Sign In | Subscribe
INSTRUCTORSCarleen EatonGrant Fraser
Start learning today, and be successful in your academic & professional career. Start Today!
Loading video...
This is a quick preview of the lesson. For full access, please Log In or Sign up.
For more information, please see full course syllabus of Algebra 2
  • Discussion

  • Study Guides

  • Practice Questions

  • Download Lecture Slides

  • Table of Contents

  • Transcription

  • Related Books

Bookmark and Share
Lecture Comments (18)

1 answer

Last reply by: Dr Carleen Eaton
Sat Nov 7, 2015 6:12 PM

Post by Peter Ke on October 24, 2015

For Example 2, shouldn't the answer be:

Log(base10)22 / Log(base10)7 ?

1 answer

Last reply by: Dr Carleen Eaton
Fri Nov 22, 2013 7:39 PM

Post by Juan Herrera on November 21, 2013

At 15:05, Log base 6 of 8 want to write as Log base 10
** It should be: Log base 10 of 8 divided by Log base 10 of 6

I believe the main purpose of this property is to change
difficult Logarithmic bases to a common (base 10) or Natural base (e)
to solve problems in a practical easy way.

1 answer

Last reply by: Dr Carleen Eaton
Fri Mar 1, 2013 10:59 PM

Post by Kenneth Montfort on February 27, 2013

Do the log properties not apply here to common logs. For example, if you have log 7/log 4, can't I just subtract 7 and 4 to find the final log value (e.g. log 3)? Or is it acceptable to just leave it in the form log 7/log 4.

1 answer

Last reply by: Dr Carleen Eaton
Sat Aug 25, 2012 12:34 PM

Post by Jorge Sardinas on August 14, 2012

in example one at 2x=log19/log3+3 your supposed to divide two by everything

1 answer

Last reply by: Dr Carleen Eaton
Thu Jun 21, 2012 9:56 PM

Post by Rob Escalera on June 19, 2012

It seems, that as long as you take the entire quantity, you can either divide or distribute and the answer will be the same.

1 answer

Last reply by: Dr Carleen Eaton
Thu Jun 21, 2012 9:55 PM

Post by Rob Escalera on June 18, 2012

In example I again, don't you need to distribute the quantity (2x-3) log3 ?

1 answer

Last reply by: Dr Carleen Eaton
Thu Jun 21, 2012 9:50 PM

Post by Rob Escalera on June 18, 2012

At 4:00, don't you need to distribute the (3x 4)log2 to become: 3xlog2 4log2 = log5?

Which should then be: 3xlog2 = log5 - 4log2 =>
x(3log2) = log5 - 4log2 => x = (log5 - 4log2)/3log2

3 answers

Last reply by: Dr Carleen Eaton
Sun Mar 11, 2012 7:21 PM

Post by Jimmi Aastrom on March 29, 2011

A short remark.
At around 5:30 I believe that a paranthases needs to be added arount the log-fraction and negative 4!?

1/3(Log5/Log2)-4 needs to be 1/3((Log5/Log2)-4)

Otherwise only part of the expression is devided by 3 which would be incorrect.

Common Logarithms

  • Solve exponential equations by taking the common log of both sides of the equation. You can then use a calculator to get a decimal approximation of the answer.
  • If you want to get a decimal approximation of a logarithmic expression, convert the log expression to a log expression to the base 10 using the change of base formula. Then use a calculator to evaluate the new expression.

