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Contents

- 1 Notes for Lesson 32: Log Properties
- 1.1 Log Input and Output
- 1.2 Sums with x
- 1.3 Sums with x
- 1.4 Guessing Logs
- 1.5 Guessing Logs
- 1.6 Finding Factors and Summing
- 1.7 Finding Factors and Summing
- 1.8 Logs of 2, 3, and 6
- 1.9 Logs of 2, 3, and 6
- 1.10 Product Property
- 1.11 Product Property
- 1.12 Predicting Logs
- 1.13 Predicting Logs
- 1.14 Quotient Property
- 1.15 Quotient Property
- 1.16 Logs of 5
- 1.17 Logs of 5
- 1.18 Get Rid of the Exponent
- 1.19 Get Rid of the Exponent
- 1.20 Power Property
- 1.21 Power Property
- 1.22 Properties of Logs
- 1.23 Properties of Logs
- 1.24 Expanding Expressions
- 1.25 Radical Exponents
- 1.26 Radical Exponents
- 1.27 Quotient Property Expansion
- 1.28 Quotient Property Expansion
- 1.29 No More Exponents
- 1.30 No More Exponents
- 1.31 Make the Coefficient Equal 1
- 1.32 Make the Coefficient Equal 1
- 1.33 Quotient Property Condensing
- 1.34 Quotient Property Condensing
- 1.35 Squish It Down
- 1.36 Squish It Down
- 1.37 MUST Have the Same Base
- 1.38 MUST Have the Same Base
- 1.39 Change of Base Formula
- 1.40 Change of Base Formula
- 1.41 Switch to Base 10
- 1.42 Switch to Base 10
- 1.43 Calculate
- 1.44 Calculate
- 1.45 Convert and Calculate
- 1.46 Convert and Calculate
- 1.47 Log Buttons
- 1.48 Log Buttons
- 1.49 From Base 2 to Base 10
- 1.50 From Base 2 to Base 10
- 1.51 Changing or Constant
- 1.52 Changing or Constant
- 1.53 Multiplying by a Positive Constant
- 1.54 Multiplying by a Positive Constant
- 1.55 True and False Equations I
- 1.56 True and False Equations I
- 1.57 True and False Equations II
- 1.58 True and False Equations II
- 1.59 True and False Equations III
- 1.60 True and False Equations III

In the last lesson, we looked at the logs that I had written here, and I asked you to see if you could come up with any links between the numbers. Now that you've had a chance to think about that for awhile, I'd like to give you a chance to document your thoughts. Please write in this box any ideas you have about the connection between the numbers we've input into the logs in the numbers that are output. Also just to remind you, these three numbers right here are rounded. So this is just a chance for you to make a guess and it's not going to be graded.

Well one thing you might have noticed, is that the three numbers here, log base outputs. The three numbers here, log base 10 of 2, log base 10 of 20, and log base 10 of 200, all end in the same decimal, which I've rounded to 0.30102. So now we've looked at the numbers to the right at the decimal point. But what about the numbers in front of the decimal? Let's just focus on these numbers for a second. Let's say that log based 10 of 2, or 0.30102, is equal to x. I'd like you to rewrite these other two numbers, log base 10 of 20 and log base 10 of 200 as sums involving x.

Well, if this number equals x, then this number equals 1 plus x, and this one is 2 plus x.

Let's look at these numbers again for a second. Remember that,looking for patterns and making educated guesses about what's going on. And then, figuring out if those guesses are right or wrong is really what math is all about. So, why do we have 1 plus x here? And why do we have 2 plus x here? Let's look at a new log for a second. Log base 10 of 3, and I'm going to tell you right now, this is about equal to 0.47712. Just to make it perfectly clear that this is rounded, I'm going to change this to an about equal sign. Now, using what log base 10 of 3 is about equal to, what do you think we can get for log base 10 of I'd like you to make a guess about what each of these is equal to. Then use a calculator to get the actual answer, and round to 5 decimal places. You won't be graded on your guesses so just try it out. Then I'd like you to also consider what log this 10 of 3000 is. And then lastly how about log based 10 of . So make your guesses and then check them with a calculator.

