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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.**

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. {"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.43056em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\">x<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["-4.42"]}}

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 | $2_{2x}=5⋅2_{x}$ |

With the Same Base | $4_{3x}=4_{2x+3}$ |

With Unlike Bases | $3_{x+4}=81_{x}$ |

With a Rational Base | $(21 )_{x}=8$ |

An exponential equation can be solved by using the definition of a logarithm. Consider the following example.
*expand_more*
*expand_more*
*expand_more*
### Alternative Solution

$8_{2x}=3 $

To solve this equation, three steps must be followed.
1

Use the Definition of a Logarithm

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

$8_{2x}=3⇔2x=g_{8}3 $

2

Solve the Resulting Equation

3

Check the Solution

The solution will be checked to make sure it is not an extraneous solution. To do so, $x≈0.264$ will be substituted in the original equation.
Since a true statement was obtained, $x≈0.264$ is a solution to the equation.

$8_{2x}=3$

$x≈0.264$

$8_{2(0.264)}≈?3$

Multiply

Multiply

$8_{0.528}≈?3$

UseCalc

Use a calculator

$2.997999…≈3✓$

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.

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(23 )_{21t} $

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 $230660.$

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

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.61508em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\">t<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">\u2248<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["20"]}}

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.

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(23 )_{21t}↓230660=4000(23 )_{21t} $

b The equation from Part A will now be solved. To do so, start by dividing both sides by $4000.$

$230660=4000(23 )_{21t}⇕57.665=(23 )_{21t} $

Now use the definition of a logarithm. This helps to remove the variable from the exponent.
$57.665=(23 )_{21t}⇕g_{23}57.665=21 t $

Finally, the equation can be solved for $t.$
$g_{23}57.665=21 t$

Solve for $t$

$g_{c}a=g_{b}cg_{b}a $

$g23 g57.665 =21 t$

UseCalc

Use a calculator

$9.999998…=21 t$

MultEqn

$LHS⋅2=RHS⋅2$

$19.999996…=t$

RearrangeEqn

Rearrange equation

$t=19.999996…$

RoundInt

Round to nearest integer

$t≈20$

$230660=4000(23 )_{21t}$

$t≈20$

$230660≈?4000(23 )_{21(20)}$

Evaluate right-hand side

MoveRightFacToNumOne

$b1 ⋅a=ba $

$230660≈?4000(23 )_{220}$

CalcQuot

Calculate quotient

$230660≈?4000(23 )_{10}$

PowQuot

$(ba )_{m}=b_{m}a_{m} $

$230660≈?4000(102459049 )$

MoveLeftFacToNum

$a⋅cb =ca⋅b $

$230660≈?1024236196000 $

UseCalc

Use a calculator

$230660≈230660.15625✓$

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.

$Ifb<1b_{x_{1}}≥b_{x_{2}}⇕x_{1}≤x_{2} Ifb>1b_{x_{1}}≥b_{x_{2}}⇕x_{1}≥x_{2} $

The statements will be proved one at a time.

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 $f(x_{1})≥f(x_{2})$ if and only if $x_{1}≥x_{2}.$ Considering the definition of $f(x),$ the statement is proven.$b_{x_{1}}≥b_{x_{2}}⇔x_{1}≥x_{2} $

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 $f(x_{1})≥f(x_{2})$ if and only if $x_{1}≤x_{2}.$ Considering the definition of $f(x),$ the statement is proven.$b_{x_{1}}≥b_{x_{2}}⇔x_{1}≤x_{2} $

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!{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["x"],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["x>-\\dfrac{3}{2}","-\\dfrac{3}{2}<x","x>-1\\dfrac{1}{2}","-1\\dfrac{1}{2}<x"]}}

Use the Exponential Property of Inequality.

To solve the inequality, all numbers involved will first be written as powers of $2.$ To start, $0.5$ will be written as $21 .$
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.

$20.5_{x} <4(2)_{x}⇕2(21 )_{x} <4(2)_{x} $

Now, all the numbers can be written as powers of $2.$
$2(21 )_{x} <4(2)_{x}$

$a1 =a_{-1}$

$2(2_{-1})_{x} <4(2)_{x}$

PowPow

$(a_{m})_{n}=a_{m⋅n}$

$22_{-x} <4(2)_{x}$

$aa_{m} =a_{m−1}$

$2_{-x−1}<4(2)_{x}$

WritePow

Write as a power

$2_{-x−1}<2_{2}(2)_{x}$

MultPow

$a_{m}⋅a_{n}=a_{m+n}$

$2_{-x−1}<2_{2+x}$

$2_{-x−1}<2_{2+x}⇔-x−1<2+x $

The inequality can finally be solved.
$-x−1<2+x$

Solve for $x$

AddIneq

$LHS+x<RHS+x$

$-1<2+2x$

SubIneq

$LHS−2<RHS−2$

$-3<2x$

DivIneq

$LHS/2<RHS/2$

$2-3 <x$

MoveNegNumToFrac

Put minus sign in front of fraction

$-23 <x$

RearrangeIneq

Rearrange inequality

$x>-23 $

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.
*expand_more*
*expand_more*
*expand_more* *about* $8.$ Therefore, the solution to the equation is $x≈8.$ This can be verified by substituting $8$ for $x$ in the given equation and checking whether a true statement is obtained.
Since a true statement was obtained, $x≈8$ is a solution to the equation. In this case, an exact solution cannot be found graphically.

