p. A-8)
6. Work with logarithms (p. A-10)

Factoring Algebraic Expressions

Factor each expression completely:

(a) \(2( x+3) ( x-2) ^{3}+( x+3) ^{2}( 3) ( x-2) ^{2}\)

(b) \(\dfrac{4}{3}x^{1/3}( 2x+1) +2x^{4/3}\)

Solution (a) In expression (a), \(( x+3)\) and \(( x-2) ^{2}\) are common factors, factors found in each term. Factor them out. \[ \begin{array}{rcl@{\qquad\hspace*{-3pt}}l} &&2 ( x+3) ( x-2) ^{3}+ ( x+3) ^{2} ( 3) ( x-2) ^{2}\\[5pt] &&\quad = ( x+3) (x-2) ^{2} [ 2 ( x-2) +3 ( x+3) ] & {\color{#0066A7}{\hbox{Factor out \(\hbox{(}x+3\hbox{)}\,\hbox{(}x-2\hbox{)}^2\).}}} \\[5pt] &&\quad = ( x+3) ( x-2) ^{2} ( 5x+5) & {\color{#0066A7}{\hbox{Simplify.}}}\\[5pt] &&\quad = 5 ( x+3) ( x-2) ^{2} (x+1) & {\color{#0066A7}{\hbox{Factor out 5.}}} \end{array} \]

(b) We begin by writing the term \(2x^{4/3}\) as a fraction with a denominator of \(3\). \[ \begin{array}{rcl@{\qquad\hspace*{-5.5pt}}l} \dfrac{4}{3}x^{1/3} ( 2x+1) +2x^{4/3} &=&\dfrac{4x^{1/3} (2x+1) }{3}+\dfrac{6x^{4/3}}{3}\\[9pt] &=&\dfrac{4x^{1/3} ( 2x+1) +6x^{4/3}}{3} &{\color{#0066A7}{\hbox{Add the two fractions.}}}\\[9pt] &=&\dfrac{2x^{1/3} [ 2 ( 2x+1) +3x] }{3} &{\color{#0066A7}{\hbox{2 and \(x^{1/3}\) are common factors.}}}\\[9pt] &=&\dfrac{2x^{1/3} (7x+2) }{3} & {\color{#0066A7}{\hbox{Simplify.}}} \end{array} \]

Simplifying Algebraic Expressions

(a) Simplify \(\dfrac{( x^{2}+1) ( 3) -( 3x+4) ( 2x) }{( x^{2}+1) ^{2}}\).

(b) Write the expression \(( x^{2}+1) ^{1/2}+x\cdot \dfrac{1}{2}( x^{2}+1) ^{-1/2}\cdot 2x\) as a single quotient in which only positive exponents appear.

Solution (a) \begin{eqnarray*} \ \dfrac{( x^{2}+1) (3) -( 3x+4) ( 2x) }{( x^{2}+1) ^{2}}&=&\dfrac{ 3x^{2}+3- ( 6x^{2}+8x) }{( x^{2}+1) ^{2}}=\dfrac{ 3x^{2}+3-6x^{2}-8x}{( x^{2}+1) ^{2}}\\[6pt] &=&\dfrac{-3x^{2}-8x+3}{( x^{2}+1) ^{2}}\underset{\underset{\color{#0066A7}{\hbox{Factor}}}{\color{#0066A7}{\uparrow }}}{=}\dfrac{- ( 3x-1) ( x+3) }{( x^{2}+1) ^{2}}\\ \end{eqnarray*}

(b) \begin{eqnarray*} ( x^{2}+1) ^{1/2}+x\cdot \dfrac{1}{2}( x^{2}+1) ^{-1/2}\cdot 2x&=&( x^{2}+1) ^{1/2}+\dfrac{x^{2}}{ ( x^{2}+1) ^{1/2}}\\[6pt] &=&\dfrac{( x^{2}+1) ^{1/2}( x^{2}+1) ^{1/2}}{( x^{2}+1) ^{1/2}}+\dfrac{x^{2}}{( x^{2}+1) ^{1/2}}\\[6pt] &=&\dfrac{( x^{2}+1) +x^{2}}{( x^{2}+1) ^{1/2}}=\dfrac{ 2x^{2}+1}{( x^{2}+1) ^{1/2}} \end{eqnarray*}

Completing the Square

Determine the number that must be added to each expression to complete the square. Then factor.