Common Logarithms

Solve 109x − 6 = 2
  • Step 1) Take the common log of both sides
  • 109x − 6 = 2 = > log(109x − 6) = log(2)
  • Step 2) Bring the exponents down using properties of logs
  • log(109x − 6) = log(2)
  • (9x − 6)log(10) = log(2)
  • Step3: Distribute using distributive property
  • 9x*log(10) − 6(log(10) = log(2)
  • Step 4 - Add 6log(10) to both sides
  • 9xlog(10) = log(2) + 6log(10)
  • Step 4 - Divide both sides by 9log(10)
  • x = [(log(2) + 6log(10))/9log(10)]
  • Use a calculator to evaluate
x = 0.7001
Solve 10 − 6x − 3 = 84
  • Step 1) Take the common log of both sides
  • 10 − 6x − 3 = 84 = > log(10 − 6x − 3) = log(84)
  • Step 2) Bring the exponents down using properties of logs
  • log(10 − 6x − 3) = log(84)
  • ( − 6x − 3)log(10) = log(84)
  • Step3: Distribute using distributive property
  • − 6x*log(10) − 3(log(10) = log(84)
  • Step 4 - Add 3log(10) to both sides
  • − 6xlog(10) = log(84) + 3log(10)
  • Step 4 - Divide both sides by − 6log(10)
  • x = [(log(84) + 3log(10))/( − 6log(10))]
  • Use a calculator to evaluate
x = − 0.8207
Solve 9*38 − 6x − 7 = 41
  • Step 1) Add seven to both sides
  • 9*38 − 6x = 48
  • Step 2) Divide both sides by nine
  • 38 − 6x = [48/9]
  • Step 3) Take the common log of both sides
  • log(38 − 6x) = log[48/9]
  • Step 4) Bring the exponents down using properties of logs
  • log(38 − 6x) = log[48/9]
  • (8 − 6x)log(3) = log([48/9])
  • Step 5) Distribute using distributive property
  • 8*log(3) − 6x(log(3)) = log([48/9])
  • Step 6) - Subtract 8log(3) from both sides
  • − 6x(log(3)) = log([48/9]) − 8*log(3)
  • Step 7) Divide both sides by − 6log(2)
  • x = [(log([48/9]) − 8*log(3))/( − 6log(3))]
  • Use a calculator to evaluate
x = 1.0794
Solve − 3*186 − 8x + 10 = − 10
  • Step 1) Subtract 10 from both sides
  • − 3*186 − 8x = − 20
  • Step 2) Divide both sides by − 3
  • 186 − 8x = [20/3]
  • Step 3) Take the common log of both sides
  • log(186 − 8x) = log[20/3]
  • Step 4) Bring the exponents down using properties of logs
  • log(186 − 8x) = log[20/3]
  • (6 − 8x)log(18) = log([20/3])
  • Step 5) Distribute using distributive property
  • 6log(18) − 8xlog(18) = log([20/3])
  • Step 6) - Subtract 6log(3) from both sides
  • − 8xlog(18) = log([20/3]) − 6log(18)
  • Step 7) Divide both sides by − 8log(18)
  • x = [(log([20/3]) − 6log(18))/( − 8log(18))]
  • Use a calculator to evaluate
x = 0.6680
Solve − 4*156x + 4 + 9 = − 65
  • Step 1) Subtract 9 from both sides
  • − 4*156x + 4 = − 74
  • Step 2) Divide both sides by − 4
  • 156x + 4 = [74/4]
  • 156x + 4 = [37/2]
  • Step 3) Take the common log of both sides
  • log(156x + 4) = log[37/2]
  • Step 4) Bring the exponents down using properties of logs
  • (6x + 4)log(15) = log[37/2]
  • Step 5) Distribute using distributive property
  • 6xlog(15) + 4log(15) = log[37/2]
  • Step 6) - Subtract 4log(15) from both sides
  • 6xlog(15) = log[37/2] − 4log(15)
  • Step 7) Divide both sides by 6log(15)
  • x = [(log[37/2] − 4log(15))/6log(15)]
  • Use a calculator to evaluate
x = − 0.4871
Solve 9*66 − 10x + 1 = 67
  • Step 1) Subtract 1 from both sides
  • 9*66 − 10x = 66
  • Step 2) Divide both sides by 9
  • 66 − 10x = [66/9]
  • 66 − 10x = [22/3]
  • Step 3) Take the common log of both sides
  • log(66 − 10x) = log[22/3]
  • Step 4) Bring the exponents down using properties of logs
  • (6 − 10x)log(6) = log[22/3]
  • Step 5) Distribute using distributive property
  • 6log(6) − 10xlog(6) = log[22/3]
  • Step 6) - Subtract 6log(6) from both sides
  • − 10xlog(6) = log[22/3] − 6log(6)
  • Step 7) Divide both sides by − 10log(6)
  • x = [(log[22/3] − 6log(6))/( − 10log(6))]
  • Use a calculator to evaluate
x = 0.4888
Write using common logarithms:
log327
  • We're going to use the change of base formula which states that
  • logax = [(logbx)/(logba)]
  • log1027 = [(log327)/(log310)]
  • Since the 10 is not written
log27 = [(log327)/(log310)]
Write using common logarithms:
log636
  • We're going to use the change of base formula which states that
  • logax = [(logbx)/(logba)]
  • log1036 = [(log636)/(log610)]
  • Since the 10 is not written
log36 = [(log636)/(log610)]
Write using common logarithms:
log452
  • We're going to use the change of base formula which states that
  • logax = [(logbx)/(logba)]
  • log1052 = [(log452)/(log410)]
  • Since the 10 is not written
log52 = [(log452)/(log410)]
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

