body{margin:0;} noscript{z-index: 10000; position: fixed; display: block; height: 100%; width: 100%; background: #fff;} noscript .fa{font-size: 40px; display:block; color: #daa520;} noscript div{margin-top: 6%; padding-left: 20px; padding-right: 20px; text-align: center;} You must have JavaScript enabled to use this site.

Recommended

Tests

An error ocurred, try again later!

eCourses /

Algebra 2 /

Chapter {{ article.chapter.number }}

{{ article.number }}.

	{{ 'ml-lesson-number-slides' \| message : article.intro.bblockCount}}
	{{ 'ml-lesson-number-exercises' \| message : article.intro.exerciseCount}}
	{{ 'ml-lesson-time-estimation' \| message }}

Image Credits

{{ item.file.title }}
{{ presentation }}

No file copyrights entries found

A logarithmic equation is an equation that contains logarithms. Similarly, a logarithmic inequality is an inequality that contains logarithms. This lesson will discuss how to solve logarithmic equations and inequalities.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Challenge

Solving Equations With the Variable in the Exponent

If both sides of an exponential equation can be written as powers with the same base, then the equation can be solved algebraically by using the Property of Equality for Exponential Equations. What happens if the sides of an exponential equation cannot be written as powers with the same base?

2^{x - 1} = 3^{x + 1}

How could this type of equation be solved? Round the answer to two decimal places.

Discussion

Exponential Equation

An exponential equation is an equation where variable expressions occur as exponents. As with any kind of equation, there are different types of exponential equations.

	Example Equation
With One Variable	$2^{x} = 32$
With the Same Variable on Both Sides	$2^{2 x} = 5 \cdot 2^{x}$
With the Same Base	$4^{3 x} = 4^{2 x + 3}$
With Unlike Bases	$3^{x + 4} = 8 1^{x}$
With a Rational Base	$(\frac{1}{2})^{x} = 8$

Method

Solving an Exponential Equation Using Logarithms

An exponential equation can be solved by using the definition of a logarithm. Consider the following example.

8^{2 x} = 3

To solve this equation, three steps must be followed.

Use the Definition of a Logarithm

First, the definition of a logarithm will be used to isolate the variable term.

8^{2 x} = 3 \Leftrightarrow 2 x = lo g_{8} 3

Solve the Resulting Equation

Now that the variable is not in an exponent any more, the obtained equation can be solved.

2 x = lo g_{8} 3

Solve for

x

$lo g_{c} a = \frac{lo g _{b} a}{lo g _{b} c}$

2 x = \frac{lo g 3}{lo g 8}

UseCalc

Use a calculator

2 x = 0.528320 \dots

DivEqn

$LHS / 2 = RHS / 2$

x = 0.264160 \dots

RoundDec

Round to $3$ decimal place(s)

x \approx 0.264

Check the Solution

The solution will be checked to make sure it is not an extraneous solution. To do so,

x \approx 0.264

will be substituted in the original equation.

8^{2 x} = 3

$x \approx 0.264$

8^{2 (0.264)} \approx ? 3

Multiply

8^{0.528} \approx ? 3

UseCalc

Use a calculator

2.997999 \dots \approx 3 ✓

Since a true statement was obtained,

x \approx 0.264

is a solution to the equation.

Alternative Solution

Both sides of the original equation are positive. Therefore, instead of using the definition of a logarithm to rewrite the equation, common logarithms can be taken on both sides. Then, the Power Property of Logarithms can be used.

8^{2 x} = 3

$lo g (LHS) = lo g (RHS)$

lo g 8^{2 x} = lo g 3

$lo g (a^{m}) = m \cdot lo g (a)$

2 x \cdot lo g 8 = lo g 3

DivEqn

$LHS / lo g 8 = RHS / lo g 8$

2 x = \frac{lo g 3}{lo g 8}

UseCalc

Use a calculator

2 x = 0.528320 \dots

DivEqn

$LHS / 2 = RHS / 2$

x = 0.264160 \dots

RoundDec

Round to $3$ decimal place(s)

x \approx 0.264

Example

Eat the Chapati Before Fungi Grows!

Emily and her friends went to the beach on a cloudy afternoon and cooked some chapati.

Since they did not smoke the chapati and did not use salt in the cooking process, Emily knows that it will not be too long before the food spoils. Food spoilage is generally caused due to the action of microorganisms like fungi. Fungal growth is exponential and given by the following exponential function.

f (t) = 4000 (\frac{3}{2})^{\frac{1}{2} t}

Here,

f (t)

is the amount of fungi in the chapati

t

hours after it has been baked. Emily knows that they cannot eat the chapati once the amount of fungi is

230660 .

a Write an exponential equation that must be solved to find out after how many hours the amount of fungi in the chapati is

230660 .

b Solve the equation from Part A. Round the answer to the nearest integer.