$ln(x−2)=2gx $

There are three steps to follow to solve logarithmic equations graphically.
1

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.

$ln(x−2)=2logx⇓y=ln(x−2)y=2logx (I)(II) $

The idea is that each function should be relatively easy to graph. 2

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 ofy=ln(x−2)x−2>0⇕x>2 Domain ofy=2logxx>0 $

With the domains in mind, the table can now be constructed. $x$ | $y=ln(x−2)$ | $y=2gx$ | ||
---|---|---|---|---|

$ln(x−2)$ | $y$ | $2gx$ | $y$ | |

$1$ | - | - | $2g1$ | $0$ |

$2$ | - | - | $2g2$ | $≈0.6$ |

$3$ | $ln(3−2)$ | $0$ | $2g3$ | $≈0.95$ |

$4$ | $ln(4−2)$ | $≈0.69$ | $2g4$ | $≈1.2$ |

$5$ | $ln(5−2)$ | $≈1.1$ | $2g5$ | $≈1.4$ |

$12$ | $ln(12−2)$ | $≈2.3$ | $2g12$ | $≈2.16$ |

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.

3

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$ln(x−2)=2gx$

$x≈8$

$ln(8−2)≈?2g8$

$1.791759…≈?1.806179…✓$

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.
### Hint

### Solution

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.

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

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=-gxsubstitute -0.5=-gx $

Next, to solve the equation, each side will be considered as a function.
$-0.5=-logx⇓{y=-0.5y=-logx (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. $x$ | $-gx$ | $y$ |
---|---|---|

$0.1$ | $-g0.1$ | $1$ |

$0.5$ | $-g0.5$ | $≈0.3$ |

$1$ | $-g1$ | $0$ |

$2$ | $-g2$ | $≈-0.3$ |

$3$ | $-g3$ | $≈-0.48$ |

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 $-0.5$ is $3,$ rounded to the nearest mole per liter. The last step is to check the solution by substituting it back into the equation. Since a true statement was obtained, $x≈3$ is a solution to the equation.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.

$g_{b}x=g_{b}y⇔x=y$

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

The biconditional statement will be proved in two parts.

$g_{b}x=g_{b}y→a=g_{b}y $

By using the definition of a logarithm, the logarithm equation can be written as an exponential equation.
$a=g_{b}y⇔b_{a}=y $

Finally, $g_{b}x$ can be substituted for $a$ and the Inverse Property of Logarithms can be used.
$x=y⇔b_{log_{b}x}=y $

Now, let $c$ be a real number such that $c=g_{b}x.$
$b_{log_{b}x}=y→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⇔c=g_{b}y $

Finally, by its own defitnion, $c$ is equal to $g_{b}x.$
$c=g_{b}y⇔log_{b}x=g_{b}y✓ $

If both sides of a logarithmic equation can be written as a single logarithm with the same base $b,$ with $b>0$ and $b =1,$ then the equation can be solved by using the Property of Equality for Logarithmic Equations. Consider the following logarithmic equation.
*expand_more*
*expand_more*
*expand_more*
*expand_more*

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

To solve the equation, there are four steps to follow.
1

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.
Next, to calculate the value of $g_{2}4,$ the definition of a logarithm will be used. Keep in mind that $2_{2}=4.$
Both sides of the equation have been written as a single logarithm with base $2.$

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

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

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

$g_{c}a=g_{b}cg_{b}a $

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

$Definition:Expression: b_{c}=a⇔g_{b}a=c2_{2}=4⇔g_{2}4=2 $

Knowing that $g_{2}4=2,$ we can continue writing both sides of the equation as a single logarithm with the same bse.
$g_{2}10x=g_{2}4g_{2}x $

Calculate logarithm

$g_{2}10x=2g_{2}x $

MultEqn

$LHS⋅2=RHS⋅2$

$2g_{2}10x=g_{2}x$

$m⋅g_{2}(a)=g_{2}(a_{m})$

$g_{2}(10x)_{2}=g_{2}x$

PowProdII

$(ab)_{m}=a_{m}b_{m}$

$g_{2}100x_{2}=g_{2}x$

2

Use the Property of Equality for Logarithmic Equations

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

$g_{2}100x_{2}=g_{2}x⇕100x_{2}=x $

3

Solve the Obtained Equation

Now, the obtained equation can be solved. This equation does not involve logarithms.
Two solutions were found, $x=0$ and $x=1001 .$

$100x_{2}=x$

Solve for $x$

SubEqn

$LHS−x=RHS−x$

$100x_{2}−x=0$

FactorOut

Factor out $x$

$x(100x−1)=0$

ZeroProdProp

Use the Zero Product Property

$x=0100x−1=0 (I)(II) $

AddEqn

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

$x=0100x=1 $

DivEqn

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

$x=0x=1001 $

4

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.
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=1001 $ will be checked now.
A true statement was obtained. Therefore, $x=1001 $ is a solution to the given equation.

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

Substitute

$x=1001 $

$g_{2}1001 +g_{2}10=?g_{4}1001 $

$g_{c}a=g_{b}cg_{b}a $

$g2g1001 +g2g10 =?g4g1001 $

UseCalc

Use a calculator

$-6.643856…+3.321928…=?-3.321928…$

AddTerms

Add terms

$-3.321928…=-3.321928…✓$

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.{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.48312em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\">x<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">\u2248<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["1.62"]}}

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

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.