Common 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
  • What are Common Logarithms? 0:10
    • Real World Applications
    • Base Not Written
    • Example: Base 10
  • Equations 1:47
    • Example: Same Base
    • Example: Different Base
  • Inequalities 6:07
    • Multiplying/Dividing Inequality
    • Example: Log Inequality
  • Change of Base 12:45
    • Base 10
    • Example: Change of Base
  • Example 1: Log Equation 15:21
  • Example 2: Common Logs 17:13
  • Example 3: Log Equation 18:22
  • Example 4: Log Inequality 21:52

Transcription: Common Logarithms

Welcome to Educator.com.0000

We are going to continue our discussion of logarithms by talking about special logarithms.0002

And the first one is common logarithms.0008

First of all, what are common logarithms?0011

Common logarithms are logarithms to the base 10; and these are used very frequently in many real-world applications.0013

For example, the scale that is used to measure the magnitude of earthquakes is base 10 scale.0020

And the base 10 is not written: logs in base 10 are just written as follows: log(x).0027

So, if I were discussing log10(7), I could instead just write that as log(7).0037

And it is assumed that, if nothing is written here, then this is base 10--0046

in the same way that, if you took a square root of 4, we know that this really means this;0050

but we don't write the number (because it is so commonly used)--we don't write the number 2.0058

If we are talking about some other root (like the third root--the cube root), then we would write it.0064

The same idea here: if you don't see anything here, you can assume that it is a base 10 log.0068

Now, one application that we are going to have with this is to use logarithms to help us solve exponential equations.0075

In previous lessons, we used exponential expressions to help us solve logarithmic equations.0081

When we had a logarithmic equation with a log on only one side, we converted that to the exponential form.0087

Well, it works the other way as well: there are times when taking an exponential equation0094

and taking the log of the equation can actually help you to solve it; so let's talk about that right now.0098

If we are working with an exponential equation, the previous techniques we used involved having the bases be the same.0108

For example, given something such as an exponential equation 3x + 1 = 94x,0115

we were only able to solve this if we had the same base.0124

And sometimes I could convert it to the same base if I didn't have the same base, because 9 is equal to 32;0131

therefore, I would end up with 3x + 1 = 38x, and then from there, you can solve,0138

because if the bases are equal, then the exponents must be equal.0152

However, getting a little bit more advanced: you are going to run into situations like this: 23x + 4 = 5.0157

These are not the same base, nor can I easily convert them to the same base.0166

In this case, if both sides cannot be written as powers of the same base0169

with an exponential equation, solve the equation by taking the common logarithm of each side.0174

