Alg ExpLog Solutions

8/13/2019 Alg ExpLog Solutions

1/46

COLLEGE ALGEBRASolutions to Practice Problems

Exponential and Logarithm Functions

Paul Dawkins


2/46

College Algebra

Table of Contents

Preface ............................................................................................................................................ 1Exponential and Logarithm Functions ........................................................................................ 2

Exponential Functions ................................................................................................................................ 2Logarithm Functions .................................................................................................................................. 7Solving Exponential Equations .................................................................................................................19Solving Logarithm Equations ...................................................................................................................27Applications ..............................................................................................................................................39

Preface

Here are the solutions to the practice problems for my Calculus I notes. Some solutions will have more orless detail than other solutions. The level of detail in each solution will depend up on several issues. Ifthe section is a review section, this mostly applies to problems in the first chapter, there will probably notbe as much detail to the solutions given that the problems really should be review. As the difficulty levelof the problems increases less detail will go into the basics of the solution under the assumption that ifyouve reached the level of working the harder problems then you will probably already understand thebasics fairly well and wont need all the explanation.

This document was written with presentation on the web in mind. On the web most solutions are brokendown into steps and many of the steps have hints. Each hint on the web is given as a popup however inthis document they are listed prior to each step. Also, on the web each step can be viewed individually byclicking on links while in this document they are all showing. Also, there are liable to be some formattingparts in this document intended for help in generating the web pages that havent been removed here.These issues may make the solutions a little difficult to follow at times, but they should still be readable.

2007 Paul Dawkins i http://tutorial.math.lamar.edu/terms.aspx


3/46

College Algebra

Exponential and Logarithm Functions

Exponential Functions

1. Given the function ( ) 4xf x = evaluate each of the following.

(a) ( )2f (b) ( )12f (c) ( )0f (d) ( )1f (e) ( )32

f

(a) ( )2f All we need to do here is plug in thexand do any quick arithmetic we need to do.

( ) 2 21 12 4 4 16f = = =

(b) ( )12f All we need to do here is plug in thexand do any quick arithmetic we need to do.

( ) 1

212

12

1 1 14

244f

= = = =

(c) ( )0f

All we need to do here is plug in thexand do any quick arithmetic we need to do.

( ) 00 4 1f = =

(d) ( )1f All we need to do here is plug in thexand do any quick arithmetic we need to do.

( ) 11 4 4f = =

(e) ( )32f All we need to do here is plug in thexand do any quick arithmetic we need to do.

( ) ( )33 1

2 2 332

4 4 2 8f = = = =

2007 Paul Dawkins 2 http://tutorial.math.lamar.edu/terms.aspx


4/46

College Algebra

2. Given the function ( ) ( )15x

f x = evaluate each of the following.

(a) ( )3f (b) ( )1f (c) ( )0f (d) ( )2f (e) ( )3f

(a) ( )3f

All we need to do here is plug in thexand do any quick arithmetic we need to do.

( )3 3 3

3

1 5 53 125

5 1 1f

= = = =

(b) ( )1f All we need to do here is plug in thexand do any quick arithmetic we need to do.

( )1 1

1 55

5 1f

= = =

(c) ( )0f All we need to do here is plug in thexand do any quick arithmetic we need to do.

( )0

10 1

5f

= =

(d) ( )2f All we need to do here is plug in thexand do any quick arithmetic we need to do.

( )2 2

2

1 1 12

5 5 25f

= = =

(e) ( )3f All we need to do here is plug in thexand do any quick arithmetic we need to do.

( )3 3

3

1 1 13

5 5 125f

= = =

3. Sketch each of the following.

(a) ( ) 6xf x = (b) ( ) 6 9xg x = (c) ( ) 16xg x +=

(a) ( ) 6xf x = We can build up a quick table of values that we can plot for the graph of this function.



5/46

College Algebra

x ( )f x

-2 ( ) 22 1 1

3662 6f = = =

-1 ( ) 1 161 6f = =

0 ( ) 00 6 1f = =

1 ( ) 11 6 6f = =

2 ( ) 22 6 36f = =

Here is a quick sketch of the graph of the function.

(b) ( ) 6 9xg x = For this part all we need to do is recall the Transformationssection from a couple of chapters ago. Using

the base function of ( ) 6xf x = the function for this part can be written as,

( ) ( )6 9 9xg x f x= =

Therefore, the graph for this part is just the graph of ( )f x shifted down by 9.

The graph of this function is shown below. The blue dashed line is the base function, ( )f x , and the

red solid line is the graph for this part, ( )g x .



6/46

College Algebra

(c) ( ) 16xg x += For this part all we need to do is recall the Transformationssection from a couple of chapters ago. Using

the base function of ( ) 6x

f x = the function for this part can be written as,

( ) ( )16 1xg x f x+= = +

Therefore, the graph for this part is just the graph of ( )f x shifted left by 1.



4. Sketch the graph of ( ) xf x = e .

Solution



7/46

College Algebra

For this problem all we need to do is recall the Transformationssection from a couple of chapters ago.

Using the base function of ( ) xf x = e the function for this part can be written as,

( ) ( )xg x f x= = e

Therefore, the graph for this part is just the graph of ( )f x reflected about they-axis.



5. Sketch the graph of ( ) 3 6xf x = +e .

SolutionFor this problem all we need to do is recall the Transformationssection from a couple of chapters ago.

Using the base function of ( ) xf x = e the function for this part can be written as,

( ) ( )3 6 3 6xf x f x= + = +e

Therefore, the graph for this part is just the graph of ( )f x shifted right by 3 and up by 6.





8/46

College Algebra

Logarithm Functions

1. Write5

7 16807= in logarithmic form.

SolutionThere really isnt all that much to do here other than refer to the definition of the logarithm function givenin the notes for this section.

