I must find 'x' and I don't know how to solve the following equation.
Does it have a solution? How can I solve it?
$$ S=\left(\lfloor\log_{10}(x)\rfloor+1\right)x - \frac{10^{\lfloor\log_{10}(x)\rfloor+1}-10}{9} $$
$$S,x\in\mathbb N$$
Asked
Active
Viewed 736 times
1
Martin Sleziak
- 53,687
Carlos
- 11
-
2That \right+1) isn't going to work. – Gerry Myerson May 19 '13 at 13:00
-
2Also, an equation like $p(x)={\rm something\ in\ }x$ isn't meant to have a solution. Do you mean ${\rm something\ ib\ }x=0$? – Gerry Myerson May 19 '13 at 13:01
-
@Gerry: I noticed that the \right was after instead of before the \rfloor (even if I didn't understand your message first... :-)) – Raymond Manzoni May 19 '13 at 13:06
-
Note that, for positive integral $x$, $ \lfloor\log(x)\rfloor +1$ can be thought of as the number of digits of $x$ when written in the same base as the logarithm. – Mark Bennet May 19 '13 at 13:06
-
I must find 'x' for a given 'p(x)', exact or approximate value. – Carlos May 19 '13 at 13:12
-
Carlos, if that's what you want, then that's what you should write in the question. Please edit. – Gerry Myerson May 19 '13 at 13:20
-
1If your $\log$ is in base $10$ (possible from Mark's remark) please specify it too... – Raymond Manzoni May 19 '13 at 13:21
-
Hey i dont know latex, so can i post a picture with the answer? – Rohinb97 Jun 09 '13 at 16:22
2 Answers
0
If you could solve it by reasoning, here is my opinion.
If you can observe, the first part of S has and depends on log x to be a natural no. and the other one has (x-1)/9 as a term. Since both S and x are natural numbers, x can only be 1,10,100, 1000...
Rohinb97
- 1,702
0
For an integer $k\ge 0$, if $10^k \le x < 10^{k+1}$, then $\lfloor \log_{10}x\rfloor = k$ so $$ S(x) = (k+1)x -\frac{10^{k+1}-10}{9}=(k+1)x-\frac{10}{9}(10^k-1) $$ So, for example, if $1\le x \le 9$, then $S(x) = x$. If $10\le x \le 99$, then $S=2x-10$. If $100\le x\le 999$, then $S=3x-110$, and so on.
The first few $S$ values are $$ \begin{array} &x = & 1 & 2 &\dots & 9 & 10 & 11 & 12 & \dots & 99 & 100 & 101 & 102 & \dots 999\\ S = & 1 & 2 &\dots & 9 & 10 & 12 & 14 & \dots & 188 & 190 & 193 & 196 & \dots 2887\\ \end{array} $$
Rick Decker
- 8,718