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How I can solve this equation with respect to the variable $t$?

$$\left\lfloor{\frac{\ln(t+1)}{\ln 2}}\right\rfloor=\left\lfloor{\frac{\ln t}{\ln2}}\right\rfloor+1$$

where $\left\lfloor {y} \right\rfloor$ is the integer part of $y$.

I have no idea to start.

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$\frac{\ln(x)}{ \ln(a)} = \log_a(x)$. So your equation is $$ \lfloor \log_2(t+1)\rfloor = \lfloor \log_2(t) \rfloor + 1 $$ That means that $t+1$ must be a little bigger than a power of 2, and $t$ must be a little less. For instance, $t = 31.5$ works fine. I guess the general solution is any value $t$ satisfying

$2^i - 1 \le t < 2^i$ for $i = 0, 1, 2, \ldots$, except for $t = 0$, where the $\log$ doesn't make sense.

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  • $\begingroup$ Beat me to it. +1 $\endgroup$ – Michael Albanese Dec 5 '14 at 17:46
  • $\begingroup$ A red-letter day for me! :) $\endgroup$ – John Hughes Dec 5 '14 at 17:47
  • $\begingroup$ Nice. I could see immediately that $t=1$ works, but of course this is more general. $\endgroup$ – user_of_math Dec 5 '14 at 17:48
  • $\begingroup$ @ John Hughes: Are you sure about your last claim. $\endgroup$ – DER Dec 5 '14 at 17:55
  • $\begingroup$ Thanks, DER...I fixed it to rule out the $t = 0$ case. Was that what you were concerned about? (Maybe I should have let Michael A beat me to it!) $\endgroup$ – John Hughes Dec 5 '14 at 18:19

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