# Amc 12b #20

• Apr 6th 2006, 04:25 PM
Jameson
Amc 12b #20
Let $x$ be chosen at random from the interval $(0,1)$. What is the probability that $[\log_{10}(4x)]-[\log_{10}(x)]=0$?

Here $[x]$ denotes the greatest integer that is less than or equal to $x$.

This is multiple choice, but I don't think posting the possibilities are necessary.
• Apr 6th 2006, 06:52 PM
ThePerfectHacker
Quote:

Originally Posted by Jameson
Let $x$ be chosen at random from the interval $(0,1)$. What is the probability that $[\log_{10}(4x)]-[\log_{10}(x)]=0$?

Here $[x]$ denotes the greatest integer that is less than or equal to $x$.

This is multiple choice, but I don't think posting the possibilities are necessary.

I do not have much time now but I was thinking maybe you can do this.
$[x]=[x+k],0
Then, it seems to me (I did not formally prove it) then,
$j
are all solutions for each integer $j$.

Thus,
$[\log (4x)]-[\log x]=0$
Thus, since $0<\log 4<1$
$[\log 4+\log x]=[\log x]$
• Apr 7th 2006, 06:12 AM
ThePerfectHacker
Let, us solve, for $0
$[\log 4x]=[\log x]$
We have,
$[\log 4+\log x]=[\log x]$
Thus, all solutions for $u=\log x$ are,
$u\in \bigcup_{j\in\mathbb{Z}}(j,1-\log 4+j]$
Thus, we that, all solutions satisfies
$j<\log x\leq 1-\log 4+j$
Iff,
$10^j
.
Note, $j$ cannot be zero or positive because it would violate the inequality $0.
Thus, $j=-1,-2,-3,-4,...$ all work.
Thus,
$.1 thus, length =.15
$.01 length=.015
$.001 length=.0015
and so on "ad infinitum" (I am so cool using latin phrases).
Thus, we have the total length of the solutions to be:
$.15+.015+.0015+....=.16666=1/6$
this is a regular infinite geometric series.

My point is that, we can "intuitively" think of probability as the length of the success (which are the solutions) divided by the total possibilities (which is the length of interval). Thus, the probability is $\frac{1}{6}$.