4. Solving Logarithmic Equations and Inequalities
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Chapter 5
4.
Solving Logarithmic Equations and Inequalities
Logarithmic equations and inequalities are mathematical expressions that involve logarithms. This lesson delves into the methods and techniques to solve these equations. One can encounter equations where variables are present in the argument of a logarithm. To tackle them, various properties of logarithms, such as the Power Property, are employed. The lesson also touches upon the graphical approach to solving, where each side of an equation is considered a function, and solutions are derived from their intersections. Additionally, the importance of understanding the domain of functions is emphasized, ensuring valid solutions. This knowledge is crucial for students and professionals alike, as it forms the foundation for more advanced mathematical topics.
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Lesson Settings & Tools
| | 15 Theory slides |
| | 13 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Solving Logarithmic Equations and Inequalities
Slide of 15
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
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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.
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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^(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 | (1/2)^x=8 |
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Solving an Exponential Equation Using Logarithms
An exponential equation can be solved by using the definition of a logarithm. Consider the following example.
8^(2x)=3
To solve this equation, three steps must be followed.
1
Use the Definition of a Logarithm
expand_more First, the definition of a logarithm will be used to isolate the variable term. 8^(2x)=3 ⇔ 2x=log_8 3
2
Solve the Resulting Equation
expand_more 3
Check the Solution
expand_more 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.
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.
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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 (3/2)^(12t) 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 230 660.
a Write an exponential equation that must be solved to find out after how many hours the amount of fungi in the chapati is 230 660.
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b Solve the equation from Part A. Round the answer to the nearest integer.
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Hint
a Does the number 230 660 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 230 660 corresponds to amount of fungi, this number will be substituted for f(t) in the given formula.
f(t) = 4000 (3/2)^(12t) ↓ 230 660 = 4000 (3/2)^(12t)
b The equation from Part A will now be solved. To do so, start by dividing both sides by 4000.
log_(32) 57.665=1/2t
▼
Solve for t
log_c a = log_b a/log_b c
log 57.665/log 32=1/2t
UseCalc
Use a calculator
9.999998...=1/2t
MultEqn
LHS * 2=RHS* 2
19.999996...=t
RearrangeEqn
Rearrange equation
t=19.999996...
RoundInt
Round to nearest integer
t≈ 20
230 660 = 4000 (3/2)^(12t)
t ≈ 20
230 660 ? ≈ 4000 (3/2)^(12( 20))
▼
Evaluate right-hand side
MoveRightFacToNumOne
1/b* a = a/b
230 660 ? ≈ 4000 (3/2)^(202)
CalcQuot
Calculate quotient
230 660 ? ≈ 4000 (3/2)^(10)
PowQuot
(a/b)^m=a^m/b^m
230 660 ? ≈ 4000 (59 049/1024)
MoveLeftFacToNum
a*b/c= a* b/c
230 660 ? ≈ 236 196 000/1024
UseCalc
Use a calculator
230 660 ≈ 230 660.15625 ✓
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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.
c|c Ifb<1 & Ifb>1 [1em] b^(x_1)≥ b^(x_2) & b^(x_1)≥ b^(x_2) ⇕ & ⇕ x_1≤ x_2 & x_1≥ x_2
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)≥ 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
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)≥ 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
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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.
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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 12.
0.5^x/2<4(2)^x ⇕ ( 12)^x/2<4(2)^x
Now, all the numbers can be written as powers of 2.
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) ⇔ - x-1<2+x
The inequality can finally be solved.
( 12)^x/2<4(2)^x
1/a=a^(- 1)
(2^(- 1))^x/2<4(2)^x
PowPow
(a^m)^n=a^(m* n)
2^(- x)/2<4(2)^x
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*a^n=a^(m+n)
2^(- x-1)<2^(2+x)
- x-1<2+x
▼
Solve for x
AddIneq
LHS+x
- 1<2+2x
SubIneq
LHS-2
- 3<2x
DivIneq
.LHS /2.<.RHS /2.
- 3/2
MoveNegNumToFrac
Put minus sign in front of fraction
- 3/2
RearrangeIneq
Rearrange inequality
x>- 3/2
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Logarithmic Equation
Method
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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) = 2log x
There are three steps to follow to solve logarithmic equations graphically.
The x-coordinate of the point of intersection is 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.
1
Create Two Functions From the Given Equation
expand_more 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)= 2log x ⇓ lcy= ln (x-2) & (I) y= 2log x & (II) The idea is that each function should be relatively easy to graph.