Hint

a Does the number

230660

correspond to amount of fungi or time?

b Divide both sides of the equation by

4000

and then use the definition of a logarithm.

Solution

a Since the number

230660

corresponds to amount of fungi, this number will be substituted for

f (t)

in the given formula.

f (t) = 4000 (\frac{3}{2})^{\frac{1}{2} t} ↓ 230660 = 4000 (\frac{3}{2})^{\frac{1}{2} t}

b The equation from Part A will now be solved. To do so, start by dividing both sides by

4000 .

230660 = 4000 (\frac{3}{2})^{\frac{1}{2} t} ⇕ 57.665 = (\frac{3}{2})^{\frac{1}{2} t}

Now use the definition of a logarithm. This helps to remove the variable from the exponent.

57.665 = (\frac{3}{2})^{\frac{1}{2} t} ⇕ lo g_{\frac{3}{2}} 57.665 = \frac{1}{2} t

Finally, the equation can be solved for

t .

lo g_{\frac{3}{2}} 57.665 = \frac{1}{2} t

Solve for

t

$lo g_{c} a = \frac{lo g _{b} a}{lo g _{b} c}$

\frac{lo g 5 7 . 6 6 5}{lo g \frac{3}{2}} = \frac{1}{2} t

UseCalc

Use a calculator

9.999998 \dots = \frac{1}{2} t

MultEqn

$LHS \cdot 2 = RHS \cdot 2$

19.999996 \dots = t

RearrangeEqn

Rearrange equation

t = 19.999996 \dots

RoundInt

Round to nearest integer

t \approx 20

Just to make sure the obtained solution is not an extraneous solution,

t \approx 20

will be substituted in the equation.

230660 = 4000 (\frac{3}{2})^{\frac{1}{2} t}

$t \approx 20$

230660 \approx ? 4000 (\frac{3}{2})^{\frac{1}{2} (20)}

Evaluate right-hand side

MoveRightFacToNumOne

$\frac{1}{b} \cdot a = \frac{a}{b}$

230660 \approx ? 4000 (\frac{3}{2})^{\frac{2 0}{2}}

CalcQuot

Calculate quotient

230660 \approx ? 4000 (\frac{3}{2})^{10}

PowQuot

$(\frac{a}{b})^{m} = \frac{a ^{m}}{b ^{m}}$

230660 \approx ? 4000 (\frac{5 9 0 4 9}{1 0 2 4})

MoveLeftFacToNum

$a \cdot \frac{b}{c} = \frac{a \cdot b}{c}$

230660 \approx ? \frac{2 3 6 1 9 6 0 0 0}{1 0 2 4}

UseCalc

Use a calculator

230660 \approx 230660.15625 ✓

Since a true statement was obtained,

t \approx 20

is the solution to the equation. This means that the chapati will become unsafe to eat due to fungi growth about

20

hours after cooking it.

Discussion

Exponential Property of Inequality

An exponential inequality is an inequality that involves exponential expressions.

Let $b$ be a positive real number different than $1 .$ The following statements hold true.

If b < 1 b^{x_{1}} \geq b^{x_{2}} ⇕ x_{1} \leq x_{2} If b > 1 b^{x_{1}} \geq b^{x_{2}} ⇕ x_{1} \geq x_{2}

These facts are also valid for strict inequalities.

Proof

The statements will be proved one at a time.

$b > 1$

If $b$ is greater than $1,$ the exponential function $f (x) = b^{x}$ is increasing for its entire domain.

In the graph, it can be seen that

f (x_{1}) \geq f (x_{2})

if and only if

x_{1} \geq x_{2} .

Considering the definition of

f (x),

the statement is proven.

b^{x_{1}} \geq b^{x_{2}} \Leftrightarrow x_{1} \geq x_{2}

$b < 1$

If $b$ is greater than $0$ and less than $1,$ then $f (x) = b^{x}$ is decreasing for its entire domain.

In the graph, it is seen that

f (x_{1}) \geq f (x_{2})

if and only if

x_{1} \leq x_{2} .

Considering the definition of

f (x),

the statement is proven.

b^{x_{1}} \geq b^{x_{2}} \Leftrightarrow x_{1} \leq x_{2}

This property is also known as Property of Inequality for Exponential Functions and is useful when solving exponential inequalities.

Example

Solving an Exponential Inequality

Back home from the beach, Emily realized that she managed to solve an exponential equation to calculate the expiration of the chapati she and her friends cooked. However, she also realized that she has not practiced solving exponential inequalities. To help her practice, she went online to find some worksheets and found an interesting inequality.

Help Emily solve the inequality!

Hint

Use the Exponential Property of Inequality.

Solution

To solve the inequality, all numbers involved will first be written as powers of

2 .

