Algebra 2 Properties of Logarithms

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We are going to continue our discussion of logarithms by exploring some properties of logarithms.0002

The first one is the product property: if you have a log_b(mn), that is equal to the log_b(m) + log_b(n).0008

And this might look familiar to you, or this general concept, from working with exponents.0022

Recall that, when you are multiplying two exponents with the same base, you add the exponents.0026

Since logs are a type of exponent, it stands to reason that there would be similarities.0032

So, you are going to see some similarities between the properties we are going to cover today and properties from working with exponents.0040

For example, log₉(10) = log₉...and I could break this out into factors, 5 times 2...0046

equals the sum of the log of this factor, plus this factor...so the sum of log₉(5) and log₉(2).0059

Or I could move, instead of from left to right, from right to left.0075

And instead of breaking it apart into 2 separate logs, I may have to combine it into a single log.0079

And combining into a single log is especially helpful when we move on to solving more complex logarithmic equations0084

than the ones we have seen in previous lessons.0091

For example, given log₆(5) + log₆(7), I may want to combine those.0095

And this would be equivalent to log₆...since these have the same base;0105

notice that they have to have the same base for this to work...of 5 times 7, which equals log₆(35).0110

This, of course, applies when we are working with variables, as well--with logs involving variables, which is a lot of what we are going to be doing.0120

So, log₈(x² - 36)...I could factor this into log₈(x - 6) (x + 6),0128

and then write it as the sum of log₈(x - 6) + log₈(x + 6).0141

And again, I can move from right to left: if I were given this, I could multiply these two and then combine this into a single logarithm.0149

The quotient property: again, you are going to notice similarities from your work with exponents.0160

If I have log_b of the quotient m/n, this is equal to the difference between the log of the dividend and the log of the divisor.0165

Given log₂(14/5), I could rewrite this as log₂(14) - log₂(5).0180

Again, this works when we are talking about logarithms with variables (algebraic logarithms).0197

log₃ of something such as (x² - 3)/y would be equal to log₃(x² - 3) - log₃y.0203

So, the log of the dividend minus the log of the divisor--that is the quotient property.0221

The third property we are going to cover is the power property.0232

This one is a bit different that what you have probably seen before.0236

But it states that log_b of m to the power p equals p times log_b(m).0238

So, if you see what happened: we took this exponent up here and moved it to the front to become a coefficient.0245

Again, you are going to find this property helpful when you are trying to simplify logarithmic equations,0262

or when you need to work with equations to solve them, to be able to move back and forth between these two forms.0270

For example, log₄(5³): I am going to take this 3 and move it to the front.0278

It is going to become the coefficient: 3log₄(5).0290

Or, another example using a variable: log₃(x⁶) =...I pull this out in front...6log₃(x).0297

Again, these properties can be used to solve logarithmic equations.0316

Earlier on, we talked about solving logarithmic equations where there was just a log on one side,0320

and situations where we had one log on each side of the equation.0326

Sometimes, however, you are given logarithmic equations where you have multiple logs on each side, sums and differences of logs...0331

And if they have the same base, then you can actually combine those, so that you end up with one log on each side.0339

And then, you can move on to use the techniques previously learned in order to solve those.0345

The product, quotient, and power properties are some extra steps that you might have to take before applying previously-learned techniques.0350

For example, given log₇(4x + 50) - 2log₇(3) = log₇(x + 1) + log₇(3),0357

I see that these all have the same bases, so I want to use the property where, if I get a log on one side0373

equal to the log of the other with the same base, then I can just say x = y.0388

But I need to get this into that form; so let's see what we can do to use these properties to combine them.0392

First, I see that I have a coefficient here; so I am going to use the power property.0400

And with the power property, we talked about how log_b(m^p) = plog_b(m).0404

So, I was moving that power to the front to yield this form.0419

I can do the opposite, though: I can take this coefficient and move it up here and make it a power again.0426