2
Graph the Functions
expand_more A table of values including both functions will be made. However, the domain of each function needs to be considered first. c|c Domain of & Domain of y=ln (x-2) & y=2log x [1em] x-2>0 & ⇕ & x>0 x>2 & With the domains in mind, the table can now be constructed.
| x | y=ln (x-2) | y=2log x | ||
|---|---|---|---|---|
| ln (x-2) | y | 2log x | y | |
| 1 | - | - | 2log 1 | 0 |
| 2 | - | - | 2log 2 | ≈ 0.6 |
| 3 | ln ( 3-2) | 0 | 2log 3 | ≈ 0.95 |
| 4 | ln ( 4-2) | ≈ 0.69 | 2log 4 | ≈ 1.2 |
| 5 | ln ( 5-2) | ≈ 1.1 | 2log 5 | ≈ 1.4 |
| 12 | ln ( 12-2) | ≈ 2.3 | 2log 12 | ≈ 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
expand_more The solutions to the equation are the x-coordinates of any points of intersection of the graphs.
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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.
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.
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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=- log x substitute - 0.5=- log x Next, to solve the equation, each side will be considered as a function. - 0.5= - log x ⇓ y= - 0.5 & (I) y= - log x & (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 | - log x | y |
|---|---|---|
| 0.1 | - log 0.1 | 1 |
| 0.5 | - log 0.5 | ≈ 0.3 |
| 1 | - log 1 | 0 |
| 2 | - log 2 | ≈ - 0.3 |
| 3 | - log 3 | ≈ - 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.
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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.
log_b x=log_b y ⇔ 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.
log_b x=log_b y ⇒ x=y
Let a be a real number such that a=log_b x. log_b x=log_b y → a=log_b y By using the definition of a logarithm, the logarithm equation can be written as an exponential equation. a=log_b y ⇔ b^a=y Finally, log_b x can be substituted for a and the Inverse Property of Logarithms can be used.x=y ⇒ log_b x=log_b y
The Inverse Property of Logarithms can be used to express x as b^(log_b x). x=y ⇔ b^(log_b x)=y Now, let c be a real number such that c=log_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=log_b y Finally, by its own defitnion, c is equal to log_b x. c=log_b y ⇔ log_b x=log_b y ✓
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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≠ 1, then the equation can be solved by using the Property of Equality for Logarithmic Equations. Consider the following logarithmic equation.
log_2 x +log_2 10 = log_4 x
To solve the equation, there are four steps to follow.
1
Rewrite Both Sides as a Single Logarithm With the Same Base
expand_more 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 log_2 4, the definition of a logarithm will be used. Keep in mind that 2^2= 4.
Definition:& b^c= a ⇔ log_b a= c Expression:& 2^2= 4 ⇔ log_2 4= 2
Knowing that log_2 4=2, we can continue writing both sides of the equation as a single logarithm with the same bse.
Both sides of the equation have been written as a single logarithm with base 2.
log_2 x +log_2 10 = log_4 x
log_2(m) + log_2(n)=log_2(mn)
log_2 10x = log_4 x
log_c a = log_b a/log_b c
log_2 10x = log_2 x/log_2 4
2
Use the Property of Equality for Logarithmic Equations
expand_more Next, the Property of Equality for Logarithmic Equations can be used to simplify the expressions. log_2 100x^2 = log_2 x ⇕ 100x^2=x
3
Solve the Obtained Equation
expand_more Now, the obtained equation can be solved. This equation does not involve logarithms.
Two solutions were found, x=0 and x= 1100.
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
lcx=0 & (I) 100x-1=0 & (II)
AddEqn
(II): LHS+1=RHS+1
lx=0 100x=1
DivEqn
(II): .LHS /100.=.RHS /100.
lx=0 x= 1100
4
Check the Solutions
expand_more 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= 1100 will be checked now.
A true statement was obtained. Therefore, x= 1100 is a solution to the given equation.
log_2 x +log_2 10 = log_4 x
Substitute
x= 1/100
log_2 1/100 +log_2 10? = log_4 1/100
log_c a = log_b a/log_b c
log 1100/log 2+log 10/log 2? =log 1100/log 4
UseCalc
Use a calculator
- 6.643856...+3.321928...? =- 3.321928...
AddTerms
Add terms
- 3.321928...=- 3.321928... ✓
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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.
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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.
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 ⇕ - 1x^2+ 1x+ 1=0
Next, a= - 1, b= 1, and c= 1 will be substituted in the Quadratic Formula.
Finally, the solutions can be verified by substituting them into the original equation. The solution x≈ - 0.618 will be checked first.
Logarithms are only defined for positive numbers, so x≈ - 0.618 is an extraneous solution. The solution x≈ 1.62 will be checked now.
