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### Base e and Natural Logarithms

- To solve exponential equations or inequalities involving base e, take the natural log (ln) of both sides.
- To solve natural log equations or inequalities, write each side as a power of e. Always check for extraneous solutions – values which result in taking the logarithm of a non-positive value in the original equation or inequality. Exclude such values from the solution set.

### Base e and Natural Logarithms

^{2x}− 36 = 0

- Step 1) Add 36 to both sides
- 6e
^{2x}= 36 - Step 2) Divide both sides by 6
- e
^{2x}= 6 - Step3 - Take the natural log of both sides
- ln(e
^{2x}) = ln6 - Step 4 ) Simplify the left side
- 2x = ln6
- Step 5) Solve

^{3x}− 55 = 0

- Step 1) Add 55 to both sides
- 5e
^{3x}= 55 - Step 2) Divide both sides by 5
- e
^{3x}= 11 - Step3 - Take the natural log of both sides
- ln(e
^{3x}) = ln11 - Step 4 ) Simplify the left side
- 3x = ln11
- Step 5) Solve

^{2}+ 24) − lnx − ln10 = 0

- Step 1) Re - write the left side of the eq. using prop. of logs
- ln(x
^{2}+ 24) − lnx − ln10 = 0 - ln([(x
^{2}+ 24)/x]) − ln10 = 0 - ln([([(x
^{2}+ 24)/x])/10]) = 0 - ln([(x
^{2}+ 24)/10x]) = 0 - Step 2)Raise both sides to e and simplify
- e
^{n([(x2 + 24)/10x])}= e^{0} - [(x
^{2}+ 24)/10x] = 1 - x
^{2}+ 24 = 10x - x
^{2}− 10x + 24 = 0 - Step 3)Solve by factoring and Zero Product Property
- x
^{2}− 10x + 24 = 0 - (x − 4)(x − 6) − 0
- x − 4 = 0
- x = 4
- x − 6 = 0
- x = 6
- Step 4 ) Check your solutions
- By inspection you can see that neither x = 4 or x = 6 will give you a negative inside any of the natural logs.
- Therefore, both solutions are valid.

^{2}+ 7) − lnx − ln8 = 0

- Step 1) Re - write the left side of the eq. using prop. of logs
- ln(x
^{2}+ 7) − lnx − ln8 = 0 - ln([(x
^{2}+ 7)/x]) − ln8 = 0 - ln([([(x
^{2}+ 7)/x])/8]) = 0 - ln([(x
^{2}+ 7)/8x]) = 0 - Step 2)Raise both sides to e and simplify
- e
^{n([(x2 + 7)/8x])}= e^{0} - [(x
^{2}+ 7)/8x] = 1 - x
^{2}+ 7 = 8x - x
^{2}− 8x + 7 = 0 - Step 3)Solve by factoring and Zero Product Property
- x
^{2}− 8x + 7 = 0 - (x − 1)(x − 7) = 0
- x − 1 = 0
- x = 1
- x − 7 = 0
- x = 7
- Step 4 ) Check your solutions
- By inspection you can see that neither x = 1 or x = 7 will give you a negative inside any of the natural logs.
- Therefore, both solutions are valid.

^{2}+ 12) − lnx − ln(7) = 0

- Step 1) Re - write the left side of the eq. using prop. of logs
- ln(x
^{2}+ 12) − lnx − ln7 = 0 - ln([(x
^{2}+ 12)/x]) − ln7 = 0 - ln([([(x
^{2}+ 12)/x])/7]) = 0 - ln([(x
^{2}+ 12)/7x]) = 0 - Step 2)Raise both sides to e and simplify
- e
^{ln([(x2 + 12)/7x])}= e^{0} - [(x
^{2}+ 12)/7x] = 1 - x
^{2}+ 12 = 7x - x
^{2}− 7x + 12 = 0 - Step 3)Solve by factoring and Zero Product Property
- x
^{2}− 7x + 12 = 0 - (x − 3)(x − 4) = 0
- x − 3 = 0
- x = 3
- x − 4 = 0
- x = 4
- Step 4 ) Check your solutions
- By inspection you can see that neither x = 3 or x = 4 will give you a negative inside any of the natural logs.
- Therefore, both solutions are valid.