Here is the logarithmic form for this expression.

7log 16807 5=

2. Write3

416 8= in logarithmic form.



16

3log 8

4=



9/46

College Algebra

3. Write

21

93

=

in logarithmic form.

SolutionThere really isnt all that much to do here other than refer to the definition of the logarithm function given

in the notes for this section.


13

log 9 2=

4. Write2log 32 5= in exponential form.


Here is the exponential form for this expression.

52 32=

5. Write 15

1625

log 5= in exponential form.



51

6255

=

6. Write 19 81

log 2= in exponential form.





10/46

College Algebra

2 1981

=

7. Without using a calculator determine the exact value of3log 81 .

Hint : Recall that converting a logarithm to exponential form can often help to evaluate these kinds oflogarithms.

Step 1Converting the logarithm to exponential form gives,

?

3log 81 ? 3 81= =

Step 2

From this we can quickly see that 43 81= and so we must have,

3log 81 4=

8. Without using a calculator determine the exact value of5log 125 .



?

5log 125 ? 5 125= =

Step 2

From this we can quickly see that 35 125= and so we must have,

5log 125 3=

9. Without using a calculator determine the exact value of 21

log 8 .





11/46

College Algebra

?

2

1 1log ? 2

8 8= =

Step 2Now, we know that if we raise an integer to a negative exponent well get a fraction and so we must have

a negative exponent and then we know that 32 8= . Therefore we can see that3 12

8 = and so we must

have,

2

1log 3

8=

10. Without using a calculator determine the exact value of1

4

log 16 .



?

1

4

1log 16 ? 16

4

= =

Step 2Now, we know that if we raise an fraction to a power and get an integer out we must have had a negative

exponent. Now, we also know that 24 16= . Therefore we can see that

2 21 4

164 1

= = and so we

must have,

1

4

log 16 2=

11. Without using a calculator determine the exact value of4ln e .

Hint : Recall that converting a logarithm to exponential form can often help to evaluate these kinds of

logarithms. Also recall what the base is for a natural logarithm.

Step 1Recalling that the base for a natural logarithm is eand converting the logarithm to exponential form gives,

4 4 ? 4ln log ?= = =e

e e e e

Step 2

From this we can quickly see that 4 4=e e and so we must have,



12/46

College Algebra

4ln 4=e

Note that an easier method of determining the value of this logarithm would have been to recall theproperties of logarithm. In particular the property that states,

log x

bb x=

Using this we can also very quickly see what the value of the logarithm is.

12. Without using a calculator determine the exact value of1

log100

.

Hint : Recall that converting a logarithm to exponential form can often help to evaluate these kinds oflogarithms. Also recall what the base is for a common logarithm.

Step 1Recalling that the base for a natural logarithm is 10 and converting the logarithm to exponential formgives,

?

10

1 1 1log log ? 10

100 100 100= = =

Step 2Now, we know that if we raise an integer to a negative exponent well get a fraction and so we must have

a negative exponent and then we know that2

10 100= . Therefore we can see that 2 1

10100

= and so we

must have,

1log 2100

=

13. Write ( )4 7log 3x y in terms of simpler logarithms.

Step 1So, were being asked here to use as many of the properties as we can to reduce this down into simplerlogarithms.

First we can use Property 5 to break up the product into individual logarithms. Note that just because theproperty only has two terms in it does not mean that it wont work for three (or more) terms. Here is theapplication of Property 5.

( ) ( ) ( ) ( )4 7 4 7log 3 log 3 log logx y x y = + +

Step 2



13/46

College Algebra

Finally we need to use Property 7 on the last two logarithms to bring the exponents out of the logarithms.Here is that work.

( ) ( ) ( ) ( )4 7log 3 log 3 4 log 7 logx y x y = +

Remember that we can only bring an exponent out of a logarithm if is on the whole argument of thelogarithm. In other words, we couldnt bring any of the exponents out of the logarithms until we haddealt with the product.

14. Write ( )2 2ln x y z+ in terms of simpler logarithms.


First we can use Property 5 to break up the product into individual logarithms. Here is that work.

( ) ( ) ( )1

2 2 2 2 2ln ln lnx y z x y z

+ = + +

Note that we converted to root to a fractional exponent at the same time to help with the next step.

Step 2Finally we need to use Property 7 on the last logarithm to bring the root exponent out of the logarithm.Here is that work.

( ) ( ) ( )2 2 2 21ln ln ln2

x y z x y z+ = + +

Remember that we can only bring an exponent out of a logarithm if is on the whole argument of thelogarithm. In other words, we couldnt bring any of the exponents out of the logarithms until we haddealt with the product. Also, in the second logarithm while each term is squared the whole argument is

not squared, i.e.its not ( )2

x y+ and so we cant bring those 2s out of the logarithm.

15. Write4 2 5

4log

x

y z

in terms of simpler logarithms.




14/46

College Algebra

First we can use Property 6 to break up the quotient into two logarithms. Here is that work.

( )1

2 54 4 42 5

4log log 4 log

xx y z

y z

=

Step 2Next we need to use Property 5 to break up the product in the second logarithm into two logarithms.

( ) ( )1

2 54 4 4 42 5

4log log 4 log log

xx y z

y z

= +

Be careful with the minus sign that was in front of the second logarithm from Step 1! Because of that weneed to have parenthesis on the product once we use Property 5. The sum of the two smaller logarithmsis the same as the product logarithm from Step 1 and so because we have a minus sign in front of theproduct logarithm we also need to have a minus sign in front of the two logarithms after using Property 5.The only way to make sure of this is to use the parenthesis as shown.

Step 3Finally well distribute the minus sign through the parenthesis and then use Property 7 on the last twologarithms to bring the exponents out of the logarithms. Here is that work.

