5. Solving Exponential Equations
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Chapter 6
5.
Solving Exponential Equations
Solving exponential equations requires a deep understanding of mathematical principles. The lesson delves into the methods and techniques to tackle these equations, both graphically and algebraically. One of the key strategies is to rewrite the equations so that both sides have a common base, allowing for easier comparison and solution. The Property of Equality for Exponential Equations plays a pivotal role, stating that if two exponential expressions with the same base are equal, their exponents must also be equal. This principle is foundational in solving such equations. The lesson also presents various examples and scenarios to illustrate the application of these methods, aiding learners in grasping the concepts effectively.
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| | 11 Theory slides |
| | 7 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Solving Exponential Equations
Slide of 11
This lesson will define and discuss equations that involve exponential expressions.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
Challenge
When Will the Cities Have the Same Population?
Dominika has a meeting with her guidance counselor on Monday afternoon to discuss her college plans. She is considering where she wants to apply in a few years. Dominika knows that she wants to study in a large city, but not one that is too big. She has two cities in mind. To decide between them, she is paying close attention to their populations, which are given by two exponential functions.
Here, x is the number of years that have passed since the year 2000. Furthermore, f and g are the populations of each city in millions of people after x years. If they keep growing like this, in what year will the populations be the same? Approximate the answer to the nearest century.
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Discussion
Exponential Equations
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 | 2x=32 |
| With the Same Variable on Both Sides | 22x=5⋅2x |
| With the Same Base | 43x=42x+3 |
| With Unlike Bases | 3x+4=81x |
| With a Rational Base | (21)x=8 |
An exponential equation can be solved graphically.
Method
Solving Exponential Equations Graphically
An example exponential equation will be considered.
The x-coordinate of the point of intersection is 1. Therefore, the solution to the equation is x=1. This can be verified by substituting 1 for x in the given equation and checking whether a true statement is obtained.
Since a true statement was obtained, x=1 is a solution to the equation. Note that if the point of intersection is not a lattice point, the exact solution may not be easy to find using this method.
2x+1=4x
There are three steps to follow to solve exponential equations 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 exponential functions.
2x+1=4x ⇒ y=2x+1y=4x(I)(II)
The idea is that each function should be relatively easy to graph. In this case, the graph of the first function is a translation one unit to the left of the graph of its parent function y=2x. The second function is a parent function itself and its graph does not undergo any transformations. 2
Graph the Functions
expand_more 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.
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. Since these graphs intersect at one point, the equation has one solution.
Example
Savings Account
Dominika's first class on Mondays is economics and personal planning. She is told that a certain savings account earns 6% annual interest compounded yearly.
500(1.06)x=800
Help Dominika solve this equation! Round the answer to the nearest integer. {"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"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":["8"]}}
Solution
The given exponential equation will be solved graphically. To do this, each side will be considered as a function.
Fortunately, the answer should be rounded to the nearest integer. Therefore, the solution to the equation is x≈8. This solution can be checked by substituting the value into the given equation.
Since a true statement was obtained, it is confirmed that the solution to the equation is x≈8. This means that Dominika will have $800 in her account in about 8 years.
500(1.06)x=800⇓y=500(1.06)xy=800(I)(II)
The first is an exponential function and the second a constant function. They will both be graphed on the same coordinate plane. The graph of the constant function is a horizontal line whose y-intercept is 800. To graph the exponential function, the initial value 500 and the constant multiplier 1.06 will be used. The graphs intersect at one point, so there is only one solution to the equation.
The x-coordinate of the point of intersection appears to be 8. However, looking closely at the graph, it can be seen that the x-coordinate of this point is a bit greater than 8.
500(1.06)x=800
x≈8
500(1.06)8≈?800
796.924037…≈800 ✓
Example
More Than One Solution
Dominika's second class on Mondays is math. As soon as she enters the classroom, she sees the following equation written on the board.
Approximate the second solution to the nearest integer.