And I am going to do that: so, this first log I am just going to keep the same for now: log₇(4x + 50).0434

Here, I am going to turn this into log₇(3), and I am going to put this back over here and raise the 3 to the second power.0444

That is the only power I have; so I am just going to work with that for now.0458

And I know that 3² is just going to be 9, so I can change that to 9 in this step: it is log₇(9).0463

And then, I am going to apply the quotient property on the left, because I have the difference of two logs that have the same base.0482

Using the quotient property, I can rewrite this as the log₇(4x + 50) divided by 9.0491

Recall that the quotient property (this was the power property up here) told me that, if I have log_b(m/n),0502

that is going to equal log_b(m) - log_b(n).0519

So here, I am moving from right to left: I am taking these and combining them--that is what I am doing right here.0525

On the right, I am going to use the product property: the product property says that, if I have a log_b of a product,0533

that is equal to the sum of the logs of those factors (log_b(m) + log_b(n)).0543

So, I am going to go ahead and combine these two into a single log, log₇(x + 1)(3).0551

Now, I have gotten this in the form that I can actually solve, because I have a log on the left to base 7 and a single log in the right to base 7.0563

And if these are equal, then this expression must equal this expression.0571

Now, I just need to go ahead and solve: 4x + 50, divided by 9, equals...I am going to pull that 3 out in front: 3(x + 1).0575

So, I am going to multiply both sides by 9 to get 4x + 50 = 27(x + 1); 4x + 50 = 27x + 27.0587

I am going to go ahead and subtract a 4x from both sides to get 50 = 23x + 27.0603

Subtract a 27 from both sides: I will get 23 = 23x, so x = 1.0611

Now, I always need to check back and make sure that the solution is valid, because I want to make sure I don't end up taking the log of a negative number.0622

Checking for validity right here: I have log₇(4x + 50).0628

Let x equal 1, and check that: log₇(4(1) + 50); that is log₇(4 + 50), or 54.0636

So, that is OK; let's look right here: the other one I have to check is log₇(x + 1), which is log₇(1 + 1), and that is just log₇(2).0647

So, this one is valid; this is a valid solution.0661

We used this technique, but we had to take a bunch of steps prior to that in order to combine these into a single log on each side of the equation.0667

Before we go on to work some logarithmic equations, let's just talk about simplifying using the properties that we have already covered.0680

And when I am asked to simplify a logarithmic expression, that means that I want to get rid of quotients.0686

I don't want to be taking the log of any quotients, of any products, or of any factors raised to powers.0692

The first thing I am going to do is work with getting rid of this fraction bar.0700

And I am going to use the quotient property for that, because log₆(x⁴y⁵),0706

divided by this polynomial expression down here, is going to be equal to...let's just say0712

that this whole thing is going to be equal to the log base 6 of the dividend (the numerator), minus log base 6 of the divisor.0723

So, log_b(quotient) equals the log of the dividend, minus the log of the divisor.0752

All right, so I got rid of that fraction bar; I don't have any more quotients; but what I do have is a product.0768

I am going to apply the product property to just get rid of this right here.0774

Instead, I am going to say, "OK, I know that this equals the sum of log base 6 of this factor, plus log base 6 of this factor."0781

So, I have to remember that I have my negative sign out here; that is going to apply to each of these terms, once I split them apart.0799

That is log₆(x - 3)³ + log₆(y + 2)⁶.0806

All right, I still have one product (I don't have any quotients), so I am going to take care of this one next.0819

This is going to be equal to log₆(x⁴) + log₆(y⁵), minus0827

log₆(x - 3)³, plus log₆(y + 2)⁶.0841

Next, I am going to apply the power property.0849

And recall that the power property says that, if I have log_b(m^p),0853

that is going to be equal to p...this is going to turn into a coefficient...times log_b(m).0863

So, I am going to take these powers and move them out front; that is going to give me 4log₆(x⁴), plus0873

5 (that is going to go to the front, also) times log₆(y), minus (3 is going to go out in front)0883

