Log probability

There's ambiguity as the asker has not specified the base of the logarithm.

The answer comes by considering the base to be $10$ and not $e$ .

We want to find the solution set of the following equation:

$\lfloor \log 4x \rfloor = \lfloor \log x \rfloor$

As $x \in (0, 1)$ , the largest integer also equal to the above equation is $-1$ .

Case 1:

$\lfloor \log 4x \rfloor = \lfloor \log x \rfloor = -1$

$\lfloor \log x \rfloor = -1$ and $\lfloor \log 4x \rfloor = -1$

$\implies -1 \le \log x < 0$ and $-1 \le \log 4x < 0$

$\implies \frac{1}{10} \le x < 1$ and $\frac{1}{40} \le x < \frac{1}{4}$

Taking the intersection of the two solution sets:

$x \in \Bigg(\dfrac{1}{10}, \dfrac{1}{4} \Bigg)$

Case 2:

$\lfloor \log 4x \rfloor = \lfloor \log x \rfloor = -2$

$\lfloor \log x \rfloor = -2$ and $\lfloor \log 4x \rfloor = -2$

$\implies -2 \le \log x < -1$ and $-2 \le \log 4x < -1$

$\implies \frac{1}{100} \le x < \frac{1}{10}$ and $\frac{1}{400} \le x < \frac{1}{40}$

Taking the intersection of the two solution sets:

$x \in \Bigg(\dfrac{1}{100}, \dfrac{1}{40} \Bigg)$

. . .

It can be seen that the total "length" of the solution set on the number line will be:

$L = \Bigg(\dfrac{1}{4} - \dfrac{1}{10}\Bigg) + \Bigg(\dfrac{1}{40} - \dfrac{1}{100}\Bigg) + \Bigg(\dfrac{1}{400} - \dfrac{1}{1000}\Bigg) + \dots$

$L = \Bigg(\dfrac{1}{4} - \dfrac{1}{10}\Bigg) \cdot \Bigg(1 + \dfrac{1}{10} + \dfrac{1}{100} + \dots\Bigg)$

$L = \dfrac{3}{20} \cdot \dfrac{10}{9} = \boxed{\dfrac{1}{6}}$

(As the length of the total sample space is 1, L is the required probability.)

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In case the base of the logarithm was $e$ , there would be no solution as $\log4 > 1$ in that case and the probability would be $0$ .