So here is what all of our numbers are equal to. I hope your guesses were close. I think looking at this last log down here, log base 10 of 3 million, gives us a clue as to what's happening. Notice that this is equal to 6.47712, and we're taking the log of 3 times 10 to the 6th. So we have a 6 here and we have a 6 right here. Noticing that this decimal here is the same as for log base 10 of 3, it looks like what we have is log base 10 of 3, accounting for the .47712, plus log base 10 of 1 million. Since this is just 10 to the 6th. That means that this is where we get that other 6 from, to make up this whole answer.

So what we saw in the answer to the last question, was that log base 10 of 3 million, which is equal to log base 10 of 3 times 10 to the 6th is equal to log base 10 of 3 plus log base 10 of 10 to the 6th. So what we've done here, is split the input to our log into factors. And then we summed the logs of those factors. Over here, I've written out the other answers we came up with in the last question. So what I'd like to know now is, does this process of finding factors of the input of the log and then summing the logs of those factors, work for these other examples? Please answer yes or no.

The answer here is yes. We know that log base 10 of 3, is 0.47712. For these three logs, what we have is that decimal, plus the number that is the power of

Let's see if this summing of the logs of factors works for other factors, apart from just powers of 10. We've already seen these 2 logs before. I've just written more decimal places out this time. We have log based 10 of 2, and log based 10 of 3. Note that both of these are still rounded numbers. Using these two values. What do you think log base 10 of 6 is? First I would like you to make a guess and then I would like you to find the actual answer with a calculator. Please find both of these to to 5 decimal places.

Log base 10 of 6, is about equal to 0.77815. And, just as we suspected, this is what we get if we add log base 10 of 2 to log base 10 of 3.

Let's take a look at this identity right here. We have log base a of AB is equal to log base a of A plus log base a of B. Now, is this true in general for any base a? And indeed, it is. The formal name of this, is the product property. And since we know this is a property of logs in general, logs of any base, we know that it is true for the natural logarithm as well. Put simply, the log of a product is the sum of the logs of the numbers that were multiplied to make that product. Let's use this rule, but in reverse to find log base 10 of 2 plus log base 10 of 5. First, simplify this into the log of one number, and then tell me what that's equal to. Please try this without a calculator first, but definitely feel free to check it to make sure it, that it works with your calculator.

If we want to add log base 10 of 2, and log base 10 of 5. Then we're going to get log base 10 of 10. Since we need to multiply 2 and 5, just like the product property says. However, we know that log base 10 of 10 is just equal to 1. Pretty cool, huh? 1 is definitely a lot simpler than all of this.

Let's look at the inverse of multiplication, division. What if we want to represent the log of a quotient? We know that division is just repeated subtraction, so using that knowledge, how do you think we might represent natural log of 3 over 7? Remember, ln is just the same as log base e, so in this slot, please make a prediction of how you would rewrite ln of 3 over 7 in terms of natural logs. Think back to the product property, and how this might be different for division instead of multiplication.

Did you predict ln of 3 minus ln of 7? If so, you were right. Let's check out the values of each of these and see what we get. First, let's find natural log of 3 over 7. So we press the ln button on the calculator, and then 3 divided by that down. So, now we've got one side of our equation into number form. Let's check out the other side now on the calculator. Okay, so we want ln of 3 minus ln of 7. I've got it plugged in right, so let's press equals. And sure enough, we get exactly the same decimals we have over here. So, our calculator has helped us confirm this prediction. ln of 3 over 7 is equal to ln of 3 minus ln of 7.

So we're in great shape right now. We just came up with another strategy to help us solve some problems using logarithms. As you might have guessed, this is called the quotient property, and yes. If it works for natural logs, then it also works for logs of any other base. The quotient property says that if we have log base a of A over B. Then that's equal to log base a of A minus log base a of B. And again, this works for any base a that we can plug in. So let's use the quotient property, but in reverse to find this. Log base 2 of 24 minus log base 2 of 3. First, write this as a single log expression, log base 2 of some number. And then tell me what that's equal to, just as a number.

Log base 2 of 24 minus log base 2 of 3 should, according to the quotient property, equal log base 2 of 24 over 3. And 24 over 3 is just 8, so this is log base 2 of 8. However, we can rewrite 8 as 2 to the 3rd power, and we know that log base 2 of 2 to the 3rd is just 3, 3 is the exponent we need to take a helped us simplify a log expression. Instead of log base 2 of 24, minus log base 2 of 3, we just have the number 3.