To start,

0.5

will be written as

\frac{1}{2} .

\frac{0 . 5 ^{x}}{2} < 4 (2)^{x} ⇕ \frac{( \frac{1}{2} ) ^{x}}{2} < 4 (2)^{x}

Now, all the numbers can be written as powers of

2 .

\frac{( \frac{1}{2} ) ^{x}}{2} < 4 (2)^{x}

$\frac{1}{a} = a^{- 1}$

\frac{( 2 ^{- 1} ) ^{x}}{2} < 4 (2)^{x}

PowPow

$(a^{m})^{n} = a^{m \cdot n}$

\frac{2 ^{- x}}{2} < 4 (2)^{x}

$\frac{a ^{m}}{a} = a^{m - 1}$

2^{- x - 1} < 4 (2)^{x}

WritePow

Write as a power

2^{- x - 1} < 2^{2} (2)^{x}

MultPow

$a^{m} \cdot a^{n} = a^{m + n}$

2^{- x - 1} < 2^{2 + x}

The base of the expressions on both sides is

2,

which is is greater than

1 .

Therefore, by the Exponential Property of Inequality, the left-hand side of the inequality is less than the right-hand side if and only if the exponent on the left-hand side is less than the exponent on the right-hand side.

2^{- x - 1} < 2^{2 + x} \Leftrightarrow - x - 1 < 2 + x

The inequality can finally be solved.

- x - 1 < 2 + x

Solve for

x

AddIneq

$LHS + x < RHS + x$

- 1 < 2 + 2 x

SubIneq

$LHS - 2 < RHS - 2$

- 3 < 2 x

DivIneq

$LHS / 2 < RHS / 2$

\frac{- 3}{2} < x

MoveNegNumToFrac

Put minus sign in front of fraction

- \frac{3}{2} < x

RearrangeIneq

Rearrange inequality

x > - \frac{3}{2}

Discussion

Logarithmic Equation

A logarithmic equation is an equation where the variable is in the argument of a logarithm. An example equation is shown below.

Method

Solving Logarithmic Equations Graphically

Logarithmic equations can be tricky to solve sometimes. However, if the exact solution is not needed, an approximated solution can be found by solving the equation graphically. Consider an example logarithmic equation.

ln (x - 2) = 2 lo g x

There are three steps to follow to solve logarithmic equations graphically.

Create Two Functions From the Given Equation

Each side of the equation can be considered as a function. In this case, both sides can be written as logarithmic functions.

l n (x - 2) = 2 l o g x ⇓ y = l n (x - 2) y = 2 l o g x (I) (II)

The idea is that each function should be relatively easy to graph.

Graph the Functions

A table of values including both functions will be made. However, the domain of each function needs to be considered first.

Domain of y = l n (x - 2) x - 2 > 0 ⇕ x > 2 Domain of y = 2 l o g x x > 0

With the domains in mind, the table can now be constructed.

$x$	$y = ln (x - 2)$		$y = 2 lo g x$
$x$	$ln (x - 2)$	$y$	$2 lo g x$	$y$
$1$	-	-	$2 lo g 1$	$0$
$2$	-	-	$2 lo g 2$	$\approx 0.6$
$3$	$ln (3 - 2)$	$0$	$2 lo g 3$	$\approx 0.95$
$4$	$ln (4 - 2)$	$\approx 0.69$	$2 lo g 4$	$\approx 1.2$
$5$	$ln (5 - 2)$	$\approx 1.1$	$2 lo g 5$	$\approx 1.4$
$12$	$ln (12 - 2)$	$\approx 2.3$	$2 lo g 12$	$\approx 2.16$

Now both functions will be graphed on the same coordinate plane.

The number of solutions to the equation is the number of points of intersection of the graphs. These graphs have one point of intersection, so there is only one solution.

Identify the

x -

coordinates of the Points of Intersection

The solutions to the equation are the $x -$ coordinates of any points of intersection of the graphs.

The

x -

coordinate of the point of intersection is about

8 .

Therefore, the solution to the equation is

x \approx 8 .

This can be verified by substituting

8

for

x

in the given equation and checking whether a true statement is obtained.

ln (x - 2) = 2 lo g x

$x \approx 8$

ln (8 - 2) \approx ? 2 lo g 8

Evaluate

SubTerm

Subtract term

ln 6 \approx ? 2 lo g 8

UseCalc

Use a calculator

1.791759 \dots \approx ? 1.806179 \dots ✓

Since a true statement was obtained,

x \approx 8

is a solution to the equation. In this case, an exact solution cannot be found graphically.

Example

Acidic, Basic, and Neutral Solutions

In her chemistry lesson, Emily is learning about the acidity and alkalinity of substances. The acidity or alkalinity of a substance is defined by the value of pH, where

x

is the hydrogen ion concentration of the substance measured in moles per liter.