A true statement was found this time. Therefore, the only solution to the equation is x≈ 1.62.
ln (x+1)=2ln x
b*ln(a)=ln(a^b)
ln (x+1)=ln x^2
x=- b±sqrt(b^2-4ac)/2a
SubstituteValues
Substitute values
x=- 1±sqrt(1^2-4( - 1)( 1))/2( - 1)
▼
Evaluate right-hand side
BaseOne
1^a=1
x=- 1±sqrt(1-4(- 1)(1))/2(- 1)
MultPosNeg
a(- b)=- a * b
x=- 1±sqrt(1-(- 4)(1))/- 2
IdPropMult
Identity Property of Multiplication
x=- 1±sqrt(1-(- 4))/- 2
SubNeg
a-(- b)=a+b
x=- 1±sqrt(5)/- 2
StateSol
State solutions
lcx_1= - 1+sqrt(5)- 2 & (I) x_2= - 1-sqrt(5)- 2 & (II)
(I), (II): Use a calculator
lx_1=- 0.618033... x_2=1.618033...
RoundSigDig
Round to 3 significant digit(s)
lx_1≈ - 0.618 x_2≈ 1.62
ln (x+1)=2ln x
x ≈ - 0.618
ln ( - 0.618+1)? ≈2ln - 0.618 *
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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≠1, then the equation can be solved by using the definition of a logarithm.
log_b m = n ⇔ b^n=m
Since logarithms are defined only for positive numbers, m must be a positive number. Consider an example equation.
4log_2 (x-1)+6=14
To solve this equation, four steps must be followed.
1
Isolate the Logarithmic Expression on One Side
expand_more The first step is to isolate the logarithmic expression. This can be done by using inverse operations.
2
Use the Definition of a Logarithm
expand_more Once the logarithmic expression is isolated, the definition of a logarithm can be used to eliminate the logarithm. log_2 (x-1)=2 ⇔ 2^2=x-1
3
Solve the Obtained Equation
expand_more The obtained equation can now be solved.
2^2=x-1
▼
Solve for x
x=5
4
Check the Solutions
expand_more Finally, to detect any extraneous solutions, the obtained value can be checked by substituting it into the original equation.
A true statement was obtained, so the solution is not extraneous.
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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=10dB log x/I_0 Here, I_0 is the intensity of the softest sound a healthy human ear can detect, the value of which is 10^(- 12) Wm^2. Emily wants to attend a rock concert, but she is concerned about how loud the music may be.
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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=10^(- 12) Wm^2. For simplicity, the units will not be considered until the end.
Next, to eliminate the logarithm, its definition will be used. Since the base is not stated, the formula involves a common logarithm.
10.5=log x/10^(- 12) ⇕ 10^(10.5)=x/10^(- 12)
Finally, the equation can be solved for x.
A rock concert with a decibel level is 105dB 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.
Since a true statement was obtained, x≈ 0.03 is a solution to the equation.
D=10log x/I_0
SubstituteII
I_0= 10^(- 12), D= 105
105=10log x/10^(- 12)
DivEqn
.LHS /10.=.RHS /10.
10.5=log x/10^(- 12)
10^(10.5)=x/10^(- 12)
x≈ 0.03
105=10log x/10^(- 12)
x ≈ 0.03
105? ≈10log 0.03/10^(- 12)
UseCalc
Use a calculator
105≈ 104.771212... ✓
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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.
c|c Ifb<1 & Ifb>1 [1em] log_b x_1≥ log_b x_2 & log_b x_1≥ log_b x_2 ⇕ & ⇕ x_1≤ x_2 & x_1≥ x_2
Proof
The statements will be proved one at a time.
b>1
If b is greater than 1, the logarithmic function f(x) = log_b x is increasing over 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. log_b x_1≥ log_b x_2 ⇔ x_1≥ x_2
b<1
If b is greater than 0 and less than 1, then f(x) = log_b x is decreasing over 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. log_b x_1≥ log_b x_2 ⇔ x_1≤ x_2
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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.
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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.
10^1=10 ⇔ 1=log 10
With this information, log 10 can be substituted for 1. Then, the Quotient Property of Logarithms can be used.
log (x-3)
Substitute
1= log 10
log (x-3)inverse operations.
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.
log(m) - log(n)=log(m/n)
log (x-3)
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.
log (x-3)x-3 < x+6/10
x<4
| log (x-3) | |
|---|---|
| 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:& x<4 Inequality II:& 3< x Inequality III:& - 6
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
Closure
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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.
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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) ⇕ log 2^(x-1) = log 3^(x+1)
Now use the Power Property of Logarithms. Then the variable x can be isolated.