CAUTION The original expression \(x^{2}+bx\) and the perfect square \( x^{2}+bx+\left( \dfrac{b}{2}\right) ^{\!\!2}\) are not equal. So when completing the square within an equation or an inequality, we must not only add \( \left( \dfrac{b}{2}\right) ^{\!\!2}\!\), we must also subtract it. That is, \[ x^{\!\!2}+bx=x^{\!\!2}+bx+{\hbox{\(\stackrel{\underbrace{\left(\dfrac{b}{2}\right) ^{\!\!2}-\left( \dfrac{b}{2}\right) ^{\!\!2}}}{=0}\)}}=\left( x+\dfrac{b}{2}\right) ^{\!\!2}-\left( \dfrac{b}{2}\right) ^{\!\!2} \]

THEOREM Quadratic Formula

Consider the quadratic equation \[ ax^{2}+bx+c=0\hbox{ } \quad a\neq 0 \]

If \(b^{2}-4ac<0\), the equation has no real solution.

If \(b^{2}-4ac\geq 0\), the real solution(s) of the equation is (are) given by the quadratic formula: \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ x=\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a} } }\]

Solving Quadratic Equations

Solve the equations:

(a) \(3x^{2}-5x+1=0\)

(b) \(x^{2}+x=-1\)

Solution (a) The discriminant: \(b^{2}-4ac=25-12=13\). We use the quadratic formula. \[ x=\dfrac{-b\pm \displaystyle\sqrt{b^{2}-4ac}}{2a}=\dfrac{5\pm \displaystyle\sqrt{13}}{6} \]

The solutions are \(x=\dfrac{5-\sqrt{13}}{6}\approx 0.232\) and \(x=\dfrac{5+ \sqrt{13}}{6}\approx 1.434\)

(b) The equation in standard form is \(x^2+x+1=0\). Its discriminant is \(b^{2}-4ac=1-4=-3\). This equation has no real solution.

NOTE

Remember to put a quadratic equation in standard form before attempting to solve it. That is, write the quadratic equation in the form \(ax^{2}+bx+c=0.\)

Solving Equations

Solve the equations:

(a) \(\dfrac{3}{x-2}=\dfrac{1}{x-1}+\dfrac{7}{( x-1) ( x-2) }\)

(b) \(x^{3}-x^{2}-4x+4=0\)

(c) \(\sqrt{x-1}=x-7\)

(d) \(\vert 1-x\vert =2\)

Solution (a) First, notice that the domain of the variable is \( \{ x~|~x\neq 1,\,x\neq 2\} \). Now clear the equation of rational expressions by multiplying both sides by \(( x-1) ( x-2) \). \[ \begin{array}{@{\hspace*{-7.1pc}}rcl@{\qquad}ll} \dfrac{3}{x-2}&=&\dfrac{1}{x-1}+\dfrac{7}{( x-1) ( x-2) }& \\ ( x-1) ( x-2) \dfrac{3}{x-2} &=&( x-1) ( x-2) \left[ \dfrac{1}{x-1}+\dfrac{7}{( x-1) ( x-2) }\right] & {\color{#0066A7}{\hbox{Multiply both sides by} }}\\ &&&{\color{#0066A7}{\hbox{\(( {x-1}) ( { x-2}).\)}}} \\ 3x-3&=&( x-1) ( x-2) \dfrac{1}{x-1} + ( x-1) ( x-2) \dfrac{7}{( x-1) ( x-2) } & {\color{#0066A7}{\hbox{Distribute on both sides.}}} \\ 3x-3&=&( x-2) +7 & {\color{#0066A7}{\hbox{Simplify.}}} \\ 3x-3&=&x+5 & \\ 2x&=&8\hbox{ } & \\ x&=&4 & \end{array} \] Since 4 is in the domain of the variable, the solution is \(4\).