3log₆(x - 3), plus...the 6 comes out in front...6log₆(y + 2).0894

The last thing I want to do is move this negative sign and apply it to each term in here.0906

And that is as far as I can go with simplifying: 4log₆...actually, this is gone now,0911

because we have pushed that out in front...plus 5log₆y, minus 3log₆(x - 3).0918

And then, the negative applies to this term, as well: minus 6log₆(y + 2).0929

And this is simplified as far as I can simplify it.0938

I look here, and I no longer have the logs of any quotients; I don't have the logs of any products;0942

and I don't have the logs of any terms or expressions raised to powers.0948

Now, instead of expanding out the expression, I am going to do the opposite.0956

I am asked to write this as a single logarithm: so I am going to instead compress this into one logarithm--the opposite of what I did in the last question.0960

Let's start with the power property: I know that, instead of 3log₄(x - 3),0970

I can rewrite this as log₄(x - 3), and I am going to turn this into an exponent and write it that way.0975

Plus log₄(x + 4): that is going to be raised to the fourth power.0985

Minus log₄(x + 7): there is no coefficient there to turn into an exponent.0995

There is here, though: log₄(x - 8); and this is going to be raised to the third power.1003

Now, I have the difference here, which means that I can combine this by using the quotient property.1010

I also, within these parentheses, have some addition; so I can use the product property to combine those.1019

And I want to start inside the parentheses; so let's go ahead and do that and use the product property1027

to combine this into log₄[(x - 3)³ (x + 4)⁴].1033

So, it's log base 4 of this times this, minus...here, also, I have a sum; so I can combine that1044

by taking the product of these two factors: log₄[(x + 7)(x - 8)³].1056

So, this is this minus this; I can actually leave these brackets off at this point, because I have a single log.1068

Now, we have to apply the quotient property, because we have a difference.1079

This is log₄ and log₄; and this is going to become the numerator--it is going to become the dividend.1089

(x - 3)³(x + 4)⁴, divided by (x + 7)(x - 8)³...this is all together.1098

I have written this as a single logarithm, first by applying the power property,1111

then by combining these two logs and these two logs, using the product property,1117

and then finally combining this log and this log, using the quotient property.1122

Now, we are going to apply what we have learned to actually solving logarithmic equations.1129

And we have a technique for solving an equation, as long as we have a single log on each side.1134

I need to combine this side, and I need to also get rid of this 1/2 out in front,1139

so that I can have something of the form log_b(x) = log_b(y),1145

and then use the property that y must equal x.1151

First, I am going to apply the power property to get rid of these coefficients.1156

log₉(2³) - log₉(2x - 1) = log₉(49^1/2).1161

So, I can do some simplifying: I know that log₉(2³)...that 2 cubed is 8, so that is log₉(8).1176

Now, 49^1/2 = √49, which equals 7; so I can write this as log₉(7)--it is already looking more manageable.1186

On the left, I need to combine these; and I can do that using the quotient property.1199

log₉...this is going to go in the numerator, and this will be the denominator.1203

So, this is log₉(x) divided by 2x - 1 equals log₉(7).1211

Once I am at this point, I have this situation, where I have the same base, and I only have one logarithmic expression on each side.1221

So, I can just say, "OK, x equals y, so 8 divided by 2x - 1 equals 7."1229

I am going to multiply both sides by 2x - 1, which is going to give me 8 = 14x - 7.1236

I am going to add 7 to both sides to get 15 = 14x, and then I am going to divide both sides by 14: so 15/14 = x, or x = 1 and 1/14.1250

It is important that I check the solution; so I want to make sure that I don't end up taking the log of a negative number.1265

And the only log here that has variables as part of it is this one, so I am checking log₉(2x - 1), and I am letting x equal 15/14.1271

log₉(2) times 15 times 14, minus 1...1284

Now, instead of figuring out this whole thing and subtracting all of that, all I have to do is say,1291