We saw before that log base 10 of 2 times 3 is equal to log base 10 of 2, plus log base 10 of 3. So that means that if we have log base 10 of 5 times 5 then, according to the product property, this should equal log base 10 of 5, plus log base 10 of 5. However, we're just adding 2 log based 10 of 5s together. And we can see that this is equal to 2 times log base 10 of 5, or 2 log base 10 of 5. Considering what we've just written, how do you think you could rewrite log base 10 of 5 times 5 times 5? Please follow this same pattern that we did up here. So what belongs in this slot should be similar to what's in this form. And this answer should be in the same form as this one.

One thing we can do to simplify this expression, is to start out by inserting one new set of parentheses around this first 5 times 5. That way, just like we've done before, we're multiplying two things together and we know how to use the product rule in a situation like this. This is just going to be equal to log base 10 of 5 times 5 plus log base 10 of 5. Conveniently, we've already found this. That's what we got up here. Log base 10 of 5 times 5 is just log base 10 of 5 plus log base 10 of 5. And then of course, we have to keep this last, log base 10 of 5 added on to the end as well. Now we have three log base

We just saw that log base 10 of 5 times 5 is equal to 2 log base 10 of 5. And log base 10 of 5 times 5 times 5 is equal to 3 log base 10 of 5. However, there's a way we could rewrite the inputs to these two logs. This first one is just equal to log base 10 of 5 squared. And this one is equal to log base 10 of How could you rewrite this without using any exponents?

Following the pattern we see up here, and should be the coefficient in front of log base 10 of 5. So we should have n log base 10 of 5. This property that we've been using is called the power property, and as we can see, it allows us to write the log without the exponent of the input.

Since ln is just another notation for log base e, the power property also works for the natural logarithm. So, let's try another problem. How should we write ln of x to the fourth without any exponents?

According the power property, this should just equal 4lnx or 4 times the natural log of x.

Here are the three properties again that we've been discussing, the Product Property, the Quotient Property, and the Power Property. Many logarithmic expressions can be rewritten or expanded or condensed using these properties. First, let's think about what it means to expand a logarithmic expression. That basically means that we're going to break down something that looks complicated into smaller pieces. This logarithmic equation is a great one to work with for doing this. We have log base 3 of 6x squared y cubed. Let's break it down, we can expand this to log base 3 of 6 plus log base 3 of x squared plus log base 3 of y cubed. Which property that you've learned about allowed us to do this expansion? Was it the product property, the log of the product is equal to the sum of the logs? Was it the quotient property, the log of the quotient is equal to the difference of logs? Or was it the power property, the log of something raised to a power is equal to the power times just the log?

The answer is the product property. 6x squared y cubed is 6 times x squared times y cubed. So, we can sum the logs of each of these factors.

You can use the power property to rewrite each of these terms. Log base 3 of x squared, according to the power property, should be 2 times log base 3 of x. In the same way log base 3 of y cubed can be written 3 times log base 3 of y.

You've had some practice expanding logarithmic expressions, so now let's try one that has a natural log. Here I have ln of the square root of 4x minus 6 over 3. Now, eventually, we can try expanding this using the quotient property, but first, all I'd like us to do, is just rewrite the numerator here using an exponent instead of the radical sign.

Remember that when we take the square root of a number, we're actually raising that number to the 1 half power. So we have natural log of 4x minus 6 to the 1 half over 3.

Now let's use the quotient property to expand this expression further.

We can rewrite this as natural log of 4x minus 6 to the 1 half, minus the natural log of 3.

We have just one more step to fully expand this expression. The only change I want us to make, is that I'd like us to get rid of this exponent here. Think about what rule you can apply to make this go away, and then use it to rewrite the expression. We're not going to change the last term at all, so I'll just write it for you already. What should the first term become?

We can rewrite this first term, natural log of 4x minus 6 to the 1 half, using the power property. That tells us that this is equal to 1 half times the natural log of 4x minus 6. That means having started off with the natural log of the square root of 4x minus 6 over 3, we've expanded this to equal 1 half ln of 4x minus 6, minus ln of 3. Awesome!

We have been expanding expressions, so now let's condense them. This is effectively the reverse of expanding. How about we play around with this expression? 3 natural log of x plus 4 minus the natural log of 5. Now a common mistake in trying to condense this, would be to start off using the quotient property, but we're not ready for that just yet. To do that we would need both terms to just be the log of something, and not have a coefficient in front. But how do we get rid of this coefficient, please rewrite this first term so that the coefficient is no longer there.