( ) ( ) ( )4 4 4 42 54 1

log log 4 2 log log5

xx y z

y z

=

Remember that we can only bring an exponent out of a logarithm if is on the whole argument of thelogarithm. In other words, we couldnt bring any of the exponents out of the logarithms until we haddealt with the quotient and product. Recall as well that we cant split up a sum/difference in a logarithm.Finally, make sure that you are careful in dealing with the minus sign we get from breaking up thequotient when dealing with the product in the denominator.

16. Combine4 4 4

12log 5log log

2x y z+ into a single logarithm with a coefficient of one.

Hint :The properties that we use to break up logarithms can be used in reverse as well.

Step 1

To convert this into a single logarithm well be using the properties that we used to break up logarithms inreverse. The first step in this process is to use the property,

( )log logrb bx r x= to make sure that all the logarithms have coefficients of one. This needs to be done first because all theproperties that allow us to combine sums/differences of logarithms require coefficients of one onindividual logarithms. So, using this property gives,



15/46

College Algebra

( ) ( )1

2 5 24 4 4

log log logx y z

+

Step 2Now, there are several ways to proceed from this point. We can use either of the two properties.

( )log log log log log logb b b b b bx

xy x y x yy

= + =

and in fact well need to use both in the end. We can use the product property on the first two logarithms(because they are a sum of logarithms) or the quotient property on the last two logarithms (because theyare a difference of logarithms).

Which we use first does not matter as well end up with the same result in the end. For this problemswell first use the product property on the first two logarithms to get,

( ) ( )2 54 4 4 4 41

2log 5log log log log2

x y z x y z+ =

Step 3Finally, we can see that we have a difference of two logarithms left and so well use the quotient propertyto combine these to get,

2 5

4 4 4 4

12log 5log log log

2

x yx y z

z

+ =

Note that the only reason we converted the fractional exponent to a root was to make the final answer alittle nicer.

17. Combine ( ) ( )3ln 5 4 ln 2 ln 1t t s+ into a single logarithm with a coefficient of one.

Hint :The properties that we use to break up logarithms can be used in reverse as well.

Step 1To convert this into a single logarithm well be using the properties that we used to break up logarithms inreverse. The first step in this process is to use the property,

( )log logrb bx r x=

to make sure that all the logarithms have coefficients of one. This needs to be done first because all theproperties that allow us to combine sums/differences of logarithms require coefficients of one onindividual logarithms. So, using this property gives,

( ) ( ) ( )3 24ln 5 ln ln 1t t s+




16/46

College Algebra


xy x y x yy

= + =

and in fact well need to use both in the end.

We should also be careful with the fact that there are two minus signs in here as that sometimes adds

confusion to the problem. They are easy to deal with however if we just factor a minus sign out of thelast two terms to get,

( ) ( ) ( ) ( ) ( )( )3 243ln 5 4 ln 2 ln 1 ln 5 ln ln 1t t s t t s+ = + +

Written in this form we can see that the last two logarithms are a sum and so we can use the productproperty to combine them to get,

( ) ( ) ( ) ( )( )3 243ln 5 4 ln 2 ln 1 ln 5 ln 1t t s t t s+ = +

Step 3We now have a difference of two logarithms and we can use the quotient property to combine them to get,

( ) ( ) ( )

( )

3

24

53ln 5 4 ln 2 ln 1 ln

1

tt t s

t s

++ =

18. Combine1

log 6 log 23

a b + into a single logarithm with a coefficient of one.

Hint :The properties that we use to break up logarithms can be used in reverse as well. For the constantsee if you figure out a way to write that as a logarithm.

Step 1To convert this into a single logarithm well be using the properties that we used to break up logarithms inreverse. The first step in this process is to use the property,

( )log logrb bx r x= to make sure that all the logarithms have coefficients of one. This needs to be done first because all theproperties that allow us to combine sums/differences of logarithms require coefficients of one onindividual logarithms. So, using this property gives,

( )1

63log log 2a b +

Step 2Now, for the 2 lets notice that we can write this in terms of a logarithm as,

22 log 10 log 100= =



17/46

College Algebra

Note that this is really just using the property,

log x

bb x=

So, we now have,

( )

1

63

log log log100a b

+



xy x y x yy

= + =

and in fact well need to use both in the end.

The first two logarithms are a difference so lets use the quotient property to first combine those to get,

( )1 3

6 236

log log log10 log log100aa bb

+ = +

We converted the fractional exponent in the first term to a root to make the answer a little nicer butdoesnt really need to be done in general.

Step 4Finally, note that we now have a sum of two logarithms and we can use the product property to combinethose to get,

( )

1 363

6

100

log log log100 log

a

a b b

+ =

19. Use the change of base formula and a calculator to find the value of12log 35 .

SolutionWe can use either the natural logarithm or the common logarithm to do this so well do both.

12

ln35 3.55534806log 35 1.43077731

ln12 2.48490665= = =

12

log35 1.54406804log 35 1.43077731

log12 1.07918125= = =

So, as we noted at the start it doesnt matter which logarithm we use well get the same answer in the end.



18/46

College Algebra

20. Use the change of base formula and a calculator to find the value of2

3

log 53 .

SolutionWe can use either the natural logarithm or the common logarithm to do this so well do both.

2

3

ln 53 3.97029191log 53 9.791944692 0.40546511

ln3

= = =

2

3

log53 1.72427587log 53 9.79194469

2 0.17609126log

3

= = =

So, as we noted at the start it doesnt matter which logarithm we use well get the same answer in the end.

21. Sketch the graph of ( ) ( )lng x x= .


Using the base function of ( ) ( )lnf x x= the function for this part can be written as,

( ) ( ) ( )lng x x f x= =

Therefore, the graph for this part is just the graph of ( )f x reflected about thex-axis.