^{ − 6x}− 8 < 28

- Step 1) Add 8 to both sides
- e
^{ − 6x}< 36 - Step 2) Take the ln of both sides and simplify
- lne
^{ − 6x}< ln36 - − 6x < ln36
- x > [ln36/( − 6)]

^{2x + 3}− 10 < 15

- Step 1) Add 10 to both sides
- e
^{2x + 3}< 25 - Step 2) Take the ln of both sides and simplify
- lne
^{2x + 3}< ln25 - 2x + 3 < ln25
- 2x < ln25 − 3
- x < [(ln25 − 3)/2]

^{ex + 2e}− 3e > 5p

- Step 1) Add 3e to both sides
- e
^{ex + 2e}> 5p + 3e - Step 2) Take the ln of both sides and simplify
- lne
^{ex + 2e}> ln(5p + 3e) - ex + 2e > ln(5p + 3e)
- ex > ln(5p + 3e) − 2e
- x > [(ln(5p + 3e) − 2e)/e]

- Step 1) Divide both sides by 5
- ln(5x − 5) ≤ 5
- Step 2) Take the e of both sides and simplify
- e
^{ln(5x − 5)}≤ e^{5} - 5x − 5 ≤ e
^{5} - 5x ≤ e
^{5}+ 5 - x ≤ [(e
^{5}+ 5)/5] - x ≤ 30.6826
- Step 3) Since you cannot have a negative inside the natural log, identify values of x that are not acceptabe
- 5x − 5 > 0
- 5x > 5
- x > 1
- Final solution would be

log

_{12}100

- We're going to use the change of base formula which states that
- log
_{a}x = [(log_{b}x)/(log_{b}a)] - log
_{10}100 = [(log_{12}100)/(log_{12}10)] - Since the 10 is not written

_{12}100)/(log

_{12}10)]

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Base e and Natural Logarithms

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Number e 0:09
- Natural Base
- Growth/Decay
- Example: Exponential Function
- Natural Logarithms 1:11
- ln x
- Inverse and Identity Function
- Example: Inverse Composition
- Equations and Inequalities 4:39
- Extraneous Solutions
- Examples: Natural Log Equations
- Example 1: Natural Log Equation 9:08
- Example 2: Natural Log Equation 10:37
- Example 3: Natural Log Inequality 16:54
- Example 4: Natural Log Inequality 18:16

### Algebra 2

### Transcription: Base e and Natural Logarithms

*Welcome to Educator.com.*0000

*Today we are going to talk about base e and natural logarithms.*0002

*And these are frequently used in real-world applications.*0006

*So, first, the number e: the number e is an irrational number that is approximately equal to 2.71828; it is called the natural base.*0010

*Exponential functions that use the base e are called natural exponential functions.*0024

*And these are used to model processes that grow or decay.*0029

*For example, something like population growth might involve the use of base e.*0033

*You can treat these like other exponents and other exponential equations, just recalling that it has a specific value.*0040

*And so, it is structured similarly to other exponential functions.*0047

*For example, f(x) = e ^{3x - 1} could be an exponential function using base e, or something like f(x) = 4(e^{x}).*0053

*When we work with logarithms to the base e, these are written with a special notation, ln(x).*0071

*We don't use log(x) in this case; we use ln(x); and this is called the natural logarithm.*0078

*So, when you see ln(x), really, the base is e.*0083

*But we don't write it that way; it is just written as ln(x).*0089

*And this is the inverse of the function f(x) = e ^{x}.*0100

*Therefore, since these are inverses, you end up with the identity functions.*0107

*Again, this is a composition of functions; if I am going to let...let's let f(x) equal ln(x), and I am going to rename this g(x) to keep things clear.*0111

*So, according to composition of functions, if these two are inverses, f composed with g should give me back x.*0130

*So, f composed with g would be f(g(x)), which would then equal f(e ^{x}); and this is ln(x), so it would be ln(e^{x}).*0140

*And the natural logarithm of x, and e to the x power, essentially cancel each other out, and we just end up with x, the natural log and this base e.*0156

*g composed with x should do the same thing--should give me g(f(x)), which is going to equal g(ln(x)), which equals e ^{ln(x)}.*0169

*And the base e and the natural log will cancel out; I will end up getting x back (it is the identity function).*0189