The common logarithm means the base 10 log, which I am just going to write without the 10 here, as is standard.0180

So, log (and that is actually base 10) of 23x + 4 = log(5).0192

Once I have it in this form, then I can solve: you treat this as you would any other equation, which is just by isolating x.0201

Now, let's first rewrite this, using the power property, so that we get rid of this exponent.0215

Let's just go ahead and write this as a coefficient: (3x + 4)log(2) = log(5).0223

I have my variable in here; I want to isolate the x, so I am going to divide both sides by log(2).0235

Keep in mind that these are just numbers--there are no variables in here.0245

log(2) has a specific value; log(5) has a specific value; I haven't found that value, but having my answer in this form is perfectly valid.0250

If you look on your calculator, there is a log button; and that log button, if it just says 'log,' is for the base 10 log,0257

although on some calculators you can specify other bases.0264

This is something you can easily find the value of.0267

So, by doing this, now I have the x much more freed up: I am going to subtract 4 from both sides.0270

And now, all I have to do is divide each side by 3.0281

And instead of writing this as a complex fraction, log(5)/log(2)/3, I can simply remember that this is the same as this.0286

So, if I said log(5)/log(2) times the reciprocal (which would be times 1/3)...I want to get rid of that complex fraction,0312

so I am going to write it like this, in a more simplified-looking form.0320

OK, therefore, I started out where I had an exponential equation where there were different bases.0330

And I couldn't convert them to the same base very easily at all.0336

So, I went ahead and just took the common log of both sides, then used the power property in reverse to get this over here as a coefficient.0339

Next, I divided both sides by log(2), because that is really just a number.0355

And I isolated the x on the left; and everything I have here, I either have a value for, or I can find a value for.0359

The same techniques that we used for solving exponential equations can also be used to solve inequalities--0367

exponential inequalities where we can't get the same base very easily at all.0376

Recall that, when you are multiplying or dividing both sides of an inequality, you need to make sure that you are not working with a negative.0381

And usually, it is obvious: in past lessons, we have known if we are dividing by -3 or multiplying by -3;0391

we need to flip the inequality sign--it is obvious.0396

But you have to be careful with a logarithmic expression, if you are taking the log of some number,0399

that you check and make sure that you are not ending up with a negative,0404

inadvertently, and then having the inequality symbol be in the wrong direction.0408

All right, so let's look at 43x - 5 > 32x - 6.0414

As we did with equations that were exponential equations with different bases, we can do the same technique with this inequality,0426

which is to go ahead and take the common log of both sides.0435

So, this new inequality is still valid, because I did the same thing to both sides of the inequality--0449

the same as if I had added a number to both sides or multiplied both sides by a number.0454

This is still a valid inequality: the relationship between the left and right sides is still holding up.0458

Next, I am going to use the power property to bring this out in front, because my goal is to isolate x.0465

And I can't isolate x when it is up there as a power.0470

3x - 5 times log(4) is greater than...bring this out in front...2x - 6 times log(3).0474

All right, next I am going to divide both sides by log(4).0487

Actually, we need to take one more step: we can't just go ahead and do that.0503

What I need to do is separate this out; I need to use the distributive property to split this out,0506

because what I really have here is 3xlog(4) (that is 3x times log(4)--we can look at this like this--it makes it much easier), minus 5log(4).0514

OK, and this is greater than 2xlog(3) - 6log(3).0540

Now that I have this, and it is split apart, I can go ahead and look at the expressions that do not contain variables.0554

This does not contain a variable, and neither does that; I want those all on the right, and my variable-containing expressions over on the left.0562

I am going to start out by adding 5log(4) to both sides.0572

My next step is going to be to subtract 2xlog(3) from both sides.0587

All right, what I have now is expressions containing variables on the left, and those with constants only are on the right.0605

And again, this is really just a number.0612

Let's go back up here and continue on.0615

And my next step is going to be to factor out x from this left side of the inequality.0618

