3.5 Exponential Equations, Logarithmic Equations, and Problem Solving 1 If b > 0 and b 1, and m and...

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3.5 Exponential Equations, Logarithmic Equations, and Problem Solving

xPreviously, we solved equations where the exponents contained the variable such as: 3 81

If b > 0 and b 1, and m and n are real numbers, then bn = bm if and only if n = m.

Property 1.

By using the property below, we are able to solve this equation by matching the bases.

x3 81x 43 3

Answer: x 4x

This method worked fine for this example because we couldmatch the bases. We will now focus on equations where thebases can not be matched. Example: 3 12

Fortunately, we have another property to take care of this type of equation.

Property of Logarithms (taking the log of both sides) .

This is often referred to as “taking the log of both sides”. Also, we will use the common logarithm for the base. We can than perform the calculations using our calculator (i.e., b = 10)

If x > 0, y > 0, b > 0, and b 1, then x = y if and only if logbx= logby.

Procedure: To solve equations of the form ax = b1. Take the log of both sides. (Just write log in front of both sides of the equation.)

2. Use the Power Rule of Logarithms to bring the exponent in front of the logarithm.

3. Solve the equation.

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x Solve 3 12 to the nearest Exampl hundree . .1 dth

1. Take the log of both sides.xlog 3 log 12

xlog 3 log 12 2. Bring the exponent in front of the logarithm.

3. Solve the equation by dividing by log 3 on both sides.

log 12x

log 3

x = 2.26 to the nearest hundredth

Check:

2.26

x

Because 3 11.975511, we say that, to the nearest hundredth, the solution set for 3 12 is 2.26 .

The check for every problem will not be shown. It is recommended to do the check for the problems.

Your Turn Problem #1

xSolve 2 14 to the nearest hundredth.

Answer: x 3.81

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a. Use parentheses around the numerator. Then divide by the denominator. The TI-30XIIS will automatically place parentheses after log. The casio fx-300ES will not. (log(15)–log(2))/log(3) using the TI-30XIIS.

(log15 – log2)/log3 using Casios or graphing calculators. You should get 1.8340…

Calculating Logarithmic ExpressionsThere are different methods for calculating these expressions. Hopefully, you have a calculator where the screen will show the operation you want to perform. A calculator such as the TI-30xa will not do this. The TI-30XIIS ($15) or the casio fx-300ES will be easier to use. Any graphing calculator would be great, but I don’t like to recommend spending $100. Examples:

log15 log2a.

log3

log8 log5b.

log3 log2

b. (log(8)–log(5))/(log(3)+log(2)). (log8–log5)/(log3+log2). You should get 0.26231…

2log3 log5c.

4log2 log3

c. (2log(3)+log(5))/(4log(2) – log(3)). (2log3+log5)/(4log2 – log3). You should get 2.274023…

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x 3 Solve 5 42 to the nearestExamp hundredth.le 2.

1. Take the log of both sides.x 3log 5 log 42

(x 3) log 5 log 42 2. Bring the exponent in front of the logarithm.

3. Solve the equation. This equation can be solved by distributing the log5 or dividing by log5 on both sides.

xlog5 3log5 log42

xlog5 log42 3log5

log42 3log5x

log5

x = 5.32 The solution set is {5.32}.

Note: This equation could have been solved by dividing by log 5 first.

(x 3) log 5 log 42

log42x 3

log5

log42x 3

log5

x = 5.32

Your Turn Problem #2x 2Solve 3 92 to the nearest hundredth.

Answer: 6.12

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3x 4 Solve 2 31 to the nearesExamp t hunle 3 d h.. redt

1. Take the log of both sides.3x 4log2 log31

(3x 4) log2 log31 2. Bring the exponent in front of the logarithm.

3. Solve the equation. This equation can be solved by distributing the log2 or dividing by log2 on both sides.

3xlog2 4log2 log31

3xlog2 log31 4log2

log31 4log2x

3log2


Note: This equation could have been solved by dividing by log2 first. (3x 4) log2 log31

log313x 4

log2

log31 4

log2x = 2.98

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Your Turn Problem #32x 3Solve 5 45 to the nearest hundredth.

Answer: 0.32

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3x 1 x 2Solve 2 7 to the nearEx esampl t hue 4. ndre . h dt

1. Take the log of both sides.3x 1 x 2log2 log7

(3x 1) log2 (x 2) log7 2. Bring the exponent in front of the

logarithms.

3. Solve the equation. This equation can be solved by distributing the log2 and

log7.

