equation, integer part, logarithm

Dec 2015
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169
Earth
\(\displaystyle x=1+\lfloor \log(x) \rfloor \; , \: x\in \mathbb{N}\).
x=?
 

greg1313

Forum Staff
Oct 2008
8,008
1,174
London, Ontario, Canada - The Forest City
$x=1$ is a solution.

\(\displaystyle f(x)=x\quad f'(x)=1\quad g(x)=1+\log x\quad g'(x)=\frac1x\)

What conclusion is reasonable?
 
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Aug 2019
88
23
India
Well I don’t have any method to solve it, but if we put anything other than 1, let’s say \(\displaystyle x \gt e \), then we always gonna end up with \(\displaystyle RHS ~\lt~LHS\). We can set up something like this
\(\displaystyle \ln (x)\lt x \) for \(\displaystyle x > 1\)
Since, \(\displaystyle [\ln (x)]\leq \ln(x) \)
Therefore, \(\displaystyle [\ln(x)]\lt x \). Now from the graph of \(\displaystyle x \) and \(\displaystyle [\ln(x)] \) (we can draw it in desmos) we find that difference, that is \(\displaystyle x - [\ln(x)] \) at x > 1 is greater than 1 and it goes on increasing. So, we can write
\(\displaystyle x - [\ln(x)] \gt 1\)
\(\displaystyle x \gt 1 + [\ln(x)] \)

Hence, x = 1 is the only solution.

P.S. : whenever I wrote \(\displaystyle [\ln(x)] \) I meant that greatest integer function.
 
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