And this is going to give me an x here, times 3log(4), minus 2log(3), is greater than -6log(3) + 5log(4).0628

So, looking at what I did going from here to here, I just factored out an x.0648

I pulled that x out of here and here, leaving me with this difference: 3log(4) - 2log(3).0653

Now that I have done this, I can isolate the x at last.0662

I just keep x on this side, and I divide (I am going to rewrite this with 5log(4) in the front0664

and the negative second, because it is more standard) by 3log(4) - 2log(3).0673

OK, so in order to solve this exponential inequality that had different bases,0686

I proceeded by first taking the common log of both sides,0692

then isolating the x (which was a little bit complicated).0697

I used the power property, and then I used the distributive property0700

to separate out this 3x and the 5, since this contains a variable and this doesn't; and the same here.0705

Then, I went ahead and factored out an x, and then just divided both sides by this expression.0713

So, I moved the constants to the right, and the variable-containing expressions to the left, and then divided.0724

Now, as I mentioned, you have to make sure that you don't end up dividing by something negative inadvertently.0728

So, at the division step (that was this), let's go ahead and take a look at this.0734

I know that I am OK, because the log of a number greater than 1 is positive.0739

So, if I take the common log of a number greater than 1, it is positive.0744

I know this is positive, and I know that the log of 4 is going to be bigger than the log of 3.0748

And I know that 3 is bigger than 2; so this is going to be larger than this, so I know I am working with something positive;0753

and I don't need to flip the inequality symbol--but that is important to check.0760

The last new concept we are covering in this lesson is going to be change of base.0766

There are times when you have a log given in one base, and you want to change it to another base.0771

Frequently, it is base 10 that you are wanting to change it to, but not always; it could be a different base.0778

And so, this change of base formula allows us to write a given logarithmic expression in a different base.0785

And it allows us to evaluate logarithmic expressions in any base by rewriting it using common logarithms.0792

And as I said, frequently what you want to change the base to is a base that is log base 10.0805

What this right here is, is the original expression.0813

This is actually the original expression; and what we do is divide that original expression by a log to the original base.0820

And what we take the log of is a value that is equal to the new base.0831

So, I take just the original expression and divide it by a log to that base;0835

but for this number, I use the new base that I want.0841

Illustrating this: let's say I had log6(8), and I want to write it as log10.0844

So, I want to write this as log10; I don't need to write the 10 here.0859

This with a new base would be equal to log6(8), the original expression,0870

divided by a log with the original base (base 6), but for the x value right here, I am going to use 10; that 10 is implied.0880

So, if I needed to, for some reason, change this to base 10, this is equivalent to log10(8).0892

You just need to learn this formula and follow it, and know that it is the original expression,0904

divided by a log of the original base; and you are taking the log of the value equal to the base you are looking for.0908

OK, in Example 1, we are looking at an exponential equation where the bases are not the same, and I can't easily get the bases the same.0922

So, I am going to use common logs to solve this.0930

I am going to start out by taking the common log of each side.0933

And I am allowed to do this, as long as I do the same thing to both sides.0938

Once I have it in this form, I want to isolate x; so I need to use the power property to get (2x - 3)log(3) = log(19).0944

To isolate x, I divide both sides by log(x), because my x is here; I want to split that away.0957

Next, I am going to add 3 to both sides, so that the 3 ends up on the right.0967

And then finally, I am going to divide both sides by 2.0980

Let's come up here in the second column: 2x = log(19)/log(3), divided by (2 + 3).0983

And to make this look at little better, we can rewrite this as...actually, the x is now isolated, because we divided by 2;0995

so x equals 1/2, because if I take log(19)/log(3), divided by 2, that would be the same as multiplying by the reciprocal, 1/2.1003

So, I was able to solve this, even though this exponential expression had different bases,1024

by taking the common log of each side, and then isolating x.1028

Write using common logarithms: log7(22).1036

I need to use my change of base formula, which is loga(x) = logb(x), divided by logb(a).1039