We can use the power property to transform the first term we have here, to become ln of (x+4) cubed. Note that the cube here applies to the entire input of the log. It goes round the whole expression (x+4).

Now it's time to finish off our combining of expressions. Please use the quotient property to combine the two natural logs that we have here. What do we end up with?

Since we're subtracting two logs right here, and they have the same base, the quotient property tells us we can combine them into one log expression, by dividing them as the input of one log. So having started out with 3 natural log of (x+4) minus the natural log of 5 we've ended up with this simpler, more compact thing-- natural log of (x+4) cubed, divided by 5.

You're doing a great job with this so let's just keep going. Here's another expression for us to play with. Now I'd like you to use all three log properties that we've learned to condense this expression as much as you can.

In order to condense this expression the first thing I'm going to do is condense within each term. To do that I need to use the power property so that we can get rid of these coefficients. Doing that gives us- log of x cubed plus log of y squared minus log of z to the one half. However I also know that z to the one half is just the square root of z. So I think I prefer to write it that way. Now we've gotten rid of our coefficients we're finally allowed to use our product and quotient properties. From the product property we know that log of x cubed plus log of y squared is equal to log of x cubed y squared. Since we're subtracting log of the square root of z from that, I can use the quotient property as well. The numerator of our fraction is going to be the input of the first log, and the denominator is going to be the input of the log we're subtracting. So that means that in the end we've condensed our original expression- to be just this little compact thing right here- log of x cubed y squared over the square root of z. Awesome!

Note that in all the examples we've done so far, we've always used the same base for our logs. The 3 properties that you've learned about. The product property, the quotient property, and the power property. Only apply when the base of the logs that are being combined is the same. This is a super important concept. And we really need to make sure to keep it in mind. That we don't do anything that's illegal. Thinking about this important concept, which of these pairs of logs down here can we combine using the product property?

Of the four pairs of logs down here, there are only two sets that we can combine. Log base e of 4 plus ln of 5, and log base 10 of 5 plus log base 10 of because the bases are the same for the two logs we're trying to add. Here, we have log base e and natural log. But we know that ln is actually shorthand for log base e. So both of these actually have the same base and we can combine them. The same is true for the logs down here. Log base 10 of 5 and log base 10 of 6, have the same base. Since remember, that's the number that goes down here as the subscript of log. The two remaining choices can't be combined because the bases of the logs we're trying to add are different. Here we have log base number. So, we can't combine these. The same concept applies here. We have a base of 2 and a base of 10, so we can't use the product property.

We know that we can not use the product property to combine either of these pairs of logs, since remember they have different basis. However, let's try grabbing our calculator and seeing if we can add them. To add log base 10 of 4 plus natural log of 5. I start out by pressing the log button, since I know this has an implicit invisible ten down here as a base. You want base 10 of 4 plus natural log, so the ln button of 5. And there we go, we have our number. Let's write it down. So, according to our calculator, log base 10 of 4 plus natural log of 5 is about equal to 2.211 so on and so forth. Let's try our second combination here. So, I've got my calculator all ready, and I just need to enter log base 2 of 5. But wait, there's no log base 2 button. What am I going to do? Well, I could use trial and error. But that would be really, really hard. There must be something easier we can do. Luckily there is. There's something called changing bases that we can do. What this means is that we'll convert log base 2 of 5 to something with a different base that we can evaluate using a calculator. So this is a formula we can use to change a log from being base a to base b. We have log base a of x equals log base b of x over log base b of a. This probably feels a little bit confusing right now and kind of abstract. So, let's try using it to see how it really works. Let's do this first off with something we know the answer, just to make sure that our method makes sense. I like us to change log base 2 of 8 to something in base

Log base 2 of 8 equals 3. Since remember 8 is equal to 2 to the third power.

Now let's actually actually use our new formula. Please write log base 2 of 8 in terms of logs of base 10.

If we look at our formula, and match up the different letters here to the different numbers over here, we have that a equals 2, the first base here. X the input of the first log is equal 8, and b, the new base we're trying to change to is equal 10. Since we want log base b of x over log base b of a, we need to write that log base 2 of 8 is equal to log base 10 of 8 over log base

Now that we've written log base 2 of 8, as the ratio of, things with log base do that, and tell me what number you get.

