Trying to solve the equation involving floor function

Question

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

Answer 1

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.

Answer 2

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 $\lfloor \log x \rfloor + x > 10$.