And here, a is the new base I want, and b equals the original base--the base that I already have.1050

I want to find an expression that is equivalent to an expression with the original base.1059

OK, log10(22): this is what I want, and I need to find an expression that is equivalent to that.1069

And this would be just the original, log7(22), divided by log with the original base;1080

and then, for a, I am going to use this base that I want.1090

So now, I have written this using common logarithms.1094

OK, again, I have an exponential equation in which there is no common base, and I can't easily get a common base.1104

So, I am going to take the common log of both sides to help me solve this.1111

Now, I need to isolate x; so I need to get the x out of the exponent--I am going to do that by using the power property.1122

And this becomes the coefficient (2x - 6)log(5) =...I am going to bring this out in front...(4x + 3)log(7).1128

Now that I have done this, I am going to write it like this to make it clear what needs to be done.1148

And we need to use the distributive property, because I want to split away these terms that have x in them.1154

I am going to multiply 2x times log(5), and then I am going to multiply -6 times log(5), 4x times log(7),1159

and remember, this is base 10 that we are talking about; and 3 times log10(7).1173

Next, I am going to get all of the variables on the left, and expressions containing constants only on the right.1183

And I am considering the log of 5 to be a constant, because we can find a specific value of that.1188

So, I am going to add 6log(5) to both sides.1192

The next thing I need to do is subtract 4log(7) from both sides, so I can get this on the left, because it does contain a variable.1209

So, 4x - 4x log(7); now I have my constants on the right and variables on the left.1218

The next thing to do: I want to isolate x, so I can factor out an x, because there is an x factor here and one here.1227

I pull that out; I leave behind 2log(5); I pull out the x; I leave 4log(7).1238

And I am almost there; all I need to do now is divide both sides by this expression, and I get x = 3log(7), plus 6...1247

and this is log...that is a 5, not an exponent of 5...all divided by 2log(5) - 4log(7).1262

Again, the technique is to take the common log of both sides, isolate x by first using the power property,1281

then the distributive property, adding and subtracting as needed to bunch all of the expressions and terms1288

containing x's on the left, constants on the right; factor out x, and divide both sides by this expression.1297

And now, I have x isolated, so I have solved for x.1307

This time, we are going to solve an inequality; and it is an inequality involving exponential expressions1314

without a common base, and where I can't easily find a common base.1322

So again, as I did with the exponential equation, I am going to take the common log of each of these.1327

My next step is going to be to isolate the x.1339

All right, so in order to do that, I need to get the x out of being an exponent, so I can use that power property.1346

4x - 5 times log(3) is less than...this becomes...(-3x + 5)log(6).1354

Further separating out the x, I need to use the distributive property: this is going to give me 4x...1365

this is equivalent to this...so 4xlog(3) - 5log(3) is less than -3xlog(6) + 5log(6).1372

As usual, we are going to move the variables to the left and constants to the right.1392

So, I am going to add 5log(3) to both sides; and now I have this x on the right, so I need to move this term to the left by adding 3xlog(6) to both sides.1397

I look on the left; I am still trying to isolate the x, and I see that these two terms have a common factor of x,1438

so I factor that out to leave behind 4log(3) + 3log(6) < 5log(6) + 5log(3).1444

Now, I am going to divide both sides by this expression; and I will have x isolated.1458

Now, I am done; but I just need to double-check about this step, because, whenever you multiply or divide1475

both sides of an inequality, you need to make sure that you are not multiplying or dividing by a negative number,1484

because if you are, you need to reverse the inequality symbol.1490

So, I am looking right here; and I divided both sides by 4log(3) + 3log(6).1494

However, I know that I am OK, because the common log of a number greater than 1 is positive.1501

So, this is positive; and I am adding that to something else positive; so I know that I am OK--that that comes out to be the log of a positive number.1508

That concludes this lesson about common logarithms on Educator.com.1517

Thank you for visiting!1523