*And we are going to use this property as we solve equations involving base e and natural logs.*0195

*Briefly reviewing what the graph is going to look like: if I look at the graph of f(x),*0202

*this is going to look like the typical graph of a logarithmic function,*0207

*where it is going to increase as x becomes more positive, and the y-axis is going to function as an asymptote.*0211

*Therefore, this graph is going to approach x = 0, but it is never actually going to reach it; and so that domain is going to be restricted to positive numbers.*0240

*Whereas, if we are looking at the graph of g(x), this is going to look like the graph of a typical exponential function.*0251

*And it is going to increase: as x becomes positive, y will increase, and here the x-axis is functioning as an asymptote.*0260

*So, g(x) = e ^{x}.*0273

*When we are working with equations and inequalities involving powers of e, we can proceed as follows.*0280

*If we are working with the powers of e in an equation or inequality, we are going to use natural logs to solve them.*0288

*We talked earlier about using logarithms to solve exponential equations that have different bases.*0294

*We would find the common log of both sides; here, if the base is e, then you use the natural log.*0303

*If you have an equation or inequality that involves natural log, you can use powers of e.*0311

*If you have powers of e, use natural logs to solve; if you have natural logs, use powers of e.*0317

*All the properties we discussed previously for logarithms are valid (like the product property, the quotient property, and the power property).*0323

*In addition, when you are working with logarithms, you always have to be careful for extraneous solutions.*0331

*So again, we are going to talk about extraneous solutions and make sure that we are looking out for those*0336

*and that we check for those before we decide that the solution that we found is valid.*0342

*For example, if I am working with 2 ^{x + 6}, and that equals 7^{3x - 2},*0349

*earlier on in the course, we said that we would just find the common log of both sides;*0361

*and that is going to give me log _{7}(3x - 2).*0370

*And then, from there, we went about isolating x and went on down.*0376

*We are doing the same thing here; only now we are going to talk about natural logs.*0383

*For example, e ^{5x - 3} = 9: since I am talking about powers, I am going to use natural logs to solve.*0388

*So, I am going to take ln(e ^{5x - 3}) = ln(9), the natural log of 9.*0397

*Remember that using the inverse property, if I am taking ln(e to some power), I am going to get x back; this is the identity function.*0407

*That is very helpful, because I will just end up with 5x - 3 = ln(9).*0420

*This is a number with a specific value; again, you can use your calculator to find it.*0429

*It is so commonly used that there is going to be a button on your calculator for this.*0434

*Now, I just need to isolate the x; 5x = ln(9) + 3; x = ln(9) + 3, divided by 5: x will be isolated.*0438

*We started out with powers of e; and we used natural logs to solve.*0460

*Conversely, if I am given natural logs, such as ln(x + 3) = 4, I can take both sides and make them base e, and raise them to the power...*0465

*here it would be ln(x + 3); on the other side, I am going to get e ^{4}.*0483

*Again, I have inverses: e ^{ln} or ln(e) were inverses; these canceled out; I got 5x - 3.*0489

*The same here: e ^{ln(x + 3)}...I end up with just x + 3 = e^{4}.*0495

*Therefore, x = e ^{4} - 3; and this is a solution, because we know that e is a specific value (a little more than 2).*0504

*We take that to the fourth power and subtract 3 from it, and we have a specific value for x.*0511

*I did start out with a logarithm, though; so I have to make sure that I am not ending up with a solution that will give me something negative.*0519

*Let's go ahead and look back to check that: ln(x + 3)...and I am letting x equal e ^{4} - 3.*0526

*So, ln(e ^{4} - 3 + 3): this is ln(e^{4}, because these cancel out.*0532

*And I know that e is positive, that it is greater than 2; so I take that to the fourth power; that is positive as well.*0540

*So, this is a valid solution.*0546

*In the first example, I have 4 times e ^{4x} = 40; so I am working with a power of e, so I am going to use natural logs to solve.*0550

*To make this simpler, I am going to first divide both sides by 4.*0563

*Next, I am going to take the natural log of both sides.*0573

*And, according to my inverse property, ln(e ^{x}) = x; because these are inverses,*0579

*the identity function is going to give me ln(e ^{x}) simply equals x.*0587

*Therefore, I end up with 4x = ln(10), the natural log of 10.*0593

*I am going to isolate x now by dividing both sides by 4; so x = ln(10), divided by 4.*0602