2(3)x=35x+37
The teacher says that the equation has two solutions and that when solved by graphing, the first solution can be exactly determined but that the second can only be approximated. Write the exact solution for Dominika. {"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"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":["-1"]}}
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Hint
Consider each side of the equation as a function. Then graph the functions and find their point of intersection.
Solution
The equation will be solved graphically. To do so, each side of the equation will be considered as a function.
It is seen that x=-1 is an exact solution. Conversely, the exact value of the second solution cannot be determined from the graph. However, approximated to the nearest integer, this solution is x≈0. These solutions can be verified by substituting into the given equation. First, x=-1 will be checked.
By following the same procedure, the solution x≈0 can be verified.
2(3)x=35x+37⇓y=2(3)xy=35x+37(I)(II)
The first function is an exponential function and the second one is a linear function. Both can be graphed on the same coordinate plane. The slope and y-intercept of the linear function will be used to graph it. To graph the exponential function, its initial value and constant multiplier will be used. The graphs intersect at two points. Therefore, the equation has two solutions, which are the x-coordinates of these points of intersection.
2(3)x=35x+37
Substitute
x=-1
2(3)-1=?35(-1)+37
▼
Evaluate
NegExponentToFrac
a-m=am1
2(311)=?35(-1)+37
ExponentOne
a1=a
2(31)=?35(-1)+37
Multiply
Multiply
32=?-35+37
AddFrac
Add fractions
32=32 ✓
| Solution | Substitute | Simplify |
|---|---|---|
| x=-1 | 2(3)-1=?35(-1)+37 | 32=32 ✓ |
| x≈0 | 2(3)0≈?35(0)+37 | 2≈37⇕2≈2.333333…✓
|
Since true statements were obtained, x=-1 and x≈0 are solutions to the equation.
Example
Soccer or Football?
Dominika has physical education right before lunch. She and her teacher are both football and soccer fans. They know that, starting from 2020, the attendance to the Major League Soccer's final game and the Super Bowl can be modeled by exponential functions.
Dominika and her teacher want to find the year in which the attendance will be the same for both events. To do so, they need to solve an exponential equation by graphing. Help them find the year in which the attendance will be the same!
MLS Final Gamey=40000(1.25)xSuper Bowly=120000(0.87)x
In both cases, x is the number of years that have passed since 2020. {"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"Year:","formTextAfter":null,"answer":{"text":["2023"]}}
Hint
Graph y=40000(1.25)x and y=120000(0.87)x on the same coordinate plane. What is the x-coordinate of the point of intersection? What does it mean in this context?
Solution
To find in which year the attendance to both events will be the same, Dominika needs to solve the exponential equation formed by the right-hand sides of the given functions.
40000(1.25)x=120000(0.87)x
To solve the equation, two exponential functions will be considered.
40000(1.25)x=120000(0.87)x⇓y=40000(1.25)xy=120000(0.87)x(I)(II)
These two functions will be graphed on the same coordinate plane. Their initial values and constant multipliers will be used to graph the functions. The graphs intersect at one point. Although it can be seen that the x-coordinate of the point of intersection is a bit greater than 3, its exact value cannot be determined by graphing.
Since x is the number of years that have passed since the year 2020, it can be stated that the attendance to both events will be roughly the same in 2020+3=2023.
Discussion
Solving Exponential Equations Algebraically
Before discussing how to solve exponential equations algebraically, an important property must be learned.
Rule
Property of Equality for Exponential Equations
Two powers with the same positive base b, where b=1, are equal if and only if their exponents are equal.
If b>0 and b=1, then bx=by if and only if x=y.
Proof
This property will be proven in two parts.
If bx=by, Then x=y
It is known that b is a positive number other than 1. This implies, among other things, that b is not zero. Consequently, by is never equal to zero and both sides of the equation can therefore be divided by by. Then, the Quotient of Powers Property can be used. If a power with base b=1 is equal to one, then the exponent is zero.bx−y=1⇒x−y=0
The equation obtained means that x and y are equal.
x−y=0⇔x=y
It has been shown that if bx=by, then x=y. Note that if b=1, the first implication is not valid because 1 raised to any power equals 1. In such a case, x−y would not be necessarily 0. This is why b must be a number other than 1! If x=y, Then bx=by
Suppose for a moment that b=0. Now, raise b to a negative exponent -n, where n is a natural number.b-n→0-n
Next, simplify the negative exponent and recall that 0n=0 for any natural number n.
