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INSTRUCTORSCarleen EatonGrant Fraser
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Lecture Comments (6)

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

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

Solve 6e2x − 36 = 0
  • Step 1) Add 36 to both sides
  • 6e2x = 36
  • Step 2) Divide both sides by 6
  • e2x = 6
  • Step3 - Take the natural log of both sides
  • ln(e2x) = ln6
  • Step 4 ) Simplify the left side
  • 2x = ln6
  • Step 5) Solve
x = [ln6/2] = 0.8959
Solve 5e3x − 55 = 0
  • Step 1) Add 55 to both sides
  • 5e3x = 55
  • Step 2) Divide both sides by 5
  • e3x = 11
  • Step3 - Take the natural log of both sides
  • ln(e3x) = ln11
  • Step 4 ) Simplify the left side
  • 3x = ln11
  • Step 5) Solve
x = [ln11/3] = 0.7993
Solve ln(x2 + 24) − lnx − ln10 = 0
  • Step 1) Re - write the left side of the eq. using prop. of logs
  • ln(x2 + 24) − lnx − ln10 = 0
  • ln([(x2 + 24)/x]) − ln10 = 0
  • ln([([(x2 + 24)/x])/10]) = 0
  • ln([(x2 + 24)/10x]) = 0
  • Step 2)Raise both sides to e and simplify
  • en([(x2 + 24)/10x]) = e0
  • [(x2 + 24)/10x] = 1
  • x2 + 24 = 10x
  • x2 − 10x + 24 = 0
  • Step 3)Solve by factoring and Zero Product Property
  • x2 − 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.
x = 4;x = 6
Solve ln(x2 + 7) − lnx − ln8 = 0
  • Step 1) Re - write the left side of the eq. using prop. of logs
  • ln(x2 + 7) − lnx − ln8 = 0
  • ln([(x2 + 7)/x]) − ln8 = 0
  • ln([([(x2 + 7)/x])/8]) = 0
  • ln([(x2 + 7)/8x]) = 0
  • Step 2)Raise both sides to e and simplify
  • en([(x2 + 7)/8x]) = e0
  • [(x2 + 7)/8x] = 1
  • x2 + 7 = 8x
  • x2 − 8x + 7 = 0
  • Step 3)Solve by factoring and Zero Product Property
  • x2 − 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.
x = 1;x = 7
Solve ln(x2 + 12) − lnx − ln(7) = 0
  • Step 1) Re - write the left side of the eq. using prop. of logs
  • ln(x2 + 12) − lnx − ln7 = 0
  • ln([(x2 + 12)/x]) − ln7 = 0
  • ln([([(x2 + 12)/x])/7]) = 0
  • ln([(x2 + 12)/7x]) = 0
  • Step 2)Raise both sides to e and simplify
  • eln([(x2 + 12)/7x]) = e0
  • [(x2 + 12)/7x] = 1
  • x2 + 12 = 7x
  • x2 − 7x + 12 = 0
  • Step 3)Solve by factoring and Zero Product Property
  • x2 − 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.
x = 3;x = 4
Solve e − 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)]
x > − 0.5973
Solve e2x + 3 − 10 < 15
  • Step 1) Add 10 to both sides
  • e2x + 3 < 25
  • Step 2) Take the ln of both sides and simplify
  • lne2x + 3 < ln25
  • 2x + 3 < ln25
  • 2x < ln25 − 3
  • x < [(ln25 − 3)/2]
x < 0.1094
Solve eex + 2e − 3e > 5p
  • Step 1) Add 3e to both sides
  • eex + 2e > 5p + 3e
  • Step 2) Take the ln of both sides and simplify
  • lneex + 2e > ln(5p + 3e)
  • ex + 2e > ln(5p + 3e)
  • ex > ln(5p + 3e) − 2e
  • x > [(ln(5p + 3e) − 2e)/e]
x > − 0.8330
Solve 5ln(5x − 5) ≤ 25
  • Step 1) Divide both sides by 5
  • ln(5x − 5) ≤ 5
  • Step 2) Take the e of both sides and simplify
  • eln(5x − 5) ≤ e5
  • 5x − 5 ≤ e5
  • 5x ≤ e5 + 5
  • x ≤ [(e5 + 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
1 < x ≤ 30.6826
Write using common logarithms:
log12100
  • We're going to use the change of base formula which states that
  • logax = [(logbx)/(logba)]
  • log10100 = [(log12100)/(log1210)]
  • Since the 10 is not written
log100 = [(log12100)/(log1210)]

*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

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) = e3x - 1 could be an exponential function using base e, or something like f(x) = 4(ex).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) = ex.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(ex); and this is ln(x), so it would be ln(ex).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 eln(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) = ex.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 those0336

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 2x + 6, and that equals 73x - 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 log7(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, e5x - 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(e5x - 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 e4.0483

Again, I have inverses: eln or ln(e) were inverses; these canceled out; I got 5x - 3.0489

The same here: eln(x + 3)...I end up with just x + 3 = e4.0495

Therefore, x = e4 - 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 e4 - 3.0526

So, ln(e4 - 3 + 3): this is ln(e4, 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 e4x = 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(ex) = x; because these are inverses,0579

the identity function is going to give me ln(ex) 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 logb(x) = logb(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 be0681

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(x2 + 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(x2 + 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(x2 + 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(x2 + 12, divided by 8x).0773

I just took this and multiplied by 1/8; that is going to give me x2 + 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 eln(x2 + 12, divided by 8x) equals e0.0842

I have inverses here; so they cancel out, and that leaves me with (x2 + 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

x2 + 12 equals...multiply both sides by 8x; I see I have a quadratic equation.0878

This is x2 - 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 x2; 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(x2 + 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(x2 + 12) = ln(22 + 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) ≥ e8.1126

Since these are inverses, they essentially cancel each other out--they negate each other--so I get 2x - 3 ≥ e8.1137

Now, all I have to do is solve for x: 2x ≥ e8, and add 3 to both sides, so this is going to give me e8 + 3.1147

Then divide both sides by 2: x is going to be greater than or equal to (e8 + 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 e8 + 3, divided by 2; and if this is OK, the whole solution set is OK.1191

This gives me ln(2((e8 + 3)/2) - 3).1202

This looks worse than it actually is, because the 2's cancel out: this gives me ln(e8 + 3 - 3).1212

These cancel out; this gives me ln(e8).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