3xlog2 log2 xlog7 2log7

2log7 log2x

3log2 log7

x = 34.34 to the nearest hundredth

The solution set is {34.34}.

3xlog2 xlog7 2log7 log2

Get the x terms on one side.

x(3log2 log7) 2log7 log2

Factor out the x.

Your Turn Problem #4x 1 2x 1Solve 3 2 to the nearest hundredth.

Answer: 6.23

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Also, ln e = 1

Recall that the natural logarithm has a base of e.

eln e = xReason:xe = e

Therefore x = 1.

The next example will be of the form ex = b. Therefore, instead of writing log in front of both sides, write ln in front of both sides.

x 2Solve e 21 to the neareEx stample hundr5 e. dth.

1. Take the ln of both sides.x 2lne ln21

(x 2) lne ln21 2. Bring the exponent in front of the logarithm. 3. Solve the equation. Use the fact that lne=1.

(x 2)(1) ln21 x ln21 2


Your Turn Problem #5x 3Solve e 5 to the nearest hundredth.

Answer: 1.39

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x 2 Solve 3e 7 19 to the neareE sxampl t hune 6 dre. dth.

Before we can take the log of both sides, ax or ex must be by itself. Therefore, subtract the 7, then divide by 3 before taking the ln of both sides.

x 23e 12 x 2e 4

x 2lne ln4

(x 2) lne ln4

The solution set is {– 0.61}.

x ln4 2

Your Turn Problem #6x 1Solve 2e 1 17 to the nearest hundredth.

Answer: 3.08

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Previously, we used the Product Rule and Quotient Rule of Logarithms to solve equations such as:

2 2 10 10log x log (x 3) 2 and log (2x 1) log (x 2) 1

With the property from this section, our equation solving capabilities have been expanded.If x > 0, y > 0, b > 0, and b 1, then x = y if and only if logbx= logby.

The previous problems done used the property by taking the log of both sides.

If x = y then logbx= logby.

Now we will use the property in the opposite direction.

If logbx= logby, then x = y.

The property is true in either direction because of the words “if and only if.”

Next Slide

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3.5 Exponential Equations, Logarithmic Equations, and Problem Solving Solve log(Example 7. x 2) log( 2 3x 1 ).

If log x = log y, then x = y. x 2 3x 12

Then solve the equation.2x 14

x 7

Recall: The domain of a logarithmic function must contain only positive numbers. Therefore, we need to check if x+2 and 3x – 12 are positive when x is replaced by 7.

The solution set is {7}.

In this example, the equation contains logarithms of positive numbers, therefore x = 7 is the solution. If the equation contained a logarithm of a negative number, x = 7 would not be a solution.


Solve log(3x 7) log 2(x 1)

Answer: 5

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3.5 Exponential Equations, Logarithmic Equations, and Problem Solving Solve log x log (xExamp 21)le 8 l 0. og1 0.

First use the Product Rule of Logarithms to create a single logarithm on each side. Then use the property, if log x = log y, then x = y.

x(x 21) 100

(x 25)(x 4) 0

x 25 or x 4

log x x 21 log100

2x 21x 100 0

The only solution is 25, and the solution set is {25}.

In this example, x = – 4 is not a solution because it leads to the logarithm of a negative number in the original equation.

log( 4) log( 4 21) log100

Note: Fortunately the quadratic equation was factorable. If it were not factorable, we would use the quadratic formula to solve.


Solve logx log(x 2) log24

Answer: {6}

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Solve log 5x log (Examp 2xle 9 1) g4. lo . First use the Quotient Rule of Logarithms to create a single logarithm on each side. Then use the property, if log x = log y, then x = y.

5x 4(2x 1)

5xlog log4

2x 1

5x4

2x 1

4x

3

5x 8x 4

3x 4

The solution set is {4/3}.

In this example, x = 4/3 a solution because it does not lead to the logarithm of a negative number in the original equation.


Solve ln(3a 4) ln(a 1) ln2

Answer: {6}

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3.5 Exponential Equations, Logarithmic Equations, and Problem Solving Solve logExample x log (x10. 15) 2.

10log x(x 15) 2

The solution set is {20}.

In this example, x = – 5 is not a solution because it leads to the logarithm of a negative number in the original equation.

x 20 or x 5

First use the Product Rule of Logarithms to create a single logarithm on each side. Then solve using the definition of a logarithm to change from logarithmic form to exponential form.

2x 15x 100 0

2x(x 15) 10 Remember the base is 10 if it is not written.