*Again, as soon as I saw that I had a power of e, I knew that I could solve this using natural logs.*0616

*First, I am just dividing by 4 to make this simpler, taking the natural log of both sides, and then realizing that,*0623

*because these are inverses, I am going to end up with 4x = ln(10), and isolating the x.*0631

*This is a logarithmic equation where I can use the properties of logarithms to solve,*0641

*just as I did with earlier equations when we used other logs, base 10 logs, or other bases.*0647

*Remember that I can solve these as long as I get them in some kind of form.*0657

*Earlier we talked about log _{b}(x) = log_{b}(y) if x = y.*0663

*Well, if I get ln(x) = ln(y), these have the same base, so x = y as well.*0673

*Even though we use a different notation, ln, it is still just a log, only the base happens to be*0681

*a certain number that we are specifying, which is e, worth a little bit more than 2.*0688

*All right, so we are going to start out using our regular properties for logarithms to get this into a form that we can work with.*0695

*First, I have ln(x ^{2} + 12) - ln(x) - ln(8).*0707

*I want to combine these first two; and I can do that using the quotient property.*0713

*because ln(x ^{2} + 12), divided by x...all of this can be combined,*0718

*because if I am subtracting one log from another (I have the difference of two logs), I can just take the log of the quotient of those two.*0730

*Again, I have subtraction; so I want to use the quotient property once again.*0741

*This is going to give me ln(x ^{2} + 12, divided by x, divided by 8).*0746

*So, I use the quotient property again, because I had all this divided by this.*0753

*I want to simplify this complex fraction; so if I am dividing this numerator by 8, that is the same as multiplying this by 1/8,*0761

*which will give me ln(x ^{2} + 12, divided by 8x).*0773

*I just took this and multiplied by 1/8; that is going to give me x ^{2} + 12, divided by 8 times x, or 8x, equals 0.*0781

*Now, I got this far; and looking at this, I actually only have a log on one side.*0797

*So, this isn't going to work; so what is my other option?*0806

*Recall that, when I am working with natural logs, the easiest thing to do is going to be to take a power of e.*0809

*So, our strategy, if we have logs on two sides, is to use this.*0816

*If we have a log only on one side, the strategy is going to be to take the equivalent exponential expression.*0820

*So, I need to take both sides and make them e and raise them to this power and this power.*0826

*Let's go up here to the second column and continue on.*0835

*This is going to give me e ^{ln(x2 + 12, divided by 8x)} equals e^{0}.*0842

*I have inverses here; so they cancel out, and that leaves me with (x ^{2} + 12)/8x.*0855

*Any number to the 0 power is simply going to be 1; it doesn't matter if it is 4 to the 0 power, or e (which it happens to be--a little bit more than 2).*0864

*It is going to end up being 1: now I have something that I can solve.*0874

*x ^{2} + 12 equals...multiply both sides by 8x; I see I have a quadratic equation.*0878

*This is x ^{2} - 8x + 12 = 0; now I can factor; I have a positive here and a negative there,*0885

*because this is...actually, I have two negatives, because this is a positive, and this is a negative; so I have two negatives.*0900

*And my factors of 12 are going to be 1 and 12, 2 and 6, and 3 and 4.*0910

*And I know that 2 and 6 together...2 + 6 equals 8, so I am going to make this 6 and this 2.*0917

*That is x ^{2}; that is -2x + -6x gives me -8x; -6 times -2 gives me positive 12; so I factored correctly.*0924

*Using the product property, I know that x - 6 = 0; that would be a solution, so that is going to give me x = 6.*0935

*Or, if x - 2 = 0, I have a solution; and that gives me x = 2.*0943

*Now, I have two solutions; but I always want to make sure that they are valid solutions, since I started out working with logs.*0949

*So, I am going to go back up here and check them: ln(x ^{2} + 12) = ln...I will check the first one:*0958

*that is going to be 6 squared plus 12, and that is going to be 36 + 12ln(48); that is valid.*0967

*The second...I need to check 6 over here; ln(6); that is also going to be valid, because that is taking the log of a positive.*0977

*Now, let's check 2: ln(x ^{2} + 12) = ln(2^{2} + 12) = 12 + 4, so that is ln(16).*0986