For example, since their pH values are less than

7,

lemon juice and vinegar are acidic solutions. Conversely, hand soap and eggs are a basic, because their pH value is greater then

7 .

Water is a neutral solution. Emily is working with a solution with a pH of

- 0.5 .

What is the hydrogen ion concentration of this solution? Round the answer to the nearest mole per liter.

Hint

Substitute $- 0.5$ for pH and solve the logarithmic equation by graphing.

Solution

To find the hydrogen ion concentration of a solution with a pH of

- 0.5,

the number must be substituted for pH in the given formula.

pH = - lo g x substitute - 0.5 = - lo g x

Next, to solve the equation, each side will be considered as a function.

- 0.5 = - l o g x ⇓ {y = - 0.5 y = - l o g x (I) (II)

The first function is a horizontal line whose

y -

intercept is

- 0.5 .

The second is a logarithmic function. Considering that its domain is

x > 0,

make a table of values.

$x$	$- lo g x$	$y$
$0.1$	$- lo g 0.1$	$1$
$0.5$	$- lo g 0.5$	$\approx 0.3$
$1$	$- lo g 1$	$0$
$2$	$- lo g 2$	$\approx - 0.3$
$3$	$- lo g 3$	$\approx - 0.48$

Now, graph the functions on the same coordinate plane.

The graphs intersect at one point. The $x -$ coordinate of the point of intersection is the hydrogen ion concentration of the solution.

The exact value of the

x -

coordinate cannot be identified. However, the answer must be rounded to the nearest mole per liter, so the hydrogen concentration of the solution with a pH of

- 0.5

3,

rounded to the nearest mole per liter. The last step is to check the solution by substituting it back into the equation.

- 0.5 = - lo g x

$x \approx 3$

- 0.5 \approx ? - lo g 3

UseCalc

Use a calculator

0.5 \approx 0.477121 \dots ✓

Since a true statement was obtained,

x \approx 3

is a solution to the equation.

Discussion

Property of Equality for Logarithmic Equations

Let $b$ be a positive real number different than $1 .$ Two logarithms with the same base $b$ are equal if and only if their arguments are equal.

$lo g_{b} x = lo g_{b} y \Leftrightarrow x = y$

Since logarithms are defined for positive numbers, $x$ and $y$ must be positive.

Proof

The biconditional statement will be proved in two parts.

$lo g_{b} x = lo g_{b} y \Rightarrow x = y$

Let

a

be a real number such that

a = lo g_{b} x .

lo g_{b} x = lo g_{b} y \to a = lo g_{b} y

By using the definition of a logarithm, the logarithm equation can be written as an exponential equation.

a = lo g_{b} y \Leftrightarrow b^{a} = y

Finally,

lo g_{b} x

can be substituted for

a

and the Inverse Property of Logarithms can be used.

b^{a} = y

Substitute

$a = l o g_{b} x$

b^{l o g_{b} x} = y

$b^{l o g_{b} (m)} = m$

x = y ✓

$x = y \Rightarrow lo g_{b} x = lo g_{b} y$

The Inverse Property of Logarithms can be used to express

x

b^{l o g_{b} x} .

x = y \Leftrightarrow b^{l o g_{b} x} = y

Now, let

c

be a real number such that

c = lo g_{b} x .

b^{l o g_{b} x} = y \to b^{c} = y

Next, the definition of a logarithm can be used to rewrite the obtained exponential equation as a logarithmic equation.

b^{c} = y \Leftrightarrow c = lo g_{b} y

Finally, by its own defitnion,

c

is equal to

lo g_{b} x .

c = lo g_{b} y \Leftrightarrow l o g_{b} x = lo g_{b} y ✓

Method

Solving Logarithmic Equations by Using the Property of Equality

If both sides of a logarithmic equation can be written as a single logarithm with the same base

b,

with

b > 0

and

b \neq = 1,

then the equation can be solved by using the Property of Equality for Logarithmic Equations. Consider the following logarithmic equation.

lo g_{2} x + lo g_{2} 10 = lo g_{4} x

To solve the equation, there are four steps to follow.

Rewrite Both Sides as a Single Logarithm With the Same Base

To rewrite the left-hand side as a single logarithm with base

2,

the Product Property of Logarithms will be used. To rewrite the right-hand side as a single logarithm with base

2,

the Change of Base Formula will be used.

lo g_{2} x + lo g_{2} 10 = lo g_{4} x

$lo g_{2} (m) + lo g_{2} (n) = lo g_{2} (m n)$

lo g_{2} 10 x = lo g_{4} x

$lo g_{c} a = \frac{lo g _{b} a}{lo g _{b} c}$

lo g_{2} 10 x = \frac{lo g _{2} x}{lo g _{2} 4}

Next, to calculate the value of

lo g_{2} 4,

the definition of a logarithm will be used. Keep in mind that

2^{2} = 4 .