This is equal to 3. Lets check it out on the calculator just to be sure. So here's my calculator, and I plug in log of 8, which means of course log base 10 of 8 divided by log base 10 of 2. And, sure enough that equals 3. This is what we got before, so our change of base formula must work. Awesome.

Let's try another problem where we need to you use our change of base formula. This time, I'd like you to switch log base 5 of 9 to something in log base 10. Then, evaluate that using a calculator, to tell me what number this is equal to.

Here's our change of base formula again. And if we use that, we can convert log base 5 of 9 to equal log base 10 of 9, over log base 10 of 5. This is calculator appropriate, so let's plug it in. Log base 10 of 9 divided by log base 10 of 5 is equal to this decimal. This is great. Now we know how to find a number that pretty much any log expression with numbers is equal to.

No our calculator has two log buttons. It has the plain log, which means log base 10, and it has an LN button for a natural log button, base E. That means that converting to natural log is also really useful to us, if we're going to change bases. Converting to base E is just as useful as converting to base 10, since we can also plug this in on our calculator. So converting from base five to base e, or in other words, converting to natural logs, what numbers belong in these spots right here. And when you use a calculator, what number does this expression evaluate to? Note that all I'm changing from this step to this step is I'm rewriting log base e as ln.

Log base 5 of 9 is equal to log base e of 9 over log base e of 5. Or in other words ln of 9 over ln of 5. Now, it's time to bring out our calculator. So we plug in ln of 9 divided by ln of 5. And here we go, we get the same decimal that we did before. 1.36521238897 and continuing. This is, of course, rounded. Great. The important thing is that this number is equal to what we had when we converted to base 10 instead base e. That means that regardless of what we set b our new base equal to, the original expression log base 5 of 9 has this value. Awesome.

We haven't talked about graphs at all, so let's think about them for a minute. How do you think graphs of logarithmeic functions of different bases relate to one another? Well, here's just an example, but this line right here shows the graph of Y equals log base 10 of X. And this one shows Y equals log base 2 of X. Since we've been talking about change of base so much. How about we try to convert log base two of X to an expression with log base 10? How would you rewrite this with this change of base?

Using our change of base formula, log base 2 of x is equal to log base 10 of x over log base 10 of 2.

Right now, we have that log base 2 of x is equal to log base 10 of x over log base 10 of 2. But there's another way that we can rewrite this that I think could help us out. Instead, we could write 1 over log base 10 of 2 times log base 10 of x. Now, what can we say about the value of 1 over log base 10 of 2? Does it change, or is it constant?

The value of 1 over log base 10 of 2 is constant, and according to our calculator it's about equal to 3.3.

Once again lets look at our two graphs. This one shows y equals log base 10 of x. And this one shows y equals log base 2 of x. Which we just saw is about equal to 3.3 times log base 10 of x. The only difference between these functions now is just the coefficient out in front of log base 10 of x here. So what does multiplying a function by a positive constant do to its graph? Does it stretch or shrink it in the y-direction? Does it move it in some direction, either horizontally or vertically? Does it reflect the graph in the x or y axis? Or does it find the graph's inverse?

The first choice is correct. Multiplying a function by some positive constant stretches or shrinks the graph in the Y direction, which is vertically here. Since 3.3 is greater than one we have a stretch in this case. We can also see that in our graph. The distance from the X axis up to the curve for Y equals log base ten of X. Is just a fraction of the distance from the x axis up to our second curve. 3.3 times log base 10 of x. This curve is identical to this one, just stretched upward. It's taller, interestingly enough then even the change of base of a log is a transformation. That is really cool.

You've learned a lot of rules with some details involved in them in this lesson. So I think it would be good if we just did a check of your understanding of these properties. Here are three equations, and by checking the values of the left and right hand side of each of them. I'd like you to decide which ones are correct.