*That is a positive number that I am taking the log of, so that is valid.*0999

*I am also going to check it here: ln(2) is positive; so both of these are valid solutions.*1003

*Both of the solutions I have are valid: x = 6 and x = 2.*1009

*Now, I am working with the power of e, so I am going to use natural logs to solve this.*1018

*But first, I am going to make it a little bit simpler by adding 9 to both sides.*1024

*So, I have an inequality; but I don't need this extraneous 9 on the left making things difficult, so e ^{-3x} < 12 + 9, which is 21.*1030

*I am going to take the natural log of both sides: ln(e ^{-3x}) < ln(21).*1042

*These are inverses; they essentially cancel each other out, leaving me with -3x < ln(21).*1054

*I am going to divide both sides by -3, and I am going to immediately switch the inequality symbol to >,*1063

*because I am dividing by a negative number, dividing both sides by -3.*1068

*So, it is going to give me -ln(21), divided by 3; so x > -ln(21)/3.*1073

*Again, I first just added 9 to both sides and made this a little bit simpler.*1084

*And then, I recognized that I am working with powers of e, so I can solve this by taking the natural log of both sides.*1088

*OK, here I have an inequality that involves logarithms; and it is the natural log, so I am going to keep in mind that I can use powers of e to solve this inequality.*1096

*First, let's just go ahead and divide both sides by 3 to make this a bit simpler, combining the constants on the right.*1105

*Now, I have ln(2x - 3) ≥ 8: I am going to take powers of e of both sides.*1118

*That is ln(2x - 3) ≥ e ^{8}.*1126

*Since these are inverses, they essentially cancel each other out--they negate each other--so I get 2x - 3 ≥ e ^{8}.*1137

*Now, all I have to do is solve for x: 2x ≥ e ^{8}, and add 3 to both sides, so this is going to give me e^{8} + 3.*1147

*Then divide both sides by 2: x is going to be greater than or equal to (e ^{8} + 3)/2.*1164

*But I am working with logs, so I need to check back and make sure I don't end up with something that is going to make this negative.*1174

*So, this says that x is greater than or equal to this; so the smallest this value of x will be is this.*1180

*So, if I check it for this, and it is OK, all of the larger values will be OK, as well.*1186

*ln(2x - 3): let's let x equal e ^{8} + 3, divided by 2; and if this is OK, the whole solution set is OK.*1191

*This gives me ln(2((e ^{8} + 3)/2) - 3).*1202

*This looks worse than it actually is, because the 2's cancel out: this gives me ln(e ^{8} + 3 - 3).*1212

*These cancel out; this gives me ln(e ^{8}).*1225

*And I know that, by the identity property, this is actually going to be 8, so I am fine.*1230

*I could also just say that I know that e is a specific number, and it is 2.*1236

*And so, if I take a number slightly more than 2 to the eighth power, I am definitely going to get something positive.*1241

*So, I am going to be taking the natural log of a positive number; and then these other values are going to be even greater, because this is the minimum value.*1249

*Today, we talked about natural logs and the base e, and talked about solving equations that involve powers of e, as well as natural logarithms.*1261

*Thanks for visiting Educator.com; I will see you next lesson!*1271

0 answers

Post by julius mogyorossy on February 20, 2014

I had a dream about a church where people went to worship mathematics.

0 answers

Post by Kenneth Montfort on February 27, 2013

I forgot to mention I was talking about example 2

1 answer

Last reply by: Dr Carleen Eaton

Fri Mar 1, 2013 11:23 PM

Post by Kenneth Montfort on February 27, 2013

How come at the beginning you didn't just take all the natural logs to the inverse e? I tried and the solutions came out as complex numbers...How do you know when to use properties and when not to use properties?

1 answer

Last reply by: Dr Carleen Eaton

Wed Jan 30, 2013 12:13 AM

Post by Jeff Mitchell on January 29, 2013

at ~13:30 you stated that (x^2+12 / x/8) is the same as (x^2+12)/(8x).

is it? I tried plugging a value of 3 for x just to test and U get;

3^2+12 / (3/8) = 21 / (3/8) = 56 (first equation)

3^2+12 / (8x) = 21 / 24 = < 1

wouldn't (x^2 + 12) / (x/8) = (x^2 + 12)*8 / x