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### 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

^{9x − 6}= 2

- Step 1) Take the common log of both sides
- 10
^{9x − 6}= 2 = > log(10^{9x − 6}) = log(2) - Step 2) Bring the exponents down using properties of logs
- log(10
^{9x − 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

^{ − 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

^{8 − 6x}− 7 = 41

- Step 1) Add seven to both sides
- 9*3
^{8 − 6x}= 48 - Step 2) Divide both sides by nine
- 3
^{8 − 6x}= [48/9] - Step 3) Take the common log of both sides
- log(3
^{8 − 6x}) = log[48/9] - Step 4) Bring the exponents down using properties of logs
- log(3
^{8 − 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

^{6 − 8x}+ 10 = − 10

- Step 1) Subtract 10 from both sides
- − 3*18
^{6 − 8x}= − 20 - Step 2) Divide both sides by − 3
- 18
^{6 − 8x}= [20/3] - Step 3) Take the common log of both sides
- log(18
^{6 − 8x}) = log[20/3] - Step 4) Bring the exponents down using properties of logs
- log(18
^{6 − 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

^{6x + 4}+ 9 = − 65

- Step 1) Subtract 9 from both sides
- − 4*15
^{6x + 4}= − 74 - Step 2) Divide both sides by − 4
- 15
^{6x + 4}= [74/4] - 15
^{6x + 4}= [37/2] - Step 3) Take the common log of both sides
- log(15
^{6x + 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

^{6 − 10x}+ 1 = 67

- Step 1) Subtract 1 from both sides
- 9*6
^{6 − 10x}= 66 - Step 2) Divide both sides by 9
- 6
^{6 − 10x}= [66/9] - 6
^{6 − 10x}= [22/3] - Step 3) Take the common log of both sides
- log(6
^{6 − 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

log

_{3}27

- We're going to use the change of base formula which states that
- log
_{a}x = [(log_{b}x)/(log_{b}a)] - log
_{10}27 = [(log_{3}27)/(log_{3}10)] - Since the 10 is not written

_{3}27)/(log

_{3}10)]

log

_{6}36

- We're going to use the change of base formula which states that
- log
_{a}x = [(log_{b}x)/(log_{b}a)] - log
_{10}36 = [(log_{6}36)/(log_{6}10)] - Since the 10 is not written

_{6}36)/(log

_{6}10)]

log

_{4}52

- We're going to use the change of base formula which states that
- log
_{a}x = [(log_{b}x)/(log_{b}a)] - log
_{10}52 = [(log_{4}52)/(log_{4}10)] - Since the 10 is not written

_{4}52)/(log

_{4}10)]

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

### 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

### Algebra 2

### 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 log _{10}(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 equation*0094

*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 3 ^{x + 1} = 9^{4x},*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 3 ^{2};*0131

*therefore, I would end up with 3 ^{x + 1} = 3^{8x}, 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: 2 ^{3x + 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 base*0169

*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 2 ^{3x + 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 4 ^{3x - 5} > 3^{2x - 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 front*0664

*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 property*0700

*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 log _{6}(8), and I want to write it as log_{10}.*0844

*So, I want to write this as log _{10}; I don't need to write the 10 here.*0859

*This with a new base would be equal to log _{6}(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 log _{10}(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: log _{7}(22).*1036

*I need to use my change of base formula, which is log _{a}(x) = log_{b}(x), divided by log_{b}(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, log _{10}(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, log _{7}(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 log _{10}(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 terms*1288

*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 expressions*1314

*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 divide*1475

*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

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.