(b) We group the terms of \(x^{3}-x^{2}-4x+4=0\), and factor by grouping. \[ \begin{array}{rclll} x^{3}-x^{2}-4x+4&=&0 & \\ [4pt] ( x^{3}-x^{2}) -( 4x-4) &=&0 & {\color{#0066A7}{\hbox{Group the terms.}}} \\[4pt] x^{2}( x-1) -4( x-1) &=&0 & {\color{#0066A7}{\hbox{Factor out the common factor}}} \\[-2pt] &&&{\color{#0066A7}{\hbox{from each group.}}}\\[4pt] ( x^{2}-4) ( x-1) &=&0 & {\color{#0066A7}{\hbox{Factor out the common factor \((x-1).\)}}} \\[4pt] ( x-2) ( x+2) ( x-1) &=&0 & {\color{#0066A7}{\hbox{\({x}^{2}-4=( {x-2}) ( {x+2})\)}}} \\[4pt] x-2=0\hbox{ or }x+2&=&0\hbox{ or }x-1=0 & {\color{#0066A7}{\hbox{Set each factor equal to 0.}}} \\[3pt] x\hbox{ }\,=\,\hbox{ }2\hbox{ }x&=&-2\hbox{ }x\hbox{ }\,=\,\hbox{ }1 & {\color{#0066A7}{\hbox{Solve.}}} \end{array} \] The solutions are \(-2\), 1, and 2.

(c) We square both sides of the equation since the index of a square root is \(2\). \[ \begin{array}{@{{-8pt}}rclll} \hbox{ }\sqrt{x-1}&=&x-7 & \\[4.5pt] ( \sqrt{x-1}) ^{2}&=& ( x-7) ^{2} & {\color{#0066A7}{\hbox{Square both sides.}}} \\[4.5pt] \hbox{}x-1&=&x^{2}-14x+49\hbox{} & \\[4.5pt] x^{2}-15x+50&=&0 & {\color{#0066A7}{\hbox{Put in standard form.}}} \\[4.5pt] ( x-10) ( x-5) &=&0 & {\color{#0066A7}{\hbox{Factor. }}} \\[4.5pt] \hbox{ }x=10\hbox{ or }x&=&5 & {\color{#0066A7}{\hbox{Set each factor equal to 0 and solve.}}} \end{array} \]

Check: \(x=10\): \(\sqrt{x-1}=\sqrt{10-1}=\sqrt{9}=3\) and \( x-7=10-7=3\)
\(x=5\): \(\sqrt{x-1}=\sqrt{5-1}=\sqrt{4}=2\) and \( x-7=5-7=-2\)

The apparent solution \(5\) is extraneous; the only solution of the equation is \(10\).

(d) \begin{array}[t]{@{{-.8pc}}rclrcl@{\quad}l} \vert 1-x \vert &=&2 & & & \\[5pt] \hbox{ }1-x&=&2 & \hbox{or}\quad 1-x&=&-2 & {\color{#0066A7}{\hbox{The expression inside the absolute }}} \\[-2pt] &&&&&&{\color{#0066A7}{\hbox{value bars equals 2 or \(-2\).}}} \\[5pt] -x&=&1 & -x&=&-3 & {\color{#0066A7}{\hbox{Simplify.}}} \\[5pt] x&=&-1 & x&=&3 & {\color{#0066A7}{\hbox{Simplify.}}} \end{array} The solutions are \(-1\) and \(3\).