0-n=0n1=01
Since division by zero is not defined, the expression 01 is not defined. This means that if b=0, then it cannot be raised to a negative exponent. However, since in this case b>0, it can be raised to any exponent x and the expression bx will always be well defined.
bx is well defined for b>0
Now, by the Symmetric Property of Equality, write b=b. Then, use the above information to raise both sides of this equation to the power of x. Finally, use the fact that x and y are equal.
It has been shown that if x=y, then bx=by. Therefore, the biconditional statement has been proven. If b>0 and b=1, then bx=by if and only if x=y.
With this property in mind, a method for solving exponential equations algebraically can be explained.
Method
Solving Exponential Equations Algebraically
Let b be a positive number other than 1 and a(x) and c(x) be two algebraic expressions in terms of the same variable. If an exponential equation is or can be written in the following form, then it can be solved algebraically by using the Property of Equality for Exponential Equations.
ba(x)=bc(x)
42x=4096
To solve the equation, four steps must be followed.
1
Rewrite the Expressions on Both Sides as Powers With the Same Base
expand_more First, the exponential expressions on both sides must be rewritten as powers with the same base. In this case, 4096 can be written as a power of 4.
42x=4096 ⇔ 42x=46
If the exponential expressions on both sides of the equation already have the same base, this step can be skipped. 2
Use the Property of Equality for Exponential Equations
expand_more Next, the Property of Equality for Exponential Equations can be used. Since the bases are equal, the exponents must be equal for the equation to be true.
42x=46 ⇔ 2x=6
3
Solve the Resulting Equation
expand_more The resulting equation can be solved for the variable.
2x=6 ⇔ x=3
4
Verify the Solution
expand_more Finally, the obtained solution can be verified by substituting it into the given equation.
Since a true statement was obtained, x=3 is a solution to the equation. It is important to verify all the obtained solutions, since sometimes this method can lead to extraneous solutions.
Example
Extra Credit Conundrums
Dominika decides to make good use of her free period after lunch to do some extra credit math problems.
Unfortunately, she is struggling with solving three exponential equations. Help her understand how to solve the equations algebraically to obtain the extra credit she needs!
a 33x=3x+1
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b 5x=252x+1
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c 2(2)x−3=(81)31x
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Hint
a The exponential expressions on both sides of the equation already have the same base. Therefore, the Property of Equality for Exponential Equations can be used.
b Rewrite the expression on the right-hand side as a power with base 5. Then, use the Property of Equality for Exponential Equations.
c Rewrite the expressions on both sides as single powers with base 2. Then, use the Property of Equality for Exponential Equations.
Solution
a Since the exponential expressions on both sides of the equation already have the same base, the Property of Equality for Exponential Equations can be applied.
33x=3x+1 ⇔ 3x=x+1
Next, the obtained equation can be solved for x.
The solution will now be verified by substituting 21 for x in the given equation.
33x=3x+1
Substitute
x=21
33(21)=?321+1
▼
Evaluate
MoveLeftFacToNumOne
a⋅b1=ba
323=?321+1
OneToFrac
Rewrite 1 as 22
323=?321+22
AddFrac
Add fractions
323=?323
CalcPow
Calculate power
5.196152…=5.196152… ✓
b The right-hand side of this equation must be written as a power of 5.
5x=252x+1 ⇔ 5x=54x+2
Therefore, the Property of Equality for Exponential Equations will be used.