"All right, this is slightly more than 1; so 2 times slightly-more-than-one is going to be slightly more than 2."1297

If I take a value of slightly more than 2 and subtract 1 from it, I am going to be fine; I will have a positive number.1305

I will not be taking the log of a negative number, so this solution is valid.1314

You don't have to get the exact value; you just have to check it far enough to be sure that what you have in here,1320

what you are taking the log of, is not negative; it is greater than 0.1324

OK, Example 4: we are asked to solve a logarithmic equation, and we almost have this form, but not quite.1335

I can't use x = y, because I have this 1/2 here.1346

In order to combine these two, they have to have the same base.1352

In order for us to use the product property to combine these two logs into one, I have to somehow make 1/2 into log₁₆.1355

So, my goal is log₁₆--that I am going to turn 1/2 into that.1366

Well, let's think about this: if I think about my definition of logarithms, I have log_b(x) = y.1376

And what I want is log₁₆ of some number (but I don't know what number) is going to equal 1/2,1389

because then, if I have this, instead of writing 1/2 here, I will just write this.1395

Thinking of my definition of logarithms and how I can use that to solve an equation like this,1401

an equation with a log in one side: now I have a separate equation that I need to solve in order to solve this one,1407

just to substitute there: well, recall that log_b(x) = y if b^y = x.1414

So, I am going to solve this by converting it into its exponential form,1424

which is going to give me 16^1/2 = x.1431

This is the same as √16, which is 4; therefore, log₁₆(4) = 1/2.1438

These are equivalent; since these are equivalent, I can write that up there--I can substitute.1450

Now, my next step is to combine these two logs on the right, using the product property.1469

log₁₆(12x - 21) = log₁₆[4(x² - 3)].1476

Now, I have it in this form, and I can say, "OK, 12x - 21 = 4(x² - 3)."1491

And then, I just solve, as usual: this is going to give me 12x - 21 = 4x² - 12.1500

I have a quadratic equation; I need to first simplify; so let's subtract 12x from both sides.1512

I am going to set the whole equation equal to 0; that is -21 = 4x² - 12x - 12.1520

I am going to add 21 to both sides to get 0 = 4x² - 12x + 9.1527

I am going to rewrite this in a more standard form with the variables on the left.1540

Now, this is just a matter of solving the quadratic equation; and this is actually a perfect square: it is (2x - 3) (2x - 3) = 0.1546

And you can check this out to see that 2x times 2x is 4x²; 2x times -3 is -6x, plus -6x, is -12x; -3 times -3 is 9.1558

According to the product property, if 2x - 3 equals 0 (and these are the same, so I only have to look at one factor), then I will have a solution.1576

So, 2x = 3; if I solve for x, x equals 3/2.1585

x = 3/2; now, my last step--this is a potential solution--is that I need to go ahead and check it right here.1590

Let's let x equal 3/2; and we are going to insert that into log₁₆(12x - 21).1599

So, x equals log₁₆(12(3/2) - 21)...not x equals...log₁₆...this cancels;1608

this becomes 1, and this is 6; this is 6 times 3, minus 21; so this is log₁₆(18 - 21).1627

And that gives me log₁₆(-18); therefore, this solution is not valid.1639

And since that is the only solution I have, there is no solution to this equation.1654

The only solution I came up with was an extraneous solution.1664

I solved this by first converting it to this form; and I did that by setting up an equation1668

where I figured out what x had to be for me to write 1/2 in the form log₁₆.1676

And I used the technique of converting this to the exponential form 16^1/2 = x and solving for x.1682

That told me that log₁₆(4) = 1/2; so instead of writing 1/2 here, I wrote log₁₆(4).1690

Then I used the product property to combine the two logarithmic expressions on the right.1699

Then I used this property, and I set 12x - 21 equal to this expression, and solved, using factoring, a quadratic equation.1705

But then, I went and checked my solution and found that it was not a valid solution.1714

Thanks for visiting Educator.com; that concludes today's lesson.1720