Results 1 to 10 of 10

Math Help - Solving for Variable in Exponent.

  1. #1
    Junior Member
    Joined
    Oct 2010
    Posts
    57

    Solving for Variable in Exponent.

    I'm trying to solve for n in the following inequality

    |4^(1/n) - 1| < x

    where n is a natural number and x is a positive real number. I feel like i should be able to do this but i can't think of how. Could someone please help me out with this?


    I'm also at a loss for how to solve for n with

    100^n / n! < x

    but I think i might be able to figure that out once i know how to work with the first one.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,802
    Thanks
    1576
    \displaystyle \left|4^{\frac{1}{n}} - 1\right| < x

    \displaystyle -x < 4^{\frac{1}{n}} - 1 < x

    \displaystyle 1 - x < 4^{\frac{1}{n}} < 1 + x

    \displaystyle \ln{(1 - x)} < \ln{\left(4^{\frac{1}{n}}\right)} < \ln{(1 + x)}

    \displaystyle \ln{(1 - x)} < \frac{1}{n}\ln{4} < \ln{(1 + x)}

    \displaystyle \ln{(1 - x)} < \frac{2\ln{2}}{n} < \ln{(1 + x)}

    \displaystyle \frac{1}{\ln{(1 + x)}} < \frac{n}{2\ln{2}} < \frac{1}{\ln{(1 - x)}}

    \displaystyle \frac{2\ln{2}}{\ln{(1 + x)}} < n < \frac{2\ln{2}}{\ln{(1 - x)}}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2010
    Posts
    57
    Quote Originally Posted by Prove It View Post
    \displaystyle \ln{(1 - x)} < \frac{2\ln{2}}{n} < \ln{(1 + x)}

    \displaystyle \frac{1}{\ln{(1 + x)}} < \frac{n}{2\ln{2}} < \frac{1}{\ln{(1 - x)}}
    How is it that you got from this first step to the second? I've tried everything i can think of but i think there is some log rule I don't know about.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member Sambit's Avatar
    Joined
    Oct 2010
    Posts
    355
    ln(1-x)<\frac{2ln2}{n}<ln(1+x)

    \frac{1}{ln(1-x)}>\frac{n}{2ln2}>\frac{1}{ln(1+x)} (taking reciprocal)

    that is \frac{1}{ln(1+x)}<\frac{n}{2ln2}<\frac{1}{ln(1-x)}-- which is nothing but what you want to prove.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,240
    Thanks
    1795
    If 0< a< b< c, then a< b so 1< b/a and 1/b< 1/a; b< c so, by the same divisions 1/c< 1/b. Together those give
    1/c< 1/b< 1/a.

    If the three numbers are not all positive, it is not true. For example, -2< 1< 2, 1/2< 1 but those are not less than -1/2.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,922
    Thanks
    1762
    Awards
    1
    There are several difficulties with this problem and the replies to it.

    If x>1 then \ln(1- x) does not exist.
    That is one difficulty the proposed solution.

    On the other hand, if 0<x<1 then \ln(1-x)<0 because 0<1-x<1.
    That means that \ln(1-x)<0<ln(1+x), another difficulty as pointed out in reply #5.

    If x=1 then the solutions are n\ge 3.
    Last edited by Plato; March 22nd 2011 at 12:44 PM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,922
    Thanks
    1762
    Awards
    1
    Post Script
    I should have finished this question. (I was put out at those mistakes.)

    First note that for all n we have 4^{\frac{1}{n}}>1.
    If x\ge 1 then 1-x<0<4^{\frac{1}{n}}<x+1.
    So this is true if n\ge 3.

    In the case 0<x<1 we have 0<1-x<1<4^{\frac{1}{n}} <1+x.

    Now we can use the \ln function.
    We get \ln(4)<n\ln(x+1) or n>\dfrac{\ln(4)}{\ln(x+1)}.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Oct 2010
    Posts
    57
    Quote Originally Posted by Plato View Post
    Post Script
    I should have finished this question. (I was put out at those mistakes.)

    First note that for all n we have 4^{\frac{1}{n}}>1.
    If x\ge 1 then 1-x<0<4^{\frac{1}{n}}<x+1.
    So this is true if n\ge 3.

    In the case 0<x<1 we have 0<1-x<1<4^{\frac{1}{n}} <1+x.

    Now we can use the \ln function.
    We get \ln(4)<n\ln(x+1) or n>\dfrac{\ln(4)}{\ln(x+1)}.
    Thanks for the help so far everyone. I've been trying to wrap my head around them for a while now

    Regarding when 0 < x < 1, i see how we get to 1 - x < 1 < 4^(1/n) < 1 + x
    but when we use ln, don't we get n > ln(4) / ln(x + 1) > (n * ln(1 - x)) / (ln(1 + x) ?
    when you used ln you dropped the last part for some reason.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,922
    Thanks
    1762
    Awards
    1
    Quote Originally Posted by Relmiw View Post
    Regarding when 0 < x < 1, i see how we get to 1 - x < 1 < 4^(1/n) < 1 + x but when we use ln, don't we get n > ln(4) / ln(x + 1) > (n * ln(1 - x)) / (ln(1 + x) ?when you used ln you dropped the last part for some reason.
    The 'last part' makes no different whatsoever.
    If \ln(4)<n\ln(x+1) then because \ln(x+1)>0 is a fixed positive number we get n>\dfrac{\ln(4)}{\ln(x+1)}.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member
    Joined
    Oct 2010
    Posts
    57
    Ok, thanks a lot for the help everybody
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finding an exponent variable.
    Posted in the Algebra Forum
    Replies: 4
    Last Post: November 29th 2011, 11:52 PM
  2. LImit with and exponent variable
    Posted in the Calculus Forum
    Replies: 6
    Last Post: February 26th 2011, 07:49 PM
  3. Solve for variable exponent...
    Posted in the Algebra Forum
    Replies: 4
    Last Post: October 29th 2009, 11:44 PM
  4. limit with exponent variable
    Posted in the Calculus Forum
    Replies: 4
    Last Post: June 9th 2009, 05:36 AM
  5. variable in the exponent
    Posted in the Algebra Forum
    Replies: 1
    Last Post: May 4th 2009, 05:46 AM

Search Tags


/mathhelpforum @mathhelpforum