NOTE

Every interval of real numbers contains both rational numbers, numbers that can be expressed as the quotient of two integers, and irrational numbers, numbers that are not rational.

spanDEFINITIONspan

Let \(a\) and \(b\) represent two real numbers with \(a < b\).

A closed interval, denoted by \([ a,b] ,\) consists of all real numbers \(x\) for which \(a\leq x\leq b\).

An open interval, denoted by \(( a,b) \), consists of all real numbers \(x\) for which \(a < x < b\).

The half-open, or half-closed, intervals are:

1. \(( a,b] \), consisting of all real numbers \(x\) for which \(a < x\leq b\), and
2. \([ a,b) \), consisting of all real numbers \(x\) for which \(a\leq x < b\).

Solving Inequalities

Solve each inequality and graph the solution:

(a) \(4x+7\geq 2x-3\)

(b) \(x^{2}-4x+3>0\)

(c) \(x^{2}+x+1 < 0\)

(d) \(\dfrac{1+x}{1-x}>0\)

Solution (a) \[ \begin{array}{@{\hspace*{-4pt}}rcll@{\quad}l} 4x+7&\geq& 2x-3 & & \\ 4x&\geq& 2x-10 & & {\color{#0066A7}{\hbox{Subtract 7 from both sides.}}} \\ 2x&\geq& -10 & & {\color{#0066A7}{\hbox{Subtract 2\(x\) from both sides.}}} \\ x&\geq& -5 & & {\color{#0066A7}{\hbox{Divide both sides by 2.}}}\quad \\ &&&&{\color{#0066A7}{\hbox{(The direction of the inequality symbol is unchanged.)}}} \end{array} \]

Figure 1 \(x\geq 5\)

The solution using interval notation is \([ -5,\infty ) \). See Figure 1 for the graph of the solution.

(b) This is a quadratic inequality. The related quadratic equation \(x^{2}-4x+3= ( x-1) ( x-3) =0\) has two solutions, \(1\) and \(3\). We use these numbers to partition the number line into three intervals. Now select a test number in each interval, and determine the value of \(x^{2}-4x+3\) at the test number. See Table 2.

Table 2

Figure 2 \(x<1\) or \(x>3\)

We conclude that \(x^{2}-4x+3>0\) on the set \(( -\infty ,1) \cup ( 3,\infty) \). See Figure 2 for the graph of the solution.

(c) The quadratic equation \(x^{2}+x+1=0\) has no real solution, since its discriminant is negative. [See Example 4(b)]. When this happens, the quadratic inequality is either positive for all real numbers or negative for all real numbers. To see which is true, evaluate \(x^{2}+x+1\) at some number, say, \(0\). At \(0,\) \(x^{2}+x+1=1,\) which is positive. So, \(x^{2}+x+1>0\) for all real numbers \(x\). The inequality \(x^{2}+x+1 < 0\) has no solution.

(d) The only solution of the rational equation \(\dfrac{1+x}{1-x} =0\) is \(x=-1;\) also, the expression \(\dfrac{1+x}{1-x}\) is not defined for \(x=1.\) We use the solution \(-1\) and the value \(1\), at which the expression is undefined, to partition the real number line into three intervals. Now select a test number in each interval, and evaluate the rational expression \(\dfrac{1+x}{1-x}\) at each test number. See Table 3.

Table 3

Figure 3 \(-1<x<1\)

We conclude that \(\dfrac{1+x}{1-x}>0\) on the interval \(( -1,1)\). See Figure 3 for the graph of the solution.

NOTE

If \(x > 0\) and \(y > 0\), the Triangle Inequality is an algebraic statement for the fact that the length of any side of a triangle is less than the sum of the lengths of the other two sides.

