anonymous
  • anonymous
Can someone tell me if this is correct .. 1. Expand as a sum or difference: Log3(xy)^3 i got log3(x^3y^3)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@zepdrix
anonymous
  • anonymous
\(\log_{3}(xy)^{3}\) ?
anonymous
  • anonymous
yes thats the original problem

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
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.
anonymous
  • anonymous
\[Log3(x3y3)\] this is my answer
anonymous
  • anonymous
That is not correct. Do you know the various properties of logarithms?
anonymous
  • anonymous
no
anonymous
  • anonymous
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
anonymous
  • anonymous
Can you kind of see how the first three properties might be applied to your problem?
anonymous
  • anonymous
it would be log3b( i dont know what be would b?)
anonymous
  • anonymous
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?
anonymous
  • anonymous
im so lost :(
anonymous
  • anonymous
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?
anonymous
  • anonymous
you separated them so for mine itd be log3(xy)3 = log4(x)+ log4(y) or something like that
anonymous
  • anonymous
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?
anonymous
  • anonymous
log3(x3)+log3(y3)
anonymous
  • anonymous
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?
anonymous
  • anonymous
wouldnt you just leave them?
anonymous
  • anonymous
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 \(y^{3}\)
anonymous
  • anonymous
so we take away them? or in this case we leave them?
anonymous
  • anonymous
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.
anonymous
  • anonymous
Meant \(\log_{3}(x^{5}) = 5 \log_{3}(x)\) In the last one. I messed up the 3 in the base, lol.
anonymous
  • anonymous
so what does it look like now?
anonymous
  • anonymous
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? :)
anonymous
  • anonymous
log3(x3)+3log3(y)
anonymous
  • anonymous
You moved the exponent of y in front of its log, but you also need the exponent of x in front of its log.
anonymous
  • anonymous
3log3(x)+3log3(y)
anonymous
  • anonymous
Right, that would be your answer \(3 \log_{3}(x) + 3 \log_{3}(y)\) That's as far as you can go :)
anonymous
  • anonymous
can you help me with more like this i just want to make sure im doing it right???
anonymous
  • anonymous
Yeah, let's see what you have. Can try and work ya through it to help ya practice.
anonymous
  • anonymous
Log2 r 1/3 t
anonymous
  • anonymous
I wanna think you do log2(r1/3) + log2(t1/3)
anonymous
  • anonymous
\(\log_{2}(r^{1/3}t)\) ?
anonymous
  • anonymous
yes
anonymous
  • anonymous
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.
anonymous
  • anonymous
so what i did was wrong?
anonymous
  • anonymous
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.
anonymous
  • anonymous
oh okay so log2(r1/3)+log2(t) what do we have to do afterwards?
anonymous
  • anonymous
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.
anonymous
  • anonymous
oh put it in front? oh okay 1/3log2(r)+log2(t)
anonymous
  • anonymous
Yes, that would be correct :3
anonymous
  • anonymous
Yay thank you !
anonymous
  • anonymous
You're welcome :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.