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suppose that x=log2^5 and y= log10^9. express log2^3 in terms of x and y

I need to see work, answer is 1/2y(x+1)

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Arthur D. | Effective Mathematics TutorEffective Mathematics Tutor
5.0 5.0 (9 lesson ratings) (9)
I noticed your problem. If you still need help repost it and I'll do it for you. Here are some hints:
log(10)9=log(2)9/log(2)10 (change of base)
log(2)[5*2]=log(2)5+log(2)2 (notice log(2)5=x in the original problem and log(2)2=1)
Good Luck !
Arthur D.


I'd like to pass along a comment to you that a very experienced tutor shared with me on the answers pages.
As tutors, it is professional courtesy for us to not answer over each other.  If you have a clarification or a correction of something that another tutor has said, it is more respectful to post your thoughts as a comment in response to the original answer, rather than re-answer the question, which replaces another tutor's response.  Some tutors use the answer forums as a potential way to attract new students, so answering over one can interfere with that.
Please consider that in the future,
-- Michael
I don't want this to become an argument but to me it looked like the student still didn't know how to do the problem so I offered some suggestions. I did not do it to upstage you in any way. I just gave the student a little bit more information to help her. If you stay on this forum long enough you will see that many, many problems are done the same way several times by several tutors. I've done problems with one solution and found it done again and again by other tutors. After seeing this I felt the same way that you probably did:
I felt like writing,"Did I do it wrong ?" just to be sarcastic. However, I don't say anything if the problem is done over and over by other tutors. That is their privilege. Like I said, with your problem you really didn't do the problem and, in my opinion, the student was still having difficulty. I just gave her a few more hints to help her solve it.Ordinarily I do the entire problem out but because you gave her some hints, I didn't do it out completely. I gave her a few more hints. Please don't take my offering hints personally. I really did not try to upstage you.
Arthur D.
I always appreciate dialogue in the answer area, so additional information or clarifications seem like a good idea to me.  As a professional courtesy, the additional clarification can be offered as a comment in response to the original tutor's answer, rather than as a completely new answer which interferes with the original tutor's efforts to reach out to the student.
The feedback I was provided on this etiquette came from an extremely experienced tutor, so I highly regarded his perspective and thought it was worth passing along.
Rather than discuss it here, I suggest I open a discussion over on the forums, about tutor etiquette on the Answers pages, and see what kind of opinions there are out there regarding best practices.  Keep an eye open for it,
-- Michael
Michael W. | Patient and Passionate Tutor for Math & Test PrepPatient and Passionate Tutor for Math & ...
5.0 5.0 (1432 lesson ratings) (1432)
Hi Jenny from Chicago...Michael from Itasca here.
So, first, I think we need to make sure we understand the problem.  Those numbers after the logs are their bases, right?
x = log25 and y = log109
And we're looking for log23.  That's our target.
Here are some hints.
1.  I notice that y is the log of 9, and our target has a log of 3 in it.  Can't we write y as some form of a log of 3, using one of the properties of logs.
2.  If we can do that, then we still have a problem, because y is written in base 10, and our target is in base 2.  But can't we fix that?  What would our target be in logs of base 10?  Can you convert the base?
3.  If we can convert the base of our target, then we should end up with a log3 somewhere, which is related to y.  If you figured out Step 1, that is.  :)
That takes care of the log3 part...but when we convert the base of our target, we're going to end up with a log2 to deal with, too.  In fact, you'll have 1/log2 if you did it right.
4.  Don't we know something about 1/logba?  It's not one of the top four properties of logs, but it's right behind the most common ones. If we can turn the log102 into a log with base 2, then it's going to start looking a lot like x, because x is a log of base 2, also.
6.  In fact, x = log25.  If we had that exact log, then we'd have x, and we'd be done.  But that never happens, does it.  :)  How about if we had some sort of variation of that, like log2(5*n).  Couldn't we use another property of logs to split out the part that looks like x?
7.  And if we could do that, wouldn't it be convenient if "n" were a number that we could deal with, like a power of 2?  Because then, we'd be able to figure out what the log2n is, too, using yet another property of logs.
Between those hints, and the fact that you know the answer, can you piece together the puzzle?
Hope this helps more than hurts,
-- Michael


I think I can I think I can I think I can. And yes you interpreted the question correctly, those are the correct logs/bases to start with.  So.... Basically going to have to use almost every rule of logs and change base twice? 
Yes.  I believe changing the base of the target will help get part of it into a form that relates to y (log3 vs log9), and then you should end up with 1/log2 somewhere.  Of course, that doesn't look like it helps you at all, but there's a property of logs that'll help you change it into something with a base-2 log...which at least has the same base as x, so it's getting you closer.
-- Michael