§ 9.5 Exponential and Logarithmic Equations. Blitzer, Intermediate Algebra, 5e – Slide #2 Section...
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Transcript of § 9.5 Exponential and Logarithmic Equations. Blitzer, Intermediate Algebra, 5e – Slide #2 Section...
§ 9.5
Exponential and Logarithmic Equations
Blitzer, Intermediate Algebra, 5e – Slide #2 Section 9.5
Exponential Equations
At age 20, you inherit $30,000. You’d like to put aside $25,000 and eventually have over half a million dollars for early retirement. Is this possible? In this section, you will see how techniques for solving equations with variable exponents provide an answer to this question.
One technique involves using logarithms to solve the exponentialequations. Remember that you are solving for x and that the x in these equations is in the exponent. So this is different than any equation we have looked at before.
And for the answer to the question on when the $25,000 would grow to $500,000 –If the money was invested at 9% compounded monthly, it would grow to the half million dollars you wanted for your retirement in 33.4 years.
Blitzer, Intermediate Algebra, 5e – Slide #3 Section 9.5
Exponential Equations
Solving Exponential Equations by Expressing Each Side as a
Power of the Same Base. then ,1 and If NMbbb NM
Express each side as a power of the same base.
Set the exponents equal to each other.
Blitzer, Intermediate Algebra, 5e – Slide #4 Section 9.5
Exponential Equations
EXAMPLEEXAMPLE
Solve:
SOLUTIONSOLUTION
.77b273a 6
212
xx
This is the given equation.
In each equation, express both sides as a power of the same base. Then set the exponents equal to each other.(a) Because 27 is , we express each side of the equation in terms of the base, 3.
123 x
273 12 x
Write each side as a power of the same base.
312 33 x
Equate the exponents.312 x
Subtract 1 from both sides.22 x
Blitzer, Intermediate Algebra, 5e – Slide #5 Section 9.5
Exponential Equations
Divide both sides by 2.1x
CONTINUECONTINUEDD
Check 1:273 12 x
273 112
2733 2727
?
?
This is the given equation.
(b) Because can both be expressed as 7 raised to a power, we will write each side of the equation in terms of 7.
77 6
2
x
7 and 7 6
2x
The solution is 1 and the solution set is {1}.true
Blitzer, Intermediate Algebra, 5e – Slide #6 Section 9.5
Exponential Equations
CONTINUECONTINUEDD Write each side as a power of
the same base.2
1
6
2
77 x
Equate the bases.2
1
6
2
x
Multiply both sides by the LCD, 6.
32 x
Add 2 to both sides.5x
Check 5:
77 6
2
x
77 6
25
?
Blitzer, Intermediate Algebra, 5e – Slide #7 Section 9.5
Exponential Equations
CONTINUECONTINUEDD
77 6
3
?
77 2
1
?
77
The solution is 5 and the solution set is {5}.
true
Blitzer, Intermediate Algebra, 5e – Slide #8 Section 9.5
Exponential Equations
Using Natural Logarithms to Solve Exponential Equations
1) Isolate the exponential expression.
2) Take the natural logarithm on both sides of the equation.
3) Simplify using one of the following properties:
4) Solve for the variable.
.lnor lnln xebxb xx
Blitzer, Intermediate Algebra, 5e – Slide #9 Section 9.5
Exponential Equations
EXAMPLEEXAMPLE
Solve:
SOLUTIONSOLUTION
.62501250 065.0 xe
This is the given equation.
We begin by dividing both sides by 1250 to isolate the exponential expression, . Then we take the natural logarithm on both sides of the equation.
xe 065.0
62501250 065.0 xe
Isolate the exponential factor by dividing both sides by 1250.
5065.0 xe
Take the natural logarithm on both sides.
5lnln 065.0 xe
Blitzer, Intermediate Algebra, 5e – Slide #10 Section 9.5
Exponential Equations
Use the inverse property5ln065.0 x
CONTINUECONTINUEDD
on the left side.xex ln
Divide both sides by 0.065.76.24065.0
5lnx
Thus, the solution of the equation is . Try checking
this approximate solution in the original equation, verifying that
is the solution set.
76.24065.0
5ln
76.24
065.0
5ln
Blitzer, Intermediate Algebra, 5e – Slide #11 Section 9.5
Logarithmic Equations
EXAMPLEEXAMPLE
Solve:
SOLUTIONSOLUTION
.514log2 x
This is the equivalent equation.