THEOREM Triangle Inequality

If \(x\) and \(y\) are real numbers, then \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ |x+y| \leq |x| + |y|} } \]

THEOREM

If \(a\) is a positive number and if \(u\) is an algebraic expression, then \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \begin{array}{@{}l@{{-3pt}}} \vert u\vert < a \hbox{ is equivalent to } -\!a < u < a \quad(1)\\ \vert u\vert > a \hbox{ is equivalent to } u < -a \quad \hbox{or}\quad u > a \quad(2) \end{array} }} \]

Similar relationships hold for the nonstrict inequalities \(\vert u\vert \leq a\) and \(\vert u\vert \geq a.\)

Solving Inequalities Involving Absolute Value

Solve each inequality and graph the solution:

(a) \(\vert 3-4x\vert < 11\)

(b) \(\vert 2x+4\vert -1\leq 9\)

(c) \( \left\vert \dfrac{4x+1}{2}-\dfrac{3}{5}\right\vert >1\)

Solution (a) The absolute value is less than the number \(11,\) so statement (1) applies. \[ \begin{array}{@{{-4.5pc}}rcccc@{\quad}l} \vert 3-4x\vert &<&{11} & \\[3pt] -11 &<&3-4x&<&11&{\color{#0066A7}{\hbox{Apply statement (1).}}} \\[3pt] -14 &<&-4x&<&8 & {\color{#0066A7}{\hbox{Subtract 3 from each part.}}} \\[3pt] \dfrac{-14}{-4} &>&x&>&\dfrac{8}{-4} & {\color{#0066A7}{\raise5pt\hbox{Divide each part by \(-4\), which}}} \\[-9pt] &&&&&{\color{#0066A7}{\raise9pt\hbox{reverses the inequality signs.}}}\\[-5pt] -2 &<&x&<&\dfrac{7}{2} & {\color{#0066A7}{\hbox{Simplify and rearrange the ordering.}}} \end{array} \]

The solutions are all the numbers in the open interval \(\!\left( -2,\dfrac{7}{2}\right)\!\). See Figure 4 for the graph of the solution.

Figure 4 \(-2<x<\dfrac{7}{2}\)

(b) We begin by putting \(\vert 2x+4\vert -1\leq 9\) into the form \(\vert u\vert \leq a\). \[ \begin{array}{@{{-4.7pc}}rclll@{\quad}l} \vert 2x+4\vert -1 &\leq &9 &\\[3pt] \vert 2x+4\vert &\leq &10 &&& {\color{#0066A7}{\hbox{Add 1 to each side.}}} \\[3pt] -10 &\leq &2x+4&\leq& 10 & {\color{#0066A7}{\hbox{Apply statement (1), but use \(\leq\).}}} \\[3pt] -14 &\leq &{2x}&\leq& 6 \\[3pt] -7 &\leq &{x}&\leq& 3 \end{array} \]

Figure 5 \(-7\leq x \leq 3\)

The solutions are all the numbers in the closed interval \([-7, 3]\) . See Figure 5 for the graph of the solution.

(c) \(\left\vert \dfrac{4x+1}{2}-\dfrac{3}{5}\right\vert > 1\) is in the form of statement (2). We begin by simplifying the expression inside the absolute value. \[ \left\vert \dfrac{4x+1}{2}-\dfrac{3}{5}\right\vert =\left\vert \dfrac{ 5( 4x+1) }{10}-\dfrac{2( 3) }{10}\right\vert =\left\vert \dfrac{20x+5-6}{10}\right\vert =\left\vert \dfrac{20x-1}{10} \right\vert \]

The original inequality is equivalent to the inequality below. \[ \left\vert \dfrac{20x-1}{10}\right\vert >1 \] \[ \begin{array}{@{{-2pc}}rcl@{\qquad}c@{\qquad}r@{\qquad}l} \dfrac{20x-1}{10} &<&-1 &\hbox{ or }& \dfrac{20x-1}{10}>1\hphantom{0} & {\color{#0066A7}{\hbox{Apply statement (2).}}} \\[8pt] 20x-1 &<&-10&\hbox{ or } &20x-1>10 \\[3pt] 20x &<&-9 &\hbox{ or }& 20x>11 \\[3pt] x\, &<&-\dfrac{9}{20} &\hbox{ or } & x>\dfrac{11}{20}\!\! \end{array} \]