Definition : Expression : b^{c} = a \Leftrightarrow lo g_{b} a = c 2^{2} = 4 \Leftrightarrow lo g_{2} 4 = 2

Knowing that

lo g_{2} 4 = 2,

we can continue writing both sides of the equation as a single logarithm with the same bse.

lo g_{2} 10 x = \frac{lo g _{2} x}{lo g _{2} 4}

Calculate logarithm

lo g_{2} 10 x = \frac{lo g _{2} x}{2}

MultEqn

$LHS \cdot 2 = RHS \cdot 2$

2 lo g_{2} 10 x = lo g_{2} x

$m \cdot lo g_{2} (a) = lo g_{2} (a^{m})$

lo g_{2} (10 x)^{2} = lo g_{2} x

PowProdII

$(a b)^{m} = a^{m} b^{m}$

lo g_{2} 100 x^{2} = lo g_{2} x

Both sides of the equation have been written as a single logarithm with base

2 .

Use the Property of Equality for Logarithmic Equations

Next, the Property of Equality for Logarithmic Equations can be used to simplify the expressions.

lo g_{2} 100 x^{2} = lo g_{2} x ⇕ 100 x^{2} = x

Solve the Obtained Equation

Now, the obtained equation can be solved. This equation does not involve logarithms.

100 x^{2} = x

Solve for

x

SubEqn

$LHS - x = RHS - x$

100 x^{2} - x = 0

FactorOut

Factor out $x$

x (100 x - 1) = 0

ZeroProdProp

Use the Zero Product Property

x = 0 100 x - 1 = 0 (I) (II)

AddEqn

$(II):$ $LHS + 1 = RHS + 1$

x = 0 100 x = 1

DivEqn

$(II):$ $LHS / 100 = RHS / 100$

x = 0 x = \frac{1}{1 0 0}

Two solutions were found,

x = 0

and

x = \frac{1}{1 0 0} .

Check the Solutions

Finally, the solutions must be checked. It is not uncommon to find extraneous solutions. The solution

x = 0

will be checked first.

lo g_{2} x + lo g_{2} 10 = lo g_{4} x

Substitute

$x = 0$

lo g_{2} 0 + lo g_{2} 10 = ? lo g_{4} 0 \times

Logarithms are not defined for non-positive numbers. Since

0

is not positive,

x = 0

is an extraneous solution. This means that it is not a solution to the given equation. The solution

x = \frac{1}{1 0 0}

will be checked now.

lo g_{2} x + lo g_{2} 10 = lo g_{4} x

Substitute

$x = \frac{1}{1 0 0}$

lo g_{2} \frac{1}{1 0 0} + lo g_{2} 10 = ? lo g_{4} \frac{1}{1 0 0}

$lo g_{c} a = \frac{lo g _{b} a}{lo g _{b} c}$

\frac{lo g \frac{1}{1 0 0}}{lo g 2} + \frac{lo g 1 0}{lo g 2} = ? \frac{lo g \frac{1}{1 0 0}}{lo g 4}

UseCalc

Use a calculator

- 6.643856 \dots + 3.321928 \dots = ? - 3.321928 \dots

AddTerms

Add terms

- 3.321928 \dots = - 3.321928 \dots ✓

A true statement was obtained. Therefore,

x = \frac{1}{1 0 0}

is a solution to the given equation.

Example

Solving Equations With Logarithms on Both Sides

Emily told her study buddy about how she used a graph to solve a logarithmic equation. Her friend is pretty competitive, so he challenged Emily to solve a logarithmic equation with logarithms on both sides but without graphing.

Help Emily show her study buddy that she has a solid understanding of logarithmic equations! Round the answer to three significant figures.

Hint

Start by rewriting the right-hand side as a single natural logarithm.

Solution

The equation involves natural logarithms. To solve it, the first step is to rewrite the right-hand side as a single logarithm by using the Power Property of Logarithms.

ln (x + 1) = 2 ln x

$b \cdot ln (a) = ln (a^{b})$

ln (x + 1) = ln x^{2}

Since both sides are single logarithms with the same base, the Property of Equality for Logarithmic Equations can be used.

ln (x + 1) = ln x^{2} ⇕ x + 1 = x^{2}

Notice that the resulting equation is a quadratic equation. Therefore, the values of

a,

b,

and

c

need to be identified.

x + 1 = x^{2} ⇕ - 1 x^{2} + 1 x + 1 = 0

Next,

a = - 1,

b = 1,

and

c = 1

will be substituted in the Quadratic Formula.