5x=54x+2 ⇔ x=4x+2
The value of x can be found by solving the equation.
x=4x+2
▼
Solve for x
SubEqn
LHS−4x=RHS−4x
-3x=2
DivEqn
LHS/(-3)=RHS/(-3)
x=-32
MoveNegDenomToFrac
Put minus sign in front of fraction
x=-32
5x=252x+1
Substitute
x=-32
5-32=?252(-32)+1
▼
Evaluate
MultPosNeg
a(-b)=-a⋅b
5-32=?25-2(32)+1
MoveLeftFacToNum
a⋅cb=ca⋅b
5-32=?25-34+1
OneToFrac
Rewrite 1 as 33
5-32=?25-34+33
AddFrac
Add fractions
5-32=?25-31
CalcPow
Calculate power
0.341995…=0.341995… ✓
c In this equation, the expressions on both sides will be written as powers of 2. To do so, some Properties of Exponents will be used. The left-hand side will be rewritten first.
(81)31x
▼
Rewrite
WritePow
Write as a power
(231)31x
FracToNegExponent
am1=a-m
(2-3)31x
PowPow
(am)n=am⋅n
2-3(31x)
DenomMultFracToNumber
3⋅3a=a
2-1x
MultNegPos
(-a)b=-ab
2-x
2(2)x−3=(81)31x ⇔ 2x−2=2-x
Therefore, the Property of Equality for Exponential Equations can be used.
2x−2=2-x ⇔ x−2=-x
Finally, the obtained equation can be solved for x.
Now, x=1 will be checked by substitution.
2(2)x−3=(81)31x
Substitute
x=1
2(2)1−3=?(81)31(1)
▼
Evaluate
SubTerm
Subtract term
2(2)-2=?(81)31(1)
IdPropMult
Identity Property of Multiplication
2(2)-2=?(81)31
CalcPow
Calculate power
2(41)=?21
MoveLeftFacToNumOne
a⋅b1=ba
42=?21
ReduceFrac
ba=b/2a/2
21=21 ✓
Pop Quiz
Solving Exponential Equations
Solve the exponential equations graphically or algebraically. Whenever necessary, round the answers to two decimal places.
Example
Estimating Bacterial Growth
Dominika finishes her Mondays with biology class. She learns that bacteria have the ability to multiply at incredible rates. After studying two bacteria populations, she concludes that their growth can be modeled by two exponential functions.
10(1.96)x<14(1.4)x
Help Dominika solve the inequality! {"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"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:0.5782em;vertical-align:-0.0391em;\"><\/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":["1"]}}
Hint
Rewrite the inequality so that each side is an exponential expression with the same base.
Solution
Exponential inequalities are solved the same way as exponential equations. To start, the inequality will be rewritten so that each side is an exponential expression with the same base.
Now that both sides of the inequality are written as exponential expressions with the same base, the exponents can be compared.
10(1.96)x<14(1.4)x
▼
Rewrite
DivIneq
LHS/10<RHS/10
1.96x<1.4(1.4)x
WritePow
Write as a power
(1.42)x<1.4(1.4)x
PowPow
(am)n=am⋅n
1.42x<1.4(1.4)x
MultBasePow
a⋅am=a1+m
1.42x<1.41+x
1.42x<1.41+x ⇔ 2x<1+x
Finally, the obtained inequality can be solved for x.
2x<1+x ⇔ x<1
It has been found that population (I) will be less than population (II) only for one hour. This can be verified by graphing each side of the inequality as an individual exponential function. It is shown that the graph of y=10(1.96)x is below the graph of y=14(1.4)x for x<1, even before the first observation was made at x=0. Therefore, the solution to the inequality is, indeed, x<1.
Closure
Finding the Year in Which the Cities Will Have the Same Population
With the topics seen in this lesson, the challenge presented at the beginning can be solved. Dominika has a meeting with her guidance counselor on Monday afternoon to discuss her college plans. Dominika wants to study in a smaller city. She knows that the populations of two cities in the US are modeled by two exponential functions.
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Hint
Solve the exponential equation 8(1.0048)x=3.5(1.0067)x.
Solution
To find the year in which the populations will be the same, an exponential equation must be solved.
