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Thread: Inequality with logarithm, help needed

  1. #1
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    Inequality with logarithm, help needed

    $\displaystyle 8n^2 < 64n* log_2 n$

    So I continue:
    $\displaystyle n< 8* log_2 n$
    $\displaystyle n< log_2 n^8$
    $\displaystyle 2^n < n^8$

    Not sure if this is correct procedure at all. Anyway, I'm stuck here...
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  2. #2
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    What are you trying to do here?

    Are you trying to solve this inequality for $\displaystyle n$? Or are you trying to prove that this statement is true for all $\displaystyle n$?
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    I'm trying to find the values of n, where the statement is true. So the answer would be in the form: n < some number

    I did set up a table in excel. I know the answer is between 43 and 44. That is for n= 43, 2^n is smaller. But for n=44, n^8 is smallest.

    But would like to know how the procedure is to get to that result.
    Last edited by Lars91; Aug 14th 2010 at 07:47 AM. Reason: forgot a word
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  4. #4
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    Quote Originally Posted by Lars91 View Post
    $\displaystyle 8n^2 < 64n* log_2 n$

    So I continue:
    $\displaystyle n< 8* log_2 n$
    $\displaystyle n< log_2 n^8$
    $\displaystyle 2^n < n^8$

    Not sure if this is correct procedure at all. Anyway, I'm stuck here...
    There are two solutions for $\displaystyle 2^n<n^8$

    One solution involves negative n.

    If the solution is for positive n, since your original equation is a logarithm,
    you could use a change of base to the natural logarithm.
    This makes a graphical solution much easier.

    $\displaystyle \displaystyle\huge\log_2n=\frac{log_en}{log_e2}$

    $\displaystyle log_e2=ln2=constant$

    Then you end up with

    $\displaystyle n<\frac{8}{log_e2}log_en$

    $\displaystyle \frac{8}{log_e2}log_en-n>0$

    You can use the Newton-Raphson method to find the two "n" for which this is zero.

    Your inequality holds for "n" between those two values.
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  5. #5
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    Thank you! I think it will suffice to just do the graph on my calculator when reaching an answer. It is computer science
    and Newton-Raphson method is outside scope of the course I am taking. The question was for what input size n one algorithm
    ran faster the the other one.
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