I am trying to solve for $m$ in this equation but got stuck here:

$\dfrac{1}{m} = \dfrac{1}{\log{n}} - \dfrac{1}{n}$

This is the original equation:


Any help is greatly appreciated! Thanks.

  • $\begingroup$ make a common denominator of the right hand side (which is the product of $n$ and $logn$) and reciprocate the resulting fraction $\endgroup$ – imranfat Jan 19 '16 at 21:20
  • $\begingroup$ Hint: if $y= \frac{1}{x}$, then $x = \frac{1}{\frac{1}{x}} = \frac{1}{y} $ $\endgroup$ – Adam Francey Jan 19 '16 at 21:22

Your original equation is $$ (n + m)\log(n) = mn. $$ This is a linear equation in $m$, and so you just take all the terms with $m$ to one side $$ n\log(n) + m\log(n) - mn = 0 \\ m[\log(n) - n] = -n\log(n) \\ m = \frac{...}{...}. $$

  • 1
    $\begingroup$ Very brilliant. Thank you $\endgroup$ – Philip Hardy Jan 19 '16 at 21:21

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