$-\ln(1-(\frac{1}{x})) < \frac{100}{x} $ for $ x > 1$ is what I want to prove. I pulled a negative sign out and I got $\ln(\frac{x}{(x-1)}) < \frac{100}{x} $ for $ x > 1$. How do I continue with this proof? Or is it actually possible to prove this

Edit : I want this proof Algebrically, Calculus is allowed

  • 2
    $\begingroup$ Nope, let's take $x=\frac{1}{1-e^{-100}}>1 , -\ln(1-\frac{1}{x})=100$, while $100/x<100$ since $x>1$ $\endgroup$ May 2, 2015 at 5:01
  • 1
    $\begingroup$ I don't think you can get the algebraic proof . Not by applying derivative methods at least. $\endgroup$
    – GrandAlpha
    May 2, 2015 at 5:22
  • $\begingroup$ I have to some point, But I don't know how to do the rest... I'll update it in question $\endgroup$ May 2, 2015 at 5:26
  • $\begingroup$ Can we use calculus? $\endgroup$
    – Someone
    May 2, 2015 at 5:27
  • $\begingroup$ @Mann I guess not. $\endgroup$ May 2, 2015 at 5:28

5 Answers 5


It is not true for $x=1+e^{-100}$. We then have $$ \log\frac{x}{x-1} = \log(x)+100 > 100 $$ but $$ \frac{100}{x} < 100 $$ because $x>1$.

  • $\begingroup$ I want proof of my expression lol I don't want where is it invalid :( $\endgroup$ May 2, 2015 at 6:00
  • $\begingroup$ @CuriousSciDude: Because your inequality is not true, there can be no proof of it. $\endgroup$ May 2, 2015 at 6:01
  • $\begingroup$ Oh.. But wolfram says it true , so I thought it was true $\endgroup$ May 2, 2015 at 6:02
  • $\begingroup$ @CuriousSciDude: It is true for $x>1+\varepsilon$, where $\varepsilon$ is so small (but larger than $e^{-100}$) that Wolfram Alpha cannot distinguish numerically between $1$ and $1+\varepsilon$. $\endgroup$ May 2, 2015 at 6:08
  • $\begingroup$ Btw where is this log x+ 100 from? $\endgroup$ May 2, 2015 at 14:39

As you almost wrote, making $x=\frac 1y$, the problem is to show that, for $y<1$, $$f(y)=100 y+\log(1-y)>0$$ (the function being undefined at $y=1$). We have $$f'(y)=100-\frac{1}{1-y}$$ which is positive as long as $y<\frac{99}{100}$ and negative for $y>\frac{99}{100}$. On the other hand $$f''(y)=-\frac{1}{(1-y)^2}$$ is always negative. So $f(y)$ increases up to a maximum value equal to $99-\log (100) >0$ and starts decreasing very fast and it exists a value of $y$ slightly below $1$ which makes $f(y)=1$ and negative for larger values. When $y=1-\epsilon$ , $f(y)= 100-100\epsilon+\log(\epsilon)$

So, as Henning Makholm answered, this is not true for all $x>1$.

  • $\begingroup$ hey wat about for x greater then equal to 2 $\endgroup$ May 2, 2015 at 14:54

$ln \left(\frac{x}{x-1}\right)<\frac{100}{x}$

$-ln \left(\frac{x-1}{x}\right)<\frac{100}{x}$

$-ln \left(1-\frac{1}{x}\right)<\frac{100}{x}$

$ln \left(1-\frac{1}{x}\right)^x>-100$



I think that $\frac{1}{e^{100}}$ is almost 0 and LHS is ofcourse $>0$ for all $x>1$

So this seems pretty reasonable.

Edit: It suddenly seems not strictly valid for all $x>1$ , but not sure. Edit:2 This is not true for only


Since, the root lies very close to $1$ , it is approximately valid. And I think, this is good enough for algebraic solution.

  • $\begingroup$ Can you check the above answEr? $\endgroup$ May 2, 2015 at 6:25
  • $\begingroup$ Define g(x)=$ln(\frac{x}{x-1})-\frac{100}{x}$ , you can see that function decreases in $(1,\frac{100}{99})$ Also the function has only root at x approximately 1 , after that it remains negative. Hence, it is positive before the root of the equation. $\endgroup$
    – Someone
    May 2, 2015 at 6:33

Solution to \begin{align} \ln\left(\frac{x}{x-1}\right) = \frac{100}{x} \end{align} in terms of Lambert W function is \begin{align} x&=\frac{100}{W(-100\exp(-100))+100} \\ &\approx 1.000000000000000000000000000000000000000000037200759760 \end{align}

Btw, WolframAlpha gives the answer to the inequality as

Solution over the reals enter image description here


Required to prove $ln \left(\frac{x}{x-1}\right)<\frac{100}{x}$

ie to prove $f(x) = {{100} \over x} - ln\left( {{x \over {x - 1}}} \right)>0$

${d \over {dx}}\left({{100} \over x} - ln\left( {{x \over {x - 1}}} \right)\right) = $


=${{ - 101x - 100} \over {\mathop x\nolimits^2 (x + 1)}}$

$f'(x)$ is negative for all $x > 1$

Meaning always decreasing.

Easy to show $\mathop {\lim }\limits_{x \to \infty } f(x) = 0$

So The function is always decreasing towards 0

implies $f(x) > 0$ for $x > 1$

Implies $ln \left(\frac{x}{x-1}\right)<\frac{100}{x}$


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