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Let I