Figure 6 \(x<-\dfrac{9}{20}\) or \(x>\dfrac{11}{20}\)

The solutions are all the numbers in the set \(\left( -\infty ,-\dfrac{9}{20}\right) \cup \left(\dfrac{11}{20},\infty \right) .\) See Figure 6 for the graph of the solution.

spanDEFINITIONspan

If \(a\) is a real number and \(n\) is a positive integer, then the symbol \(a^{n}\) represents the product of \(n\) factors of \(a\). That is, \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ a^{n}=\underset{\hbox{\(n\) factors}}{\underbrace{a\cdot a \cdot \ldots \cdot a}}}} \]

Here, it is understood that \(a^{1}=a\).

spanDEFINITIONspan

If \(a\neq 0\), then \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ a^{0}=1 }}\]

If \(n\) is a positive integer, then \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ a^{-n}=\dfrac{1}{a^{n}}\qquad a\neq 0 }}\]

THEOREM Laws of Exponents

In each of these properties \(a\) and \(b\) are real numbers, and \(u\) and \(v\) are integers. \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \begin{array}{rcl@{\quad}c@{\!\!\!\!\!\quad}c@{\,\,\qquad\qquad}rcll} a^{u}a^{v}&=&a^{u+v} && ( a^{u})^{v}=a^{uv} & ( ab ) ^{u}&=&a^{u}b^{u} \\[6pt] \dfrac{a^{u}}{a^{v}}=a^{u-v}&=&\dfrac{1}{a^{v-u}} &\hbox{if }a\neq 0 &&\left( \dfrac{a}{b}\right) ^{u} &=&\dfrac{a^{u}}{b^{u}}_{\vphantom{0_0}} & {\rm \ if\ }b\neq 0 \end{array} }} \]

IN WORDS

The symbol \(\sqrt[{n}]{a}\) means “the number that, when raised to the \(n\)th power, equals \(a\).”

spanDEFINITIONspan

The principal \(n{\rm th}\) root of a real number \(a\), where \(n\geq 2\) is an integer, symbolized by \(\sqrt[{ n }]{a}\), is defined as the solution of the equation \(b^{n}=a.\) \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \sqrt[{ n}]{a}=b \hbox{ is equivalent to } a=b^{n} }} \]

If \(n\) is even, then both \(a\geq 0\) and \(b\geq 0,\) and if \(n~\) is odd, then \(a\) and \(b\) are any real numbers and both have the same sign.

spanDEFINITIONspan

If \(a\) is a real number and \(n\geq 2\) is an integer, then \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ a^{1/n}=\sqrt[{ n}]{a}} } \] provided that \(a^{1/n}=\sqrt[n]{a}\) exists.

spanDEFINITIONspan

If \(a\) is a real number and \(m\) and \(n\) are integers with \(n\geq 2,\) then \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ a^{m/n}= ( a^{1/n}) ^{m} }} \] provided that \(\sqrt[{ n}]{a}\) exists.

spanDEFINITIONspan

Suppose \(y=a^{x}\!\), \(a>0\), \(a\neq 1\), and \(x\) is a real number. The logarithm with base \( a\) of \( y\), symbolized by \(\log _{a} y,\) is the exponent to which \(a\) must be raised to obtain \(y.\) That is, \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \log _{a} y=x \hbox{ is equivalent to } y=a^{x} } }\]

Working with Logarithms

(a) If \(x=\log _{3}y\), then \(y=3^{x}\). For example, \(4=\log _{3}81\) is equivalent to \(81=3^{4}\).