19/46

College Algebra

22. Sketch the graph of ( ) ( )ln 5g x x= + .

Solution

For this problem all we need to do is recall the Transformationssection from a couple of chapters ago.Using the base function of ( ) ( )lnf x x= the function for this part can be written as,

( ) ( ) ( )ln 5 5g x x f x= + = +

Therefore, the graph for this part is just the graph of ( )f x shifted left by 5.



Do not get excited about the fact that we plugged negative values ofxinto the function! The problemwith negative values is not the values we plug into a logarithm. Instead the problem with negative valuesis when we go to evaluate the logarithm.

It is perfectly fine to plug negative values into a logarithm as long as we dont end up actually evaluating

a negative number. So, in this case we can see that as long as we require 5x> then 5 0x+ > and sothose are acceptable values ofxto plug in since we arent going to evaluate negative number in thelogarithm.

Note however that we do have avoid 5x< since that would mean evaluating logarithms at negativenumbers.

23. Sketch the graph of ( ) ( )ln 4g x x= .



20/46

College Algebra


Using the base function of ( ) ( )lnf x x= the function for this part can be written as,

( ) ( ) ( )ln 4 4g x x f x= =

Therefore, the graph for this part is just the graph of ( )f x shifted down by 4.



Solving Exponential Equations

1. Solve the following equation.2 1 3

6 6x x=

Step 1

Recall the property that says if x yb b= then x y= . Since each exponential has the same base, 6 in this

case, we can use this property to just set the exponents equal. Doing this gives,

2 1 3x x=

Step 2Now all we need to do is solve the equation from Step 1 and that is a simple linear equation. Here is thesolution work.



21/46

College Algebra

2 1 3

15 1

5

x x

x x

=

= =

So, the solution to the equation is then : 15

x= .

2. Solve the following equation.15 25x =

Step 1

Recall the property that says if x yb b= then x y= . In this case it looks like we cant use this property.

However, recall that2

25 5= and if we write the right side of our equation using this we get,

1 25 5x

=

Now each exponential has the same base, 5 to be exact, so we can use this property to just set theexponents equal. Doing this gives,

1 2x =


1 2

1 1

x

x x

=

= =

So, the solution to the equation is then : 1x= .

3. Solve the following equation.2 3 108 8

x x+=

Step 1

Recall the property that says ifx y

b b= then x y= . Since each exponential has the same base, 8 in thiscase, we can use this property to just set the exponents equal. Doing this gives,

2 3 10x x= +

Step 2Now all we need to do is solve the equation from Step 1 and that is a quadratic equation that we should beable to quickly solve. Here is the solution work.



22/46

College Algebra

( ) ( )

2

2

3 10

3 10 0

5 2 0 2, 5

x x

x x

x x x x

= +

=

+ = = =

So, the solutions to the equation are then : 2x= and 5x= .

4. Solve the following equation.4 47 7x x =

Step 1

Recall the property that says if x yb b= then x y= . Since each exponential has the same base, 7 in thiscase, we can use this property to just set the exponents equal. Doing this gives,

4 4x x =


4 4

44 5

5

x x

x x

=

= =

So, the solution to the equation is then : 45x= .

5. Solve the following equation.32 10x =

Step 1For this equation there is no way to easily get both sides with the same base. Therefore, well need totake the logarithm of both sides.

We can use any logarithm and the natural logarithm and common logarithm are usually good choicessince most calculators can handle them. Because one of the bases in this equation is a 10 the commonlogarithm will probably be the better choice (although we can use the natural logarithm if we wanted to).

Taking the logarithm (using the common logarithm) of both sides gives,

3log 2 log10x =



23/46

College Algebra

Step 2Now we can easily compute the right side (which is also why we chose the common logarithm for thiscase) and we can use the logarithm property that says,

log logrb bx r x=

to move the 3xout of the exponent from the logarithm on the left. Doing this gives,

( ) ( )3 log 2 3log 2 1x x= =

We did a little rearranging of the left side to put all the numbers together in order to make the next step alittle easier.

Step 3Finally, all we need to do is solve forx. Recall that the equations at this step tend to look messier than weare used to dealing with. However, the logarithms in the equation at this point are just numbers and so wetreat them as we treat all numbers with these kinds of equations.

In other words, all we need to do is divide both sides by the coefficient of thexand then user ourcalculators to get a decimal answer.

Here is the rest of the work for this problem.

( )( )

1 13log 2 1 1.10730936

3log 2 3 0.301029996x x= = = =

6. Solve the following equation.1 3 1

7 4x x +

=


We can use any logarithm and the natural logarithm and common logarithm are usually good choicessince most calculators can handle them. In this case there really isnt any reason to use one or the other sowell use the natural logarithm (its easier to write two letters ln versus three letters log afterall).

Taking the logarithm (using the natural logarithm) of both sides gives,

1 3 1ln 7 ln 4

x x +

=

Step 2Now we can use the logarithm property that says,

log logr

b bx r x=



24/46

College Algebra

to move the exponents out of each of the logarithms. Doing this gives,

( ) ( )1 ln 7 3 1 ln 4x x = +

Step 3

Finally, all we need to do is solve forx. Recall that the equations at this step tend to look messier than weare used to dealing with. However, the logarithms in the equations at this point are just numbers and sowe treat them as we treat all numbers with these kinds of equations. The work will be messier than weare used to but just keep in mind that the logarithms are just numbers!


