# Trying to solve the equation involving floor function

I am trying to solve the following equation:

Floor[logx+1]+x=11

Where Floor function returns the greatest integer smaller than the value in bracket.
e.g. Floor[3.3] = 3

And the logarithm is to the base 10

I tried using WolframAlpha, but it's timing out.

EDIT: I gather from the responses that a solution does not have a method. ----- If I change the equation to the following:

log(x) + x = 11


Here logarithm is to the base 10. Is it solvable now? I am more interested in method.

Thanks, Siddharth

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There's no solution. – Raskolnikov Nov 17 '11 at 11:53
On the other hand if you mean $\lfloor\log(x+1)\rfloor+x=11$, the answer is $10$. – André Nicolas Nov 17 '11 at 12:18
Maple output – pedja Nov 17 '11 at 12:31
@Raskolnikov There is a solution and it's 10, as mentioned by Andre Nicolas. I am interested in the methodology. – Sid Nov 17 '11 at 12:32
@SidCool: The original question was not typeset in LaTeX, and there was in my mind some ambiguity about what was intended. The two plausible guesses are $\lfloor \log(x)+1\rfloor +x=11$ and $\lfloor \log(x+1)\rfloor +x=11$. Which one was intended? – André Nicolas Nov 17 '11 at 15:51

## 2 Answers

It's clear that the expression on the left hand side of the equation is defined for $x>0$ and in this domain it's monotonically increasing. It's also clear that for very small $x$ it will be negative, and for $x=10$ it's already larger than $11$. So if a solution exists it will be in $(0,10)$.

You can split this domain into the integer values that $[\log x + 1]$ gets: on $[10^{n-1},10^n)$ it takes value $n$. So immediately we can rule out $[1,10)$ to contain a solution, since it takes value $1$, requiring $1 + x = 11$ which leaves only $x=10$ which is not a solution.

Similarly for all whole $n<1$ a solution has to satisfy that $x \in [10^{n-1},10^n) \implies n + x = 11 \implies x = 11-n \implies x>10$ which is obviously a contradiction.

Therefore there are no solutions.

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Thanks for the response, @davin. If we remove the floor function and just have it logx + x = 11 Is there a solution? – Sid Nov 17 '11 at 12:34
@SidCool, absolutely, and that is easily proved with IVP: $\log x + x$ is continuous on $(0,\infty)$ with negative values for small $x$ and positive values for $x \geq 1$ so there must be a solution on $(0,1)$. It's also monotonic, so that will be the only solution. – davin Nov 17 '11 at 12:41
Thanks @Davin. Is there a standard method for solving such equations? – Sid Nov 17 '11 at 16:27
@SidCool, I'm not aware of any standard method. Everything I wrote is standard material in a real analysis course, but I'm not sure of anything more specific. – davin Nov 17 '11 at 16:56
Thanks for your help, @davin – Sid Nov 18 '11 at 1:47

Left side is an increasing function (as sum of a nondecreasing and increasing), so there is at most one solution. Also you can get rid of "+1" from floor:

$\lfloor \log x\rfloor + 1 + x = 11$

$\lfloor \log x\rfloor + x = 10$

If $x<10$ then $\log x<1$ so $\lfloor \log x \rfloor \leq 0$ so $\lfloor \log x \rfloor + x < 10$

If $x\geq 10$ then $\log x \geq 1$ so $\lfloor \log x \rfloor \geq 1$ so $\lfloor \log x \rfloor + x > 10$.

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