(x 20)(x 5) 0


Solve logx log(x 48) 2

Answer: {50}

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Previously, we used the definition of the logarithm to evaluate a logarithm when then base is a positive number other than 10.

Example: Evaluate log216

We evaluated this by setting the log216 equal to x and then using the definition of a logarithm to change the equation from logarithmic form to exponential form. Then the equation can be solved by matching the bases.

log216 = x

2x=16

2x=24

x = 4 Thus, we have log216 = 4

Next Slide

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Set the logarithm equal to x and then change the equation from logarithmic form to exponential form.

With the new property given in this section, we now have the capability to solve equations where the bases can not be matched.

Example 11. Evaluate log315. (Round to 3 decimal places)

Solution: log315 = x

3x=15

Now take the log of both sides.log3x = log15

By taking the log of both sides, we can now make use of power rule of logarithms; bring the exponent in front of the logarithm.

xlog3 = log15

Then divide by log3 on both sides and calculate.

log15x =

log3

Thus, log315 = 2.465


Evaluate log512. (Round to 3 decimal places)

Answer: 1.544

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Notice the pattern in Example 11 and Your Turn Problem 11.

3

log15log 15

log3

5

log12log 12

log5and

This leads us the the change-of-base formula for logarithms. (It is just a shortcut to the previous examples.)

Change-of-Base Rule

ba

b

log rlog r .

log aIf a > 0, b > 0, r > 0, with a 1 and b 1, then

Note: any positive number other than 1 can be used for the base ‘b’ in the change-of-base rule, but usually the only practical bases are ‘e’ and 10 since calculators give logarithms only for these two bases.

Next Slide

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Use the change-of-base formula to evaluate Also, use the common logarithm for the base so the calculation can be performed on the calculator.

Example 12. Evaluate log521. (Round to 3 decimal places)

Solution:5

log21log 21 =

log5

Thus, log521 = 1.892 rounded to three decimal places.


Evaluate log47.4 (Round to 3 decimal places)

Answer: 1.444

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A = accumulated amount

P = principle invested

r = interest rate

n = number of times compounded per year

t = years

Next Slide

nt

Recall the formular

A P 1n

:

where ‘A’ is the amount of money accumulated at the end of ‘t’ years if ‘P’ dollars is invested at rate of interest ‘r’ compounded ‘n’ times per year.

Previously, we covered Applications of Exponential Functions. We will now revisit application problems with expanded capabilities using the properties and rules of this section.

Solving Application Problems

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Example 13. How long will it take $2000 to be worth $2800 if it is invested at 8% compounded quarterly?

Solution:1. Write down the formula and

the given information.4t

0.082800 2000 1

4

2. Now, substitute the values into the given equation.

A = $2800

P = $2000

r = .08

n = 4t = ?

ntr

A P 1n

4t

2800 2000 1 0.02

3. Simplify inside the parentheses, then divide by 2000 on both sides. 4t

2800 2000 1.02

4. Take the log of both sides to solve.

4tlog1.4 log 1.02

log1.4 4tlog 1.02

4t1.4 1.02

5. Solve for t by dividing by 4log(1.02) on both sides.

log1.4

t4log 1.02

t = 4.2 years to the nearest tenth


How long will it take $2000 to be worth $4500 if it is invested at 12% compounded quarterly? (Round to the nearest tenth)

Answer: 6.9 years

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Example 14. The number of grams of a certain radioactive substance present after t hours is given by the equation A = A0e-0.025t, where A0 represents the initial number of grams. How long would it take 5000 grams to be reduced to 2500 grams? (i.e., What is the half-life?)

Solution:1. Write down the formula and

the given information.0.025t2500 5000e

2. Now, substitute the values into the given equation.

A0 = 5000

A = 2500

t = ?

0.025t0A A e

3. Divide by 5000 on both sides.

5. Solve for t by dividing by -0.025 on both sides.

t = 27.7 hours to the nearest tenth

0.025t0.5 e

0.025tln0.5 lne

ln0.5t

0.025

4. Take the ln of both sides to solve.

ln0.5 0.025tlne


For a certain strain of bacteria, the number of bacteria present after t hours is given by the equation A = A0e0.45t, where A0 represents the initial number of bacteria. How long would it take 600 bacteria to increase to 3000 bacteria? Answer: 3.6 years

The EndB.R.1-14-07

3.5 Exponential Equations, Logarithmic Equations, and Problem Solving 1 If b > 0 and b 1, and m and...

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