( ) ( )

( )

( )

1 ln 7 3 1 ln 4

ln 7 ln 7 3 ln 4 ln 4

ln 7 ln 4 3 ln 4 ln 7

ln 7 ln 4 3ln 4 ln 7ln 7 ln 4 1.945910149 1.386294361

0.0916682623ln4 ln 7 3 1.386294361 1.945910149

x x

x x

x x

x

x

= +

= +

= +

= +

= = =

+ +

Again, the work is messier than we are used to but it is not really different from work weve donepreviously in solving equations. The answer is also going to be messier in the sense that it is a decimaland is liable to almost always be a decimal for most of these types of problems so dont worry about that.

7. Solve the following equation.

4 69 10 x+=


We can use any logarithm and the natural logarithm and common logarithm are usually good choicessince most calculators can handle them. In this case one of the bases is a 10 and so the commonlogarithm is probably the better choice.

Taking the logarithm (using the common logarithm) of both sides gives,

4 6log9 log10 x+=


( ) ( )log10f x f x=



25/46

College Algebra

to simplify the right side of the equation. Doing this gives,

log 9 4 6x= +

Step 3Finally, all we need to do is solve forx. Recall that the equations at this step tend to look messier than weare used to dealing with. However, the logarithms in the equations at this point are just numbers and sowe treat them as we treat all numbers with these kinds of equations. The work will be messier than weare used to but just keep in mind that the logarithms are just numbers!


log 9 4 6

log 9 4 6

log9 4 0.9542425094 40.5076262484

6 6

x

x

x

= +

=

= = =


Also, be careful when evaluating the numerator in the final answer. The 4 was outside of the logarithmand so cannot be moved into the logarithm. We probably should have been a little more careful withparenthesis and written the answer as,

( )log 9 46

x

=

which makes it a little more clear that the 4 isnt inside the logarithm. However, we typically dont putthe parenthesis on the logarithm when it is just a number.

8. Solve the following equation.7 2

3 0x+ =e

Step 1Before we put any logarithms into this problem we first need to get the exponential on one side by itselfso lets do that first.

7 2 3x+ =e

Step 2Now we can take the logarithm of both sides and because we have a base of ein this problem the naturallogarithm is probably the best choice. So, taking the logarithm (using the natural logarithm) of both sidesgives,



26/46

College Algebra

7 2ln ln 3x+ =e


( )

( )ln f x

f x=e

to simplify the left side of the equation. Doing this gives,

7 2 ln 3x+ =



7 2 ln 3

2 ln 3 7

ln 3 7 1.098612289 72.950693856

2 2

x

x

x

+ =

=

= = =



( )ln 3 72

x

=


9. Solve the following equation.4 7 11 20x + =e

Step 1Before we put any logarithms into this problem we first need to get the exponential on one side by itselfso lets do that first.



27/46

College Algebra

4 7 9x =e

Step 2Now we can take the logarithm of both sides and because we have a base of ein this problem the naturallogarithm is probably the best choice. So, taking the logarithm (using the natural logarithm) of both sidesgives,

4 7ln ln 9x =e


( ) ( )ln f x f x=e

to simplify the left side of the equation. Doing this gives,

4 7 ln 9x =



4 7 ln 9

7 ln 9 4

ln9 4 2.197224577 40.2575393461

7 7

x

x

x

=

=

= = =



( )ln 9 47

x =




28/46

College Algebra

Solving Logarithm Equations

1. Solve the following equation.( ) ( )24 4log 2 log 5 12x x x =

Hint : We had a very nice property from the notes on how to solve equations that contained exactly twologarithms with the same base! Also, dont forget that the values with get when we are done solvinglogarithm equations dont always correspond to actual solutions to the equation so be careful!

Step 1

Recall the property that says if log logb bx y= then x y= . Since each logarithm is on opposite sides of

the equal sign and each has the same base, 4 in this case, we can use this property to just set thearguments of each equal. Doing this gives,

2 2 5 12x x x =

Step 2Now all we need to do is solve the equation from Step 1 and that is a quadratic equation that we knowhow to solve. Here is the solution work.

( ) ( )

2

2

2 5 12

7 12 0

3 4 0 3, 4

x x x

x x

x x x x

=

+ =

= = =

Step 3As the final step we need to take each of the numbers from the above step and plug them into the originalequation from the problem statement to make sure we dont end up taking the logarithm of zero ornegative numbers!

Here is the checking work for each of the numbers.

3 :x=

( ) ( )( ) ( )( )( ) ( )

2

4 4

4 4

log 3 2 3 log 5 3 12

log 3 log 3 OKAY

=

=

4 :x=

( ) ( )( ) ( )( )( ) ( )

2

4 4

4 4

log 4 2 4 log 5 4 12

log 8 log 8 OKAY

=

=

In this case, both numbers do not produce negative numbers in the logarithms and so they are in fact bothsolutions (wont happen with every problem so dont always expect this to happen!).



29/46

College Algebra

Therefore, the solutions to the equation are then : 3x= and 4x= .

2. Solve the following equation.( ) ( ) ( )log 6 log 4 log 3x x =

Hint : We had a very nice property from the notes on how to solve equations that contained exactly twologarithms with the same base! Also, dont forget that the values with get when we are done solvinglogarithm equations dont always correspond to actual solutions to the equation so be careful!

Step 1

Recall the property that says if log logb b

x y= then x y= . That doesnt appear to have any use here

since there are three logarithms in the equation. However, recall that we can combine a difference oflogarithms (provide the coefficient of each is a one of course) as follows,

( )6

log log 34

x

x

=

We now have only two logarithms and each logarithm is on opposite sides of the equal sign and each hasthe same base, 10 in this case. Therefore, we can use this property to just set the arguments of each equal.Doing this gives,

63

4

x

x=

Step 2Now all we need to do is solve the equation from Step 1 and that is an equation that we know how tosolve. Here is the solution work.