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

SubstituteValues

Substitute values

x = \frac{- 1 \pm 1 ^{2} - 4 ( - 1 ) ( 1 )}{2 ( - 1 )}

Evaluate right-hand side

BaseOne

$1^{a} = 1$

x = \frac{- 1 \pm 1 - 4 ( - 1 ) ( 1 )}{2 ( - 1 )}

MultPosNeg

$a (- b) = - a \cdot b$

x = \frac{- 1 \pm 1 - ( - 4 ) ( 1 )}{- 2}

IdPropMult

\IdPropMult

x = \frac{- 1 \pm 1 - ( - 4 )}{- 2}

SubNeg

$a - (- b) = a + b$

x = \frac{- 1 \pm 5}{- 2}

StateSol

State solutions

x_{1} = \frac{- 1 + 5}{- 2} x_{2} = \frac{- 1 - 5}{- 2} (I) (II)

$(I), (II):$ Use a calculator

x_{1} = - 0.618033 \dots x_{2} = 1.618033 \dots

RoundSigDig

\RoundSigDig{3}

x_{1} \approx - 0.618 x_{2} \approx 1.62

Finally, the solutions can be verified by substituting them into the original equation. The solution

x \approx - 0.618

will be checked first.

ln (x + 1) = 2 ln x

$x \approx - 0.618$

ln (- 0.618 + 1) \approx ? 2 ln - 0.618 \times

Logarithms are only defined for positive numbers, so

x \approx - 0.618

is an extraneous solution. The solution

x \approx 1.62

will be checked now.

ln (x + 1) = 2 ln x

$x \approx 1.62$

ln (1.62 + 1) \approx ? 2 ln 1.62

AddTerms

Add terms

ln 2.62 \approx ? 2 ln 1.62

$b \cdot ln (a) = ln (a^{b})$

ln 2.62 \approx ? ln 1.6 2^{2}

CalcPow

Calculate power

ln 2.62 \approx ? ln 2.6244

UseCalc

Use a calculator

0.963174 \dots \approx 0.964852 \dots ✓

A true statement was found this time. Therefore, the only solution to the equation is

x \approx 1.62 .

Discussion

Solving Equations With a Logarithm on One Side

If only one side of a logarithmic equation can be written as a single logarithm with base

b,

with

b > 0

and

b \neq = 1,

then the equation can be solved by using the definition of a logarithm.

lo g_{b} m = n \Leftrightarrow b^{n} = m

Since logarithms are defined only for positive numbers,

m

must be a positive number. Consider an example equation.

4 lo g_{2} (x - 1) + 6 = 14

To solve this equation, four steps must be followed.

Isolate the Logarithmic Expression on One Side

The first step is to isolate the logarithmic expression. This can be done by using inverse operations.

4 lo g_{2} (x - 1) + 6 = 14

SubEqn

$LHS - 6 = RHS - 6$

4 lo g_{2} (x - 1) = 8

DivEqn

$LHS / 4 = RHS / 4$

lo g_{2} (x - 1) = 2

Use the Definition of a Logarithm

Once the logarithmic expression is isolated, the definition of a logarithm can be used to eliminate the logarithm.

lo g_{2} (x - 1) = 2 \Leftrightarrow 2^{2} = x - 1

Solve the Obtained Equation

The obtained equation can now be solved.

2^{2} = x - 1

Solve for

x

CalcPow

Calculate power

4 = x - 1

AddEqn

$LHS + 1 = RHS + 1$

5 = x

RearrangeEqn

Rearrange equation

x = 5

Check the Solutions

Finally, to detect any extraneous solutions, the obtained value can be checked by substituting it into the original equation.

4 lo g_{2} (x - 1) + 6 = 14

Substitute

$x = 5$

4 lo g_{2} (5 - 1) + 6 = ? 14

Evaluate left-hand side

SubTerm

Subtract term

4 lo g_{2} 4 + 6 = ? 14

$lo g_{c} a = \frac{lo g _{b} a}{lo g _{b} c}$

4 (\frac{lo g 4}{lo g 2}) + 6 = ? 14

UseCalc

Use a calculator

4 (2) + 6 = ? 14

Multiply

8 + 6 = ? 14

AddTerms

Add terms

14 = 14 ✓

A true statement was obtained, so the solution is not extraneous.

Example

Logarithms and the Decibels at a Rock Concert

Decibels (dB) are units of measurement that are used to describe a sound's loudness. The decibels of a sound with an intensity of

x

watts per square meter are given by the following formula.

D = 10 dB lo g \frac{x}{I _{0}}

Here,

I_{0}

is the intensity of the softest sound a healthy human ear can detect, the value of which is

1 0^{- 12} \frac{W}{m ^{2}} .

Emily wants to attend a rock concert, but she is concerned about how loud the music may be.

A typical rock concert decibel level is about

105 dB .

