Question

Can someone tell me if this is correct ..
1. Expand as a sum or difference: Log3(xy)^3
i got log3(x^3y^3)

Answer 1

@zepdrix

Answer 2

\(\log_{3}(xy)^{3}\) ?

Answer 3

yes thats the original problem

Answer 4

Well, all you have done is use a property of exponents, but you have not expressed the expression as a sum or difference. You're trying to use properties specific to logarithms.

Answer 5

\[Log3(x3y3)\] this is my answer

Answer 6

That is not correct. Do you know the various properties of logarithms?

Answer 7

no

Answer 8

There are properties very specific to logarithms that allow you to turn break apart the argument of a logarithm into sums or differences of various parts of the argument. The main properties you need to know (there are others which are special cases of these)

\(\log_{b}(xy) = \log_{b}(x) + \log_{b}(y)\)

\(\log_{b}(\frac{x}{y}) = \log_{b}(x) - \log_{b}(y)\)

\(\log_{b}(x^{a}) = a \log_{b}(x)\)

\(\log_{b}(x) = a \implies x^{a} = b\)

AN important note is that all of these assume that x and y are greater than 0 and that b is greater than 0 and not equal to 1

The first three are what you need for this problem.

Sorry for slow reply, I'm kinda lagging x_x

Answer 9

Can you kind of see how the first three properties might be applied to your problem?

Answer 10

it would be log3b( i dont know what be would b?)

Answer 11

The a, b, x, and y are just arbitrary values. Its just to show how the properties work. The b is the base (hence the letter b). In your problem, the base is 3. It's important to state the base because in all of these rules, the base has to stay the same/be the same in order to work.

Another way to state the first 3 is that the first one says that if two things are being multiplied together inside of a logarithm, then you can take each other and put it into its own separate logarithm, separated by a plus sign. The 2nd one says that if two things are being divided, each thing can be put into its own logarithm, separated by a minus sign. The 3rd says that if you have EVERYTHING inside of a logarithm raised to the same exponent that you can bring it down in front of the logarithm.

Some examples to show the usages of each:

\(\log_{3}(4xy) = \log_{3}(4) + \log_{3}(x) + \log_{3}(y)\)

\(\log{7}(5/y) = \log_{7}(5) - \log_{7}(y)\)

\(\log_{2}(x^{3}) = 3 \log_{2}(x)\)

All of these assume x and y are greater than 0.

And an example that uses all of those properties would be something like this:

\(\log_{2}(x^{2}/y) = \log_{2}(x^{2}) - \log_{2}(y) = 2 \log_{2}(x) - \log_{2}(y) \)

Do those examples make sense?

Answer 12

im so lost :(

Answer 13

Well, lets look at the first example I put.

\(\log_{3}(4xy) = \log_{3}(4) + \log_{3}(x) +\log_{3}(y)\)

Can you see what happened here?

Answer 14

you separated them so for mine itd be
log3(xy)3 = log4(x)+ log4(y) or something like that

Answer 15

Well, you will use that separation into pluses, but we want to deal with the exponent first. Doing what you did for your initial answer works fine.

\(\log_{3}(x^{3}y^{3})\)

With this we can split it up into addition. Do you know how?

Answer 16

log3(x3)+log3(y3)

Answer 17

Exactly

\(\log_{3}(x^{3}y^{3}) = \log_{3}(x^{3}) + \log_{3}(y^{3})\)

Now the last thing to do is to take care of the exponents. We don't want exponents in an expanded form. So we use this property:

\(\log_{b}(x^{a}) = a \log_{b}(x)\)

So with that, can you see what to do with your exponents?

Answer 18

wouldnt you just leave them?

Answer 19

Well, it's not usually what is wanted. When you expand logarithms, the idea is to have no exponents and to have no multiplication or division inside the logarithm. So we wouldn't want \(x^{3}\) or \(y^{3}\)

Answer 20

so we take away them? or in this case we leave them?

Answer 21

We want to move them. We don't get rid of them, that changes the expression. The exponent property lets us move exponents to the front.

\(\log_{2}(3^{4}) = 4 \log_{2}(3)\)

\(\log_{3}(x^{5}) = 5 \log{3}(x)\)

You notice that I dont get rid of the exponents, I just pull them in front.

Answer 22

Meant \(\log_{3}(x^{5}) = 5 \log_{3}(x)\) In the last one. I messed up the 3 in the base, lol.

Answer 23

so what does it look like now?

Answer 24

Well, right now we have

\(\log_{3}(x^{3}) + \log_{3}(y^{3})\)

So how would it look if the exponents were pulled in front? :)

Answer 25

log3(x3)+3log3(y)

Answer 26

You moved the exponent of y in front of its log, but you also need the exponent of x in front of its log.

Answer 27

3log3(x)+3log3(y)

Answer 28

Right, that would be your answer

\(3 \log_{3}(x) + 3 \log_{3}(y)\)

That's as far as you can go :)

Answer 29

can you help me with more like this i just want to make sure im doing it right???

Answer 30

Yeah, let's see what you have. Can try and work ya through it to help ya practice.

Answer 31

Log2 r 1/3 t

Answer 32

I wanna think you do
log2(r1/3) + log2(t1/3)

Answer 33

\(\log_{2}(r^{1/3}t)\) ?

Answer 34

yes

Answer 35

It'll work basically the same way. You have your exponent distributed to each variable, so now you can split things into two logs that are being added and then pull the exponent of r to the front of its log.

Answer 36

so what i did was wrong?

Answer 37

I didnt notice your post, I think it lagged and I missed it.

You shouldn't have an exponent on t. t started off without an exponent on it, so it won't gain one.

\(\log_{2}(r^{1/3}) + \log_{2}(t)\) would be correct. There is just one more step after that.

Answer 38

oh okay so log2(r1/3)+log2(t) what do we have to do afterwards?

Answer 39

Its the same last step as the previous question. Moving the exponent to the front. Remember, we don't want any exponents when we are done.

Answer 40

oh put it in front? oh okay
1/3log2(r)+log2(t)

Answer 41

Yes, that would be correct :3

Answer 42

Yay thank you !

Answer 43

You're welcome :)