( )

63

4

6 3 4 12 3

12 49 12

9 3

x

x

x x x

x x

=

= =

= = =

Step 3As the final step we need to take the number from the above step and plug it into the original equationfrom the problem statement to make sure we dont end up taking the logarithm of zero or negativenumbers!

Here is the checking work for the number.

43:x=



30/46


31/46

College Algebra


10 :x=

( ) ( ) ( )( )

( ) ( ) ( )

ln 10 ln 10 3 ln 20 5 10

ln 10 ln 7 ln 70 NOT OKAY

+ + =

+ =

2 :x=

( ) ( ) ( )( )

( ) ( ) ( )

ln 2 ln 2 3 ln 20 5 2

ln 2 ln 5 ln 10 OKAY

+ + =

+ =

In this case, the only one number did not produce negative numbers in the logarithms so that is the onlynumber that will be a solution. The number that produced negative numbers in the logarithm is not asolution.

Therefore, the only solution to the equation is then : 2x= .

Note that it is vitally important that you do the check in the original equation. In the first step (where wecombined two of the logarithms) we changed the equation and in the process introduced a number that isnot in fact a solution.

Had we checked in any other equation in the solution work it would appear that 10x= would be asolution to the equation. However, that is only because we were checking in a modified equation andnot the original equation which is what we were being asked to solve.

This is always the danger of modifying equations during the solution process. Unfortunately, with manylogarithm equations that is our only solution path and so is something that we need to be prepared to dealwith.


( )23log 25 2x =

Hint : We had a very nice property from the notes on how to solve equations that contained exactly twologarithms with the same base and yes we can use that property here! Also, dont forget that the valueswith get when we are done solving logarithm equations dont always correspond to actual solutions to theequation so be careful!

Step 1

Recall the property that says if log logb bx y= then x y= . That doesnt appear to have any use here

since there is only one logarithm in the equation. Note however that we could write the right side, i.e.the2, as,

( )232 log 3=



32/46

College Algebra

Doing this means we can write the equation as,

( ) ( ) ( )2 23 3 3log 25 log 3 log 9x = =

We now have two logarithms and each logarithm is on opposite sides of the equal sign and each has thesame base, 3 in this case. Therefore, we can use this property to just set the arguments of each equal.Doing this gives,

225 9x =

Step 2Now all we need to do is solve the equation from Step 1 and that is a quadratic equation that we knowhow to solve. Here is the solution work.

2

2

25 9

16 16 4

x

x x

=

= = =



4 :x=

( )( )( )

2

3

3

log 25 4 2

log 9 2 OKAY

=

=

4 :x=

( )( )( )

2

3

3

log 25 4 2

log 9 2 OKAY

=

=


Therefore, the solutions to the equation are then : 4x= and 4x= .

Be careful to not make the mistake of assuming that just because a value ofxis negative that it will

automatically not be a solution to the equation. As weve shown here, even though 4x= is negative itdid not produce any negative values in the logarithms and so is perfectly acceptable as a solution.



33/46

College Algebra


( ) ( )2 2log 1 log 2 3x x+ =

Hint : If we can reduce all the logarithms to a single logarithm it would be quite easy to convert to

exponential form. Also, dont forget that the values with get when we are done solving dont alwayscorrespond to actual solutions so be careful!

Step 1First lets notice that we can combine the two logarithms on the left side to get,

2

1log 3

2

x

x

+ =

Step 2Now, we can easily convert this to exponential form.

31 2 82

x

x

+ = =


( )

18

2

1 8 2 16 8

15 59 159 3

x

x

x x x

x x

+=

+ = =

= = =

Step 4As the final step we need to take the number from the above step and plug it into the original equationfrom the problem statement to make sure we dont end up taking the logarithm of zero or negativenumbers!

Here is the checking work for the number.

53:x=

( ) ( )( ) ( )

5 52 23 3

8 12 23 3

log 1 log 2 3

log log 3 OKAY+ =

=

In this case, the number did not produce negative numbers in the logarithms so it is in fact a solution(wont happen with every problem so dont always expect this to happen!).



34/46


35/46

College Algebra

In this case, the only one number did not produce negative numbers in the logarithms so that is the onlynumber that will be a solution. The number that produced negative numbers in the logarithm is not asolution.

Therefore, the only solution to the equation is then : 2x= .

Be careful to not make the mistake of assuming that just because a value ofxis negative that it willautomatically not be a solution to the equation and just because a value ofxis positive it willautomatically be a solution. As weve shown here we can have negative values ofxthat are solutions andpositive values ofxthat are not solutions.

Also note that it is vitally important that you do the check in the original equation. In the first step (wherewe combined two of the logarithms) we changed the equation and in the process introduced a number thatis not in fact a solution.

Had we checked in any other equation in the solution work it would appear that 8x= would be asolution to the equation. However, that is only because we were checking in a modified equation and

not the original equation which is what we were being asked to solve.



( ) ( )log 2 log 21x x=

Hint : If we can reduce all the logarithms to a single logarithm it would be quite easy to convert to

exponential form. Also, dont forget that the values with get when we are done solving dont alwayscorrespond to actual solutions so be careful!

Step 1First lets notice that if we move the logarithm on the right side to the left side we can combine the twologarithms on the left side to get,

( ) ( )

( ) ( )

( )( )

log 2 log 21

log log 21 2

log 21 2

x x

x x

x x

=

+ =

=

Step 2Now, we can easily convert this to exponential form (recall that because there is no base given it isassumed to be 10!).

( ) 221 10 100x x = =

Step 3



36/46


37/46

College Algebra

Hint : If we can reduce all the logarithms to a single logarithm it would be quite easy to convert toexponential form. Also, dont forget that the values with get when we are done solving dont alwayscorrespond to actual solutions so be careful!