Calculate the watts per square meter of a typical rock concert. Round the answer to the nearest hundreds.

Hint

Substitute $105$ for $D$ in the given formula and solve for $x .$

Solution

To calculate the watts per square meter of a typical rock concert,

105

will be substituted for

D

in the given equation. The logarithmic equation will then be solved algebraically. Recall that

I_{0} = 1 0^{- 12} \frac{W}{m ^{2}} .

For simplicity, the units will not be considered until the end.

D = 10 lo g \frac{x}{I _{0}}

SubstituteII

$I_{0} = 1 0^{- 12}$ , $D = 105$

105 = 10 lo g \frac{x}{1 0 ^{- 12}}

DivEqn

$LHS / 10 = RHS / 10$

10.5 = lo g \frac{x}{1 0 ^{- 12}}

Next, to eliminate the logarithm, its definition will be used. Since the base is not stated, the formula involves a common logarithm.

10.5 = lo g \frac{x}{1 0 ^{- 12}} ⇕ 1 0^{10.5} = \frac{x}{1 0 ^{- 12}}

Finally, the equation can be solved for

x .

1 0^{10.5} = \frac{x}{1 0 ^{- 12}}

Solve for

x

MultEqn

$LHS \cdot 1 0^{- 12} = RHS \cdot 1 0^{- 12}$

1 0^{10.5} 1 0^{- 12} = x

MultPow

$a^{m} \cdot a^{n} = a^{m + n}$

1 0^{10.5 + (- 12)} = x

AddNeg

$a + (- b) = a - b$

1 0^{- 1.5} = x

RearrangeEqn

Rearrange equation

x = 1 0^{- 1.5}

UseCalc

Use a calculator

x = 0.031622 \dots

RoundDec

Round to $2$ decimal place(s)

x \approx 0.03

A rock concert with a decibel level is

105 dB

has a sound with an intensity of

0.03

watts per square meter. To make sure that this is not an extraneous solution, it can be confirmed by substituting the value back into the equation.

105 = 10 lo g \frac{x}{1 0 ^{- 12}}

$x \approx 0.03$

105 \approx ? 10 lo g \frac{0 . 0 3}{1 0 ^{- 12}}

UseCalc

Use a calculator

105 \approx 104.771212 \dots ✓

Since a true statement was obtained,

x \approx 0.03

is a solution to the equation.

Discussion

Logarithmic Property of Inequality

A logarithmic inequality is an inequality that involves logarithms.

Let $b$ be a positive real number different than $1 .$ The following statements hold true.

If b < 1 lo g_{b} x_{1} \geq lo g_{b} x_{2} ⇕ x_{1} \leq x_{2} If b > 1 lo g_{b} x_{1} \geq lo g_{b} x_{2} ⇕ x_{1} \geq x_{2}

These facts are also valid for strict inequalities.

Proof

The statements will be proved one at a time.

$b > 1$

If $b$ is greater than $1,$ the logarithmic function $f (x) = lo g_{b} x$ is increasing over its entire domain.

In the graph, it can be seen that

f (x_{1}) \geq f (x_{2})

if and only if

x_{1} \geq x_{2} .

Considering the definition of

f (x),

the statement is proven.

lo g_{b} x_{1} \geq lo g_{b} x_{2} \Leftrightarrow x_{1} \geq x_{2}

$b < 1$

If $b$ is greater than $0$ and less than $1,$ then $f (x) = lo g_{b} x$ is decreasing over its entire domain.

In the graph, it can be seen that

f (x_{1}) \geq f (x_{2})

if and only if

x_{1} \leq x_{2} .

Considering the definition of

f (x),

the statement is proven.

lo g_{b} x_{1} \geq lo g_{b} x_{2} \Leftrightarrow x_{1} \leq x_{2}

This property is useful when solving logarithmic inequalities.

Example

Solving a Logarithmic Inequality

After going to the rock concert and using logarithms to calculate the watts per square meter, Emily wants to finish this topic on a high note. Her teacher asked her to solve a logarithmic inequality for extra credit.

Write the answer in interval notation.

Hint

Rewrite the right-hand side of the inequality as a single logarithm with base $10 .$ Then, use the Logarithmic Property of Inequality.

Solution

To solve the inequality, the first step is writing the right-hand side as a single logarithm with base

10 .

Start by writing

1

as a logarithmic expression by using the definition of a logarithm.

1 0^{1} = 10 \Leftrightarrow 1 = lo g 10

With this information,

lo g 10

can be substituted for

1 .

Then, the Quotient Property of Logarithms can be used.

lo g (x - 3) < lo g (x + 6) - 1

Substitute

$1 = l o g 10$

lo g (x - 3) < lo g (x + 6) - l o g 10

$lo g (m) - lo g (n) = lo g (\frac{m}{n})$

lo g (x - 3) < lo g (\frac{x + 6}{1 0})

Both sides are written as single logarithms with base

10,

which is greater than

1 .