The solution to the exponential equation is x≈ - 4.42. This can verified by substituting it into the original equation.
log 2^(x-1) = log 3^(x+1)
log(a^m)= m*log(a)
(x-1)log 2 = (x+1)log 3
Distr
Distribute log 2 & log 3
xlog 2-log 2 = xlog 3+log 3
AddEqn
LHS+log 2=RHS+log 2
xlog 2= xlog 3+log 3+log 2
SubEqn
LHS-xlog 3=RHS-xlog 3
xlog 2- xlog 3=log 3+log 2
FactorOut
Factor out x
x(log 2- log 3)=log 3+log 2
DivEqn
.LHS /(log 2- log 3).=.RHS /(log 2- log 3).
x=log 3+log 2/log 2- log 3
UseCalc
Use a calculator
x=- 4.419022...
RoundDec
Round to 2 decimal place(s)
x≈ - 4.42
2^(x-1)=3^(x+1)
x ≈ - 4.42
2^(- 4.42-1)? ≈3^(- 4.42+1)
AddSubTerms
Add and subtract terms
2^(- 5.42)? ≈3^(- 3.42)
CalcPow
Calculate power
0.023357... ≈ 0.023347... ✓
Solving Logarithmic Equations and Inequalities
Exercise 2.1
Complete the statement.
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If ln (x+2) = ln 10, then x+2 = . |
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The Property of Equality for Logarithmic Equations states that two logarithmic expressions with the same base are equal if and only if their arguments are equal. log_b a = log_b c ⇔ a=c In our logarithmic equation, we have two natural logarithms that are equal to each other. Since both sides have the same base e, the arguments are equal. ln (x+2)=ln 10 ⇔ x+2=10 We can now complete the statement.
If ln (x+2) = ln 10, then x+2 = 10.
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Complete the statement with =, <, or >.
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If 5^(2x) < 5^(5.5), then 2x 5.5. |
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Recall the Exponential Property of Inequality. c|c Ifb<1 & Ifb>1 [1em] b^(x_1)≥ b^(x_2) & b^(x_1)≥ b^(x_2) ⇕ & ⇕ x_1≤ x_2 & x_1≥ x_2 In our case, the base of both sides of the inequality is 5, which is greater than 1. 5^(2x) < 5^(5.5) ⇔ 2x<5.5 Note that we are dealing with a biconditional statement. We can use this information to complete our statement.
If 5^(2x) < 5^(5.5), then 2x < 5.5.
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Complete the statement with =, <, or >.
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If 0.2^(3x) < 0.2^4, then 3x 4. |
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":" "},"formTextBefore":"3x","formTextAfter":"4","answer":{"text":[">"]}}
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Recall the Exponential Property of Inequality. c|c Ifb<1 & Ifb>1 [1em] b^(x_1)≥ b^(x_2) & b^(x_1)≥ b^(x_2) ⇕ & ⇕ x_1≤ x_2 & x_1≥ x_2 In our case, the base of both sides of the inequality is 0.2, which is less than 1. 0.2^(3x) < 0.2^4 ⇔ 3x>4 Note that we have a biconditional statement. We can use this information to complete our statement.
If 0.2^(3x) < 0.2^4, then 3x > 4.
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Complete the statement with =, <, or >.
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If log_7 (3x+2) > log_7 50, then 3x+2 50. |
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Recall the Property of Inequality for Logarithms. c|c Ifb<1 & Ifb>1 [1em] log_b x_1≥ log_b x_2 & log_b x_1≥ log_b x_2 ⇕ & ⇕ x_1≤ x_2 & x_1≥ x_2 In our case, the base of both sides of the inequality is 7, which is greater than 1. log_7 (3x+2) > log_7 50 ⇔ 3x+2>50 Note that we have a biconditional statement. We can use this information to complete our statement.
If log_7 (3x+2) > log_7 50, then 3x+2 > 50.
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Complete the statement with =, <, or >.
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If log_(14) (5-x) > log_(14) 10, then 5-x 10. |
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Recall the Property of Inequality for Logarithms. c|c Ifb<1 & Ifb>1 [1em] log_b x_1≥ log_b x_2 & log_b x_1≥ log_b x_2 ⇕ & ⇕ x_1≤ x_2 & x_1≥ x_2 In our case, the base of both sides of the inequality is 14, which is less than 1. log_(14) (5-x) > log_(14) 10 ⇔ 5-x<10 Note that we have a biconditional statement. We can use this information to complete our statement.
If log_(14) (5-x) > log_(14) 10, then 5-x < 10.
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Solving Logarithmic Equations and Inequalities
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