Step 1First lets notice that if we move the logarithm on the right side to the left side we can combine the two

logarithms on the left side to get,

( ) ( )

( ) ( )

ln 1 1 ln 3 2

ln 1 ln 3 2 1

1log 1

3 2

x x

x x

x

x

= + +

+ =

=

+

Step 2Now, we can easily convert this to exponential form (recall that because we are working with the naturallogarithm the base is e!).

11

3 2

x

x

= =

+ e e


( )

( )

1

3 2

1 3 2

1 3 2

3 1 2

1 21 3 1 2 0.89961

1 3

x

x

x x

x x

x x

x x

=

+

= +

= +

= +

+ = + = =

e

e

e e

e e

e

e e

e

Do not get excited about the ein the equation. It works the same as if it was just a 4 or 5 or any othernumber. The only real difference is that the answer is a little messier that we usually get with these kindsof problems. Also, for the next step it is probably best to convert these kinds of numbers into decimalform.

Step 4

As the final step we need to take the number from the above step and plug it into the original equationfrom the problem statement to make sure we dont end up taking the logarithm of zero or negativenumbers!


0.89961:x=



38/46

College Algebra

( ) ( )( )

( ) ( )

ln 0.89961 1 1 ln 3 0.89961 2

ln 1.89961 1 ln 0.69883 NOT OKAY

= + +

= +

So, in this case the only number we got from Step 3 produced negative numbers in the logarithms and socant be a solution. What this means for us is that there is no solutionto this equation. This happens onoccasion and we shouldnt worry about it when it does.





( ) ( )2log log 7 1 0x x =

Hint : If we can reduce all the logarithms to a single logarithm it would be quite easy to convert toexponential form. Also, dont forget that the values with get when we are done solving dont alwayscorrespond to actual solutions so be careful!

Step 1First lets notice that we can move the 2 in front of the first logarithm into the logarithm as follows,

( ) ( )2log log 7 1 0x x =

We can now combine the two logarithms to get,

2

log 07 1

x

x

=

Step 2Now, we can easily convert this to exponential form (recall that because there is no base given it isassumed to be 10!).

2010 1

7 1

x

x= =

Step 3



39/46

College Algebra

Now all we need to do is solve the equation from Step 2 and that is a quadratic equation that we knowhow to solve. Here is the solution work.

2

2

2

17 1

7 1

7 1 0

x

x

x x

x x

=

= + =

We cant factor this but we can use the quadratic formula on it. Doing that gives the following twonumbers.

( ) ( )

( )

27 7 4 1 1 7 450.1459, 6.8541

2 1 2x x x

= = = =

Dont worry about the fact that we needed to use the quadratic formula to solve this. This will happen on

occasion and we need to be able to deal with it when it happens.



0.1459:x=

( ) ( )( )

( ) ( )

2log 0.1459 log 7 0.1459 1 0

2log 0.1459 log 0.0213 0 OKAY

=

=

6.8541:x=

( ) ( )( )

( ) ( )

2log 6.8541 log 7 6.8541 1 0

2 log 6.8541 log 46.9787 0 OKAY

=

=


Therefore, the solutions to the equation are then : 0.1459x= and 6.8541x= .




40/46

College Algebra


This is always the danger of modifying equations during the solution process. Unfortunately, with manylogarithm equations that is our only solution path and so is something that we need to be prepared to deal

with.

Applications

1. We have $10,000 to invest for 44 months. How much money will we have if we put the money into anaccount that has an annual interest rate of 5.5% and interest is compounded,

(a) quarterly (b) monthly (c) continuously

(a) quarterlyFrom the problem statement we can see that,

5.5 44 1110000 0.055

100 12 3P r t= = = = =

Remember that the value of rmust be given as a decimal, i.e.the percentage divided by 100. Alsoremember that tmust be in years and so well need to convert the months we are given to years.

For this part we are compounding interest rate quarterly and that means it will compound 4 times per yearand so we also then know that,

4m=

At this point all that we need to do is plug into the equation and run the numbers through a calculator tocompute the amount of money that well have.

( )

( ) ( )

114

4433

0.05510000 1 10000 1.01375 10000 1.221760422 12217.60

4A

= + = = =

So, well have $12,217.60 in the account after 44 months.

(b) monthly

From the problem statement we can see that,5.5 44 1110000 0.055

100 12 3P r t= = = = =


For this part we are compounding interest rate monthly and that means it will compound 12 times peryear and so we also then know that,



41/46

College Algebra

12m=

At this point all that we need to do is plug into the equation and run the numbers through a calculator tocompute the amount of money that well have.

( )

( ) ( )

1112

3 440.055

10000 1 10000 1.00453333 10000 1.222876562 12228.7712A

= + = = =


(c) continuouslyFrom the problem statement we can see that,

5.5 44 1110000 0.055

100 12 3P r t= = = = =


For this part we are compounding continuously and so we wont have an mand will be using theother equation and all we have all we need to do the computation so,

( )

( )11

0.0550.2016666667310000 10000 10000 1.223440127 12234.40A

= = = =e e


2. We are starting with $5000 and were going to put it into an account that earns an annual interest rateof 12%. How long should we leave the money in the account in order to double our money if interest iscompounded,

(a) quarterly (b) monthly (c) continuously

Hint : Identify the given quantities, plug into the appropriate equation and use the techniques from anearlier section to solve for t.

(a) quarterlyFrom the problem statement we can see that,

1210000 5000 0.12

100A P r= = = =

Remember that the value of rmust be given as a decimal, i.e.the percentage divided by 100. Also, forthis part we are compounding interest rate quarterly and that means it will compound 4 times per year andso we also then know that,

4m=

Plugging into the equation gives us,

( )4

40.1210000 5000 1 5000 1.03

4

tt

= + =



42/46

College Algebra

Using the techniques from the Solve Exponential Equationssection we can solve for t.