Therefore, by the Logarithmic Property of Inequality, the left-hand side of the inequality is less than the right-hand side if an only if the argument of the left-hand side is less than the argument of the right-hand side.

lo g (x - 3) < lo g (\frac{x + 6}{1 0}) ⇕ x - 3 < \frac{x + 6}{1 0}

Now solve this new inequality by using inverse operations.

x - 3 < \frac{x + 6}{1 0}

Solve for

x

MultIneq

$LHS \cdot 10 < RHS \cdot 10$

(x - 3) 10 < x + 6

Distr

Distribute $10$

10 x - 30 < x + 6

AddIneq

$LHS + 30 < RHS + 30$

10 x < x + 36

SubIneq

$LHS - x < RHS - x$

9 x < 36

DivIneq

$LHS / 9 < RHS / 9$

x < 4

Finally, recall that logarithms are only defined for positive numbers. Therefore, the arguments of the logarithms in the original inequality must be greater than zero.

$lo g (x - 3) < lo g (x + 6) - 1$
Argument on the LHS	Argument on the RHS
$x - 3 > 0$ $⇕$ $x > 3$	$x + 6 > 0$ $⇕$ $x > - 6$

The solution set is the intersection of three inequalities.

Inequality I : Inequality II : Inequality III : - 6 < x < 4 - 3 < x - 6 < x

To find the solution set, the inequalities can be graphed on the same number line. The solution set consists of the numbers on the line that belong to the three individual solutions. In other words, the solution set is the intersection of all three solutions.

The solution to the inequality are all real numbers greater than

3

and less than

4

because the three lines overlap between

3

and

4 .

This solution can be written as a compound inequality or in interval notation.

3 < x < 4 or (3, 4)

Closure

Exponential Expressions With Different Bases

With the topics covered in this lesson, the challenge presented at the beginning can now be solved. Consider the given exponential equation again.

2^{x - 1} = 3^{x + 1}

Here, each side is an exponential expression. What is the value of

x ?

Round the answer to two decimal places.

Hint

Take common logarithms on both sides.

Solution

Since the expressions have different bases, the Property of Equality for Exponential Equations cannot be used. However, both

2^{x - 1}

and

3^{x + 1}

are greater than zero for any value of

x .

Therefore, common logarithms can be taken on both sides.

2^{x - 1} = 3^{x + 1} ⇕ lo g 2^{x - 1} = lo g 3^{x + 1}

Now use the Power Property of Logarithms. Then the variable

x

can be isolated.

lo g 2^{x - 1} = lo g 3^{x + 1}

$lo g (a^{m}) = m \cdot lo g (a)$

(x - 1) lo g 2 = (x + 1) lo g 3

Distr

Distribute $lo g 2 & lo g 3$

x lo g 2 - lo g 2 = x lo g 3 + lo g 3

AddEqn

$LHS + lo g 2 = RHS + lo g 2$

x lo g 2 = x lo g 3 + lo g 3 + lo g 2

SubEqn

$LHS - x lo g 3 = RHS - x lo g 3$

x lo g 2 - x lo g 3 = lo g 3 + lo g 2

FactorOut

Factor out $x$

x (lo g 2 - lo g 3) = lo g 3 + lo g 2

DivEqn

$LHS / (lo g 2 - lo g 3) = RHS / (lo g 2 - lo g 3)$

x = \frac{lo g 3 + lo g 2}{lo g 2 - lo g 3}

UseCalc

Use a calculator

x = - 4.419022 \dots

RoundDec

Round to $2$ decimal place(s)

x \approx - 4.42

The solution to the exponential equation is

x \approx - 4.42 .

This can verified by substituting it into the original equation.

2^{x - 1} = 3^{x + 1}

$x \approx - 4.42$

2^{- 4.42 - 1} \approx ? 3^{- 4.42 + 1}

AddSubTerms

Add and subtract terms

2^{- 5.42} \approx ? 3^{- 3.42}

CalcPow

Calculate power

0.023357 \dots \approx 0.023347 \dots ✓

{{ article.displayTitle }}

Catch-Up and Review

Alternative Solution

Hint

Solution

Proof

b>1

b<1

Hint

Solution

Hint

Solution

Proof

logb​x=logb​y⇒ x=y

x=y⇒ logb​x=logb​y

Hint

Solution

Hint

Solution

Proof

b>1

b<1

Hint

Solution

Hint

Solution

$b > 1$

$b < 1$

$lo g_{b} x = lo g_{b} y \Rightarrow x = y$

$x = y \Rightarrow lo g_{b} x = lo g_{b} y$

$b > 1$

$b < 1$