We first rewrite the equation as an equivalent equation in exponential form using the fact that . means log xbcx c
b
.142 means 514log 52 xx
Now we solve the equivalent equation for x.
1425 xEvaluate the exponent.1432 x
Subtract 1 from both sides.x431
Divide both sides by 4.x4
31
Blitzer, Intermediate Algebra, 5e – Slide #12 Section 9.5
Logarithmic Equations
This is the given logarithmic equation.
CONTINUECONTINUEDD
Check 31/4:
514log2 x
Replace x with 31/4.514
314log2
Multiply. 5131log2
Add. 532log2
55 .322 because 532log 52
?
?
?
true
This true statement indicates that the solution is 31/4 and the solution set is {31/4}.
Blitzer, Intermediate Algebra, 5e – Slide #13 Section 9.5
Logarithmic Equations
EXAMPLEEXAMPLE
Solve:
SOLUTIONSOLUTION
.2log12log(b)31log1log(a) 22 xxxx
This is the given equation. 31log1log(a) 22 xx
Use the product rule to obtain a single logarithm.
311log2 xx
Multiply. 31log 22 x
.12 means 31log 2322 xx12 23 x
18 2 x Evaluate the exponent.
90 2 x Subtract 8 from both sides. 033 xx Factor.
Blitzer, Intermediate Algebra, 5e – Slide #14 Section 9.5
Logarithmic Equations
03or 03 xx Set each factor equal to 0.
CONTINUECONTINUEDD
3x Solve for x.3x
Check 3: Check -3:
31log1log 22 xx 31log1log 22 xx
313log13log 22 313log13log 22
34log2log 22 32log4log 22
Negative numbers do not have logarithms.
false
342log2
38log2
823
??
?
?
?
?
Blitzer, Intermediate Algebra, 5e – Slide #15 Section 9.5
Logarithmic Equations
CONTINUECONTINUEDD 88 true
The solutions are 3 and -3. The solution set is {3, -3}.
2log12log(b) xx This is the given equation.
212
log
x
x Use the quotient rule to obtain a single logarithm.
.12
10 means 212
log 2
x
x
x
x
x
x 12102
x
x 12100
Evaluate the exponent.
12100 xx Multiply both sides by the LCD, x.
Blitzer, Intermediate Algebra, 5e – Slide #16 Section 9.5
Logarithmic Equations
CONTINUECONTINUEDD
198 x Subtract 2x from both sides.
98
1x Divide both sides by 98.
Check -1/98:
2log12log xx
298
1log1
98
12log
Negative numbers do not have logarithms. Therefore there is no solution and the solution set is .
false
Blitzer, Intermediate Algebra, 5e – Slide #17 Section 9.5
Logarithmic Equations
EXAMPLEEXAMPLE
Solve:
SOLUTIONSOLUTION
.6ln37 x
This is the given equation.6ln37 x
Subtract 7 from both sides.1ln3 x
Divide both sides by 3.3
1ln x
Exponentiate both sides.3
1ln
ee x
Simplify the left side.3
1
ex
Evaluate the exponent.72.0x
Blitzer, Intermediate Algebra, 5e – Slide #18 Section 9.5
Logarithmic Equations
Check :
CONTINUECONTINUEDD 3
1
e
This is the given equation.6ln37 x
Replace x with6ln37 3
1
e .3
1
e
Because , we conclude63
137
xex ln
.3
1ln 3
1
e
Multiply.617
Subtract.66
This true statement indicates that the solution is and the
solution set is
3
1
e
.3
1
e
?
?
?
true
Blitzer, Intermediate Algebra, 5e – Slide #19 Section 9.5
Exponential Equations in Application
EXAMPLEEXAMPLE
The formula models the population of Florida, A, in millions, t years after 2000. When will the population of Florida reach 19.2 million?
SOLUTIONSOLUTION
This is the given equation.
teA 0235.09.15
We will replace A with 19.2 and then solve for t.teA 0235.09.15
Replace A with 19.2.te 0235.09.152.19
Divide both sides by 15.9.te 0235.02.1
Take the natural logarithm of both sides.
te 0235.0ln2.1ln
Blitzer, Intermediate Algebra, 5e – Slide #20 Section 9.5
Exponential Equations in Application
Use the inverse property on the right side.
t0235.02.1ln
CONTINUECONTINUEDD
xex ln
Divide both sides by 0.0235.76.70235.0
2.1lnt
The state of Florida will have a population of 19.2 million people approximately 7.76 years after the year 2000.