Let's look at what each side of each of these equations simplifies to, to figure out which equations are true. How about we start with this equation up here. And let's go with the left hand side first. Inside the parentheses here, we have 100 times 1000 which is 100,000, or 10 to the 5th power That means that this side of the equation is equal to log base 10 of 10 to the 5th, and we know that that is just equal to 5. Great. Now, on the other side we have log base 10 of 100 plus log base 10 of 1000. So, this is the same as log base 10 of 10 squared. Plus log based ten of ten cubed. That's just two plus three, or five. So, this equation does check out. This side is equal to this side, which means it is a true statement. Awesome. Let's go for our second one. We know that log based ten of 100 is 2 and we know that log based ten of 1,000 is three. So that means that this side is equal to 2 times 3 or 6. Now over here, we're adding those two numbers we just multiplied. Log base 10 of 100, plus log base10 of to 5, so this is not a true statement. And our third equation, finally. On the left hand side we have log based 10 of 1,100. Let's see using our calculator, what that equals. Log of 1,100 equals 3.04139 and so on. Let's write that down, I'll just write some of the numbers, we know that it's rounded on the calculator anyway. On to the right hand side. Now notice that this expression is exactly the same as the one we already evaluated. On the left-hand side of the previous equation. So we know that this, log base 10 of 100 times log base is not true either. The first equation we had is the only example of the product property being used properly. The two numbers that we multiplied and then plugged in as the input of our first log expression, each act as the inputs for these separate logs over here that are then added together. We have to be super careful about which numbers are multiplying and which quantities we're adding when we use the product property. Great job.

Here's another problem similar to the last one. Just to keep checking your understanding of these log properties. Once again, figure out the value of the expression on either side of each equation. And then use that information to figure out which of these equations is true.

Lets evaluate each side of each equation, starting with this one up here. Log base 10 of 100 is 2, and log base 10 of 1000 is 3. So, we have 2 over 3 for this side. On the right-hand side we have log base 10 of 100 again, or 2, and then, minus 3 for log base 10 of 1000, this equals negative 1. Unfortunately for this equation, two thirds is not equal to negative 1, so this is not a true statement, we need to put a not equals sign here. Let's move over here to this equation. The left-hand side right now is log base 10 of 100 over 1000. So lets look at whats inside the parentheses and simplify that. 100 over 1000 is 1 over negative 1 is just negative 1. Great. What about the right-hand side? Well this is the same as what we had over here. And we got negative 1 for that. Sure enough negative 1 is equal to negative 1. So, this equation is true. Awesome. Now what about this last one here? Well, once again lets start with what's inside the parentheses here. We have log base 10 of 100 minus 1000. Which is the same as log base 10 of negative 900. Hm, I'm not sure what this is so how about I pull out the calculator and we'll see what this equals. So we have log base 10 of negative 900, and look at that we get an error. I wonder why that is? Well actually, we know why. We're trying to take the log of a negative number, and you learned before that, that is not allowed. This is just not equal to a real number. On the right-hand side, we have what we had over here, two thirds. Well, two thirds is a real number, so that means this equation is not true. This second one, log base 10 of 100 over 1000, is equal to log base being used properly of these three equations. This is a great check to show that the quotient property actually works. And I hope it's also pointed out some easy ways that you might accidently make a mistake when you're trying to use it.

And finally one more question for you. Once again, which of these equations is correct? Figure this out by checking the value of the left hand side and the right hand side for each one of them.

Let's start with our first equation over here. If log based 10 of 100 all taken to the third power, equals 3 times log base 10 of 100. So let's see what each side is equal to. Well first you need to figure out what log base 10 of 100 is. And it's two, since 100 is 10 squared. That means on the left-hand side, we have 2 cubed, or 8. And on the right-hand side, we have 3 times 2, or 6. 8 is definitely not equal to 8, so this equation is not true. Let's move on to our second equation. Log base 10 of 100 cubed is equal to 3 times log base 10 of have log based 10 of 10 squared taken to the 3rd power. We know from our rules of exponents that this is just log bas 10 of 10 to the 6th. We need to multiply the 2 and the 3, and that just simplifies to 6. Awesome. What about the right-hand side? Well once again we make use of the fact that log base 10 of for this last one over here. Log base 10 of 100 cubed, equals 3 times log base parentheses here around the 100 cubed. However in this case the parentheses don't actually change the way we're evaluating the expression. The first thing we did over here was find what 100 cubed was equal to and that was 10 to the Once again the right-hand side is the same as it's been the other 2 times and we know that we got 6 before. So look at that, this time we have 2 true equations. Wonderful. Both of these show proper uses of the product rule. And in both cases the exponent is only applying to the input of the log expression. It doesn't go around the entire log expression, which is why this first one was not correct.