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How to solve this logarithmic equation?

  1. Nov 15, 2015 #1
    1. The problem statement, all variables and given/known data
    For clarification a have posted the equation below as a picture file.


    2. Relevant equations
    log(a*b)= loga + logb
    log(a/b) = loga-logb
    log(a^n) = nloga


    3. The attempt at a solution
    I dont know how to start. I cant remember the rule for powering the logarithms if there is any. Any hint would do...
    Its got to do something with the fact that the numbers inside the logarithms go by some sort of a rule, 3 (+3), 6 (+6), 12 (+12), 24
     

    Attached Files:

    Last edited: Nov 15, 2015
  2. jcsd
  3. Nov 15, 2015 #2
    Hey diredragon, maybe you could try to post the picture file once more, I don't seem to be able to see it.
     
  4. Nov 15, 2015 #3
    My mistake sorry it seems it read error the last time. Can you see it now?
     
  5. Nov 15, 2015 #4
    Yes I can see it now, looks interesting, I'll have a go at it.
     
  6. Nov 15, 2015 #5

    Samy_A

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    Second one is a typo, should be
    log(a/b) = loga-logb

    Hints:
    ##2^{log_2x}=x##

    ##log_26=log_23+log_22=log_23+1##
    ##log_212=log_23+log_24=log_23+2##
    and so on ...
    If you calculate the right side expression in terms of ##log_23##, it simplifies nicely.
     
    Last edited: Nov 15, 2015
  7. Nov 15, 2015 #6
    Yup, missed that
     
  8. Nov 15, 2015 #7
    Yeah, i think i got it, the numbers do play a role as i can take out log2(3) for the other members as they all then can be if i apply the product rule like log2(6) = log2(2) + log2(3) = 1 + log(3). Ill have a go at it a bit later but ill appreciate it if you can try it so i can compare my results.
     
  9. Nov 15, 2015 #8

    Samy_A

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    In case you want to compare with my result:
    I got ##2^A=72## by simply calculating the right side expression using the hints given in my previous post. Maybe there is a smarter (and quicker) way to solve it.
     
  10. Nov 17, 2015 #9
    I'm curious, could you share exactly what you did when you say you got your answer by 'calculating the right side expression'? I feel like I've missed a short-cut and taken the long way around: I replaced log2(3) with x, so: 1/6( (x)^3 - (x + 1)^3 - (x + 2)^3 + (x + 3)^3 ), evaluated that, and ended up with 2x + 3.

    So that left me with 2^[2(log2(3)) + 3]. I feel like that can be simplified, but I don't know how. I plug it into a calculator and I got the answer you got, so I know everything I've done so far is correct (but probably inefficient...)
     
    Last edited: Nov 17, 2015
  11. Nov 17, 2015 #10
    2^(2log2(3) + 3) = (2^3)*2(log2(3^2)) = 8*3^2 = 72
    If it helps n*logb(c) = logb(c^n)
     
  12. Nov 18, 2015 #11

    Samy_A

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    That's exactly what I did (there may be a shortcut that escaped me).

    You end up with ##A=2*\log_23+3##
    Then, using the properties of the logarithm given previously, and ##2^{log_2y}=y##:
    ##2^A=2^{(2*\log_23+3)}=2^{\log_23^2}*2^3=3^2*2^3=9*8=72##
     
  13. Nov 18, 2015 #12
    It does seem like there is a shorter way out. Maybe there is a formula that calculates the sum of n consecutive cubes?
     
  14. Nov 18, 2015 #13
    For some reason I couldn't see that 2^log2(3^2) was the same thing as 3^2... It was easier to see when I turned the 3^2 into 9.

    I just needed to remember that logarithms and exponents with the same base cancel each other out (which is the definition of a logarithm... Kind of important).

    Thank you!!

    Still curious if there's a faster way, though.
     
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