( ) ( )

( ) ( )( )( )

4

4

2 1.03

ln 2 ln 1.03

ln 2 4 ln 1.03

ln 25.8624

4ln 1.03

t

t

t

t

=

=

=

= =

So, well double our money in approximately 5.8624 years.

(b) monthlyFrom the problem statement we can see that,

1210000 5000 0.12

100A P r= = = =

Remember that the value of rmust be given as a decimal, i.e.the percentage divided by 100. Also, forthis part we are compounding interest rate monthly and that means it will compound 12 times per yearand so we also then know that,

12m=

Plugging into the equation gives us,

( )12

120.1210000 5000 1 5000 1.01

12

tt

= + =

Using the techniques from the Solve Exponential Equationssection we can solve for t.

( ) ( )( ) ( )

( )( )

12

12

2 1.01

ln 2 ln 1.01

ln 2 12 ln 1.01

ln 25.8051

12ln 1.01

t

t

t

t

==

=

= =


(c) continuously

From the problem statement we can see that,12

10000 5000 0.12100

A P r= = = =

Remember that the value of rmust be given as a decimal, i.e.the percentage divided by 100. For this partwe are compounding continuously and so we wont have an mand will be using the other equation.

Plugging into the continuously compounding interest equation gives,



43/46

College Algebra

0.1210000 5000 t= e

Now, solving this using the techniques from the Solve Exponential Equationssection gives,

( ) ( )( )

( )

0.12

0.12

2

ln 2 ln

ln 2 0.12

ln 25.7762

0.12

t

t

t

t

=

=

=

= =

e

e


3. A population of bacteria initially has 250 present and in 5 days there will be 1600 bacteria present.

(a) Determine the exponential growth equation for this population.(b) How long will it take for the population to grow from its initial population of 250 toa population of 2000?

(a) Determine the exponential growth equation for this population.

We can start off here by acknowledging that we know the initial number of bacteria is 250 and so

0 250Q = . Therefore the equation is then,

( ) 250 k tQ t = e

Now, we also know that ( )5 1600Q = and plugging this into the equation above gives,( ) 51600 5 250 kQ= = e

We can use techniques from the Solve Logarithm Equationssection to determine the value of k.5

5

1600 250

1600

250

32ln 5

5

1 32ln 0.37125965 5

k

k

k

k

=

=

=

= =

e

e

Depending upon your preferences we can use either the exact value or the decimal value. Note howeverthat because kis in the exponent of an exponential function well need to use quite a few decimal placesto avoid potentially large differences in the value that wed get if we rounded off too much.

Putting all of this together the exponential growth equation for this population is,



44/46

College Algebra

1 32ln

5 5250t

Q

= e

Hint : Identify the given quantities, plug into the appropriate equation and use the techniques from an

earlier section to solve for t.

(b) How long will it take for the population to grow from its initial population of 250 to a population

of 2000?

What were really being asked to do here is to solve the equation,

( )1 32

ln5 52000 250

t

Q t

= = e

and we know from the Solve Logarithm Equationssection how to do that. Here is the solution work forthis part.

( )

( )( )

1 32ln

5 5

325

2000

250

1 32ln 8 ln

5 5

5ln 85.6010

ln

t

t

t

=

=

= =

e

It will take 5.601 days for the population to reach 2000.

4. We initially have 100 grams of a radioactive element and in 1250 years there will be 80 grams left.(a) Determine the exponential decay equation for this element.(b) How long will it take for half of the element to decay?(c) How long will it take until there is only 1 gram of the element left?

(a) Determine the exponential decay equation for this element.

We can start off here by acknowledging that we know the initial amount of the radioactive element is 100

and so0 100Q = . Therefore the equation is then,

( ) 100 k tQ t = e

Now, we also know that ( )1250 80Q = and plugging this into the equation above gives,

( ) 125080 1250 100 kQ= = e

We can use techniques from the Solve Logarithm Equationssection to determine the value of k.



45/46

College Algebra

1250

1250

80 100

80

100

4ln 1250

51 4

ln 0.0001785151250 5

k

k

k

k

=

=

=

= =

e

e

Depending upon your preferences we can use either the exact value or the decimal value. Note howeverthat because kis in the exponent of an exponential function well need to use quite a few decimal placesto avoid potentially large differences in the value that wed get if we rounded off too much.

Putting all of this together the exponential decay equation for this population is,

1 4ln

1250 5100t

Q

= e

Hint : Identify the given quantities, plug into the appropriate equation and use the techniques from anearlier section to solve for t.

(b) How long will it take for half of the element to decay?

What were really being asked to do here is to solve the equation,

( )1 4

ln1250 550 100

t

Q t

= = e

and we know from the Solve Logarithm Equationssection how to do that. Here is the solution work forthis part.

( )( )

1 4ln

1250 5

12

45

50

100

1 1 4ln ln

2 1250 5

1250ln3882.8546

ln

t

t

t

=

=

= =

e

It will take 3882.8546 years for half of the element to decay. On a side note this time is called the half-

lifeof the element.

Hint : This part is pretty much the same as the previous part.

(c) How long will it take until there is only 1 gram of the element left?

In this part were being asked to solve the equation,



46/46

College Algebra

( )1 4

ln1250 51 100

t

Q t

= = e

The solution process for this part is the same as that for the previous part. Here is the solution work forthis part.

( )( )

1 4

ln1250 5

1100

45

1100

1 1 4ln ln

100 1250 5

1250ln25797.1279

ln

t

t

t

=

=

= =

e

There will only be 1 gram of the element left after 25,797.1279 years.

Alg ExpLog Solutions

Documents

Transcript of Alg ExpLog Solutions