# equation, integer part, logarithm

#### idontknow

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

#### greg1313

Forum Staff
$x=1$ 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?

idontknow

#### Adesh Mishra

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.

idontknow
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