Homework Help: Log sequence

1. Oct 18, 2012

zzinfinity

1. The problem statement, all variables and given/known data
Number 3 in the attached file

2. Relevant equations

Not sure

3. The attempt at a solution
Been staring at it for about an hour now and still don't have a clue. Can someone point me in the right direction? Thanks.

Attached Files:

• HW7.pdf
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Last edited: Oct 18, 2012
2. Oct 18, 2012

Staff: Mentor

-1- There is no attached file

-2- You *must* show some effort before you can get tutorial help here. That's in the Rules link at the top of the page.

3. Oct 18, 2012

zzinfinity

1. Sorry, I forgot to attach the file. Here it is .

2. Be nice berkeman. I showed a lot of effort. I tried my change of base formula modeling it in matlab and all sorts of other things. Just nothing yielded results worth noting here. I just need a hint.

4. Oct 18, 2012

Staff: Mentor

I am being nice

It would help if you showed some of that work here in the forum, so folks can see how you are approaching those questions...

5. Oct 18, 2012

Dick

If you want a hint, do what you say you are doing. Change all of the logs to the same base, say e. Show what you get. It will work.

6. Oct 18, 2012

zzinfinity

Thanks Dick, that was helpful! I was trying to change the bases to all be in terms of a but I like you idea of using natural log. I changed base of the first three using the equation,
logd(x)= ln(x)/ln(d). I then multiplied the first three terms and, after canceling every thing was left with ln(2a+3)/ln(2a).

Then for the last term ( the one with "b" in it) I changed the base to e and got ln(2b)/ln(2b-1). I pulled my exponents to the front on both my fractions and multiplied

ln(2a+3)/(a*ln(2)) * (bln(2)/ln(2b-1)). I canceled my ln(2) terms and was left with

b*ln(2a+3)/(a*ln(2b-1)

Can you tell if it simplifies further than this? I feel like it might but am not sure where. Thanks again for your help!

7. Oct 18, 2012

Dick

Ok, so you've got the first three factors and the last factor and you've figured out there is a lot of cancellation. Now you just have figure out what they mean by '.....'. What's the term BEFORE the last one if you write it out explicitly?

8. Oct 18, 2012

zzinfinity

I don't have any idea what is meant by '.......'. That's why I just multiplied it through. But I guess its not that simple, huh? :)

Basically I know the next to the last term is log2a+?(2a+?+1). So that means to me that 2b=2a+?

Am I thinking about that correctly? Even if I am, I'm still not really sure where to go from there. Is there a relationship between b and a I'm not seeing? I'm not sure how to write it out explicitly. I feel like I'm getting close though!

9. Oct 18, 2012

Dick

They left it up you to infer what '.....' means. I would say the next to last term must be $log_{2^b-2}(2^b-1)$ Maybe the cancellation runs all the way? Pick an a<b like a=1 and b=3 and run it through matlab.

10. Oct 19, 2012

zzinfinity

I get what you're saying. A friend of mine showed me this problem and I had no idea what was going on. Its been bothering me for a few days, but it has all come together. Thanks again for your help!