(b) If \(x=\log _{5}y\), then \(y=5^{x}\). For example, \[ -1=\log _{5}\!\left(\dfrac{1}{5}\right) \hbox{ is equivalent to } \dfrac{1}{5}=5^{-1} \]

THEOREM Properties of Logarithms

In the properties given next, \(u\) and \(a\) are positive real numbers, \( a \neq 1\), and \(r\) is any real number. The number \(\log _{a} u\) is the exponent to which \(a\) must be raised to obtain \(u\). That is, \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ a^{\log _{a} u}=u }} \]

The logarithm with base \(a\) of \(a\) raised to a power equals that power. That is, \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \log _{a} a^{r}=r }}\]

Using Properties of Logarithms

(a) \(2^{\log _{2}\pi }=\pi\)

(b) \(\log _{0.2} 0.2^{(-\sqrt{2})}=-\sqrt{2}\)

(c) \(\log _{1/5}\! \left( \dfrac{1}{5}\right) ^{\!\!kt}=kt\)

THEOREM Properties of Logarithms

In the following properties, \(u\), \(v\), and \(a\) are positive real numbers, \(a\neq 1\), and \(r\) is any real number:

1. The Log of a Product Equals the Sum of the Logs \[ \fbox{\({ \log _{a}( uv) =\log _{a}u+\log _{a}v}\)}\tag{3} \]
2. The Log of a Quotient Equals the Difference of the Logs \[ \fbox{\({ \log _{a}\!\left( \dfrac{u}{v}\right) =\log _{a}u-\log _{a}v }\)}\tag{4} \]
3. The Log of a Power Equals the Product of the Power and the Log \[ \fbox{\({ \log _{a}u^{ r}=r\log _{a}u }\)}\tag{5} \]

Using Properties (3), (4), and (5) of Logarithms

(a) \( \begin{array}{@{}rcll@{\quad}lcl} \log _{a} \big( x\displaystyle\sqrt{x^{2}+1} \big) &=&\log _{a}x+\log _{a}\displaystyle\sqrt{x^{2}+1} & & {\color{#0066A7}{\log _{a} ({uv})}}{\color{#0066A7}{=}}\,{\color{#0066A7}{\log _{a}u+\log_av}} \\[3pt] &=&\log _{a}x+\log _{a}( x^{2}+1) ^{1/2} & & \\[2pt] &=&\log _{a}x+\dfrac{1}{2}\log _{a}( x^{2}+1) & & {\color{#0066A7}{\log_{a}{u}^{r}}}{\color{#0066A7}{=}}\,{\color{#0066A7}{r\log _{a}u}} \end{array} \)

(b)

(c) \( \begin{array}[t]{lll} \log _{a}x+\log _{a}9+\log _{a}( x^{2}+1) -\log _{a}5 & =\log _{a} (9x) +\log _{a}( x^{2}+1) -\log _{a}5 & \\[5pt] & =\log _{a} [ 9x( x^{2}+1)] -\log _{a}5 & \\[5pt] & =\log _{a}\!\left[ \dfrac{9x( x^{2}+1) }{5}\right] \end{array} \)

CAUTION

In using properties (3) through (5), be careful about the values that the variable may assume. For example, the domain of the variable for \(\log _{a}{x}\) is \({x>0}\), and for \(\log _{a}( {x-1})\) it is \({x>1}\). That is, the equality \(\log _{a} x+\log _{a}(x-1) =\log _{a} [{x}({x-1})]\) is true only for \({x>1}\).

THEOREM Change-of-Base Formula

If \(a\neq 1,\) \(b\neq 1\), and \(u\) are positive real numbers, then \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \log _{a}u=\dfrac{\log _{b}u}{\log _{b}a}}} \]

For example, to approximate \(\log _{2}15,\) we use the change-of-base formula. Then we use a calculator. \begin{eqnarray*} \log _{2}15 \underset{\underset{\color{#0066A7}{\hbox{Change-of-Base Formula}}}{\color{#0066A7}{\uparrow }}} {=} \dfrac{\log 15}{\log 2}\approx 3.907\\[-10pt] \end{eqnarray*}