8(1.0048)x=3.5(1.0067)x
Since the expressions on each side cannot be written as exponential expressions with the same base, the equation will be solved graphically. To do so, both functions will be graphed on the same coordinate plane by using their initial values and constant multipliers. The x-coordinate of the point of intersection cannot be determined precisely on the graph. However, it does show that, to the nearest hundred years, the x-coordinate is 400. This means that the cities will have the same population around the year 2400, many years after Dominika has finished college.
Since Dominika wants to study in a smaller city, she will talk to her counselor about the city with the smaller population during the period of time when she will be in college. Based on the graph, this means she will apply to a college in City B.
Solve the following exponential equations.
52x+1=59
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3x−2=33
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Since the bases on both sides of the equation are the same, we can use the Property of Equality for Exponential Equations to solve the given exponential equation. This property says that if the bases are the same, the exponents must also be equal for the equation to be true. 5^(2x+1) = 5^9 ⇔ 2x+1 = 9 Let's solve this linear equation created from the exponents.
To check our answer, we can substitute 4 for x in the given equation and simplify.
Since substituting 4 for x in the given equation produces a true statement, x=4 is the solution to our equation.
This exponential equation has the same base in both sides of the equation as well. Because of this, we know that the exponents must also be equal.
3^(x-2)=3^3 ⇔ x-2=3
Let's solve this linear equation!
x-2=3 ⇔ x=5
We can substitute 5 for x in the equation in order to check our answer.
Since substituting 5 for x in the given equation produces a true statement, x=5 is the solution to our equation.
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Solve the following exponential equations.
2x+1=8
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42x−1=64
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To solve the given exponential equation, we will begin by rewriting the equation so that both sides have the same base. 2^(x+1)=8 ⇔ 2^(x+1)=2^3 We now have two equivalent expressions with the same base. By the Property of Equality for Exponential Equations, the equation holds true if and only if the exponents are equal. 2^(x+1) = 2^3 ⇔ x+1 = 3 Let's solve the linear equation created by the exponents. x+1 = 3 ⇔ x=2 To check our answer, we will substitute x for 2 in the given equation.
Since substituting 2 for x in the given equation produces a true statement, x=2 is the solution to our equation.
As we did in Part A, let's rewrite the right-hand side of the equation so that both sides have the same base.
4^(2x-1)=64 ⇔ 4^(2x-1)=4^3
Now both sides of the equation have the same base. By the Property of Equality for Exponential Equations, the equation holds true if and only if the exponents are equal.
4^(2x-1)=4^3 ⇔ 2x-1=3
Let's solve this linear equation!
Finally, we will substitute 2 for x in the equation in order to check our answer.
Since substituting 2 for x in the given equation produces a true statement, x=2 is the solution to our equation.
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Consider the following equation.
Select one of the following.
3x=23
Which of the following graphs can be used to solve this equation?
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We want to match the exponential equation 3^x= 32 with one of the given graphs. Let's start by finding the y-intercept of the function y=3^x. We can do this by substituting 0 for x and evaluating the right-hand side.
This tells us that the y-intercept for one of the lines occurs at (0,1). The only graphs with this point are A and C.
This means that we can disregard choices B and D. Next, we can think of the right-hand side of the equation as the function y= 32. Note that this equation represents a horizontal line at y= 32.
The graph that contains both pieces of information required to solve the equation is graph in choice A. We did not even need to solve the equation in order to find this answer!
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Solve the following exponential equations.
15⋅52x+3=75
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935−x=1
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We will start by rewriting the exponential equation so that both sides have a common base.
Now we have two equivalent expressions with the same base. By the Property of Equality for Exponential Equations, the equation holds true if and only if the exponents are the same. 5^(2x+3) = 5^1 ⇔ 2x+3 = 1 Let's solve the linear equation created by the exponents.
Therefore, the answer to the given equation is x=-1.
Just like we did in Part A, we will rewrite the equation so that both sides have the same base.
By the Property of Equality for Exponential Equations, the above equation holds true if and only if the exponents are equal. 3^(5-x) = 3^2 ⇔ 5-x = 2 Let's solve this equation for x.
The answer to the given equation is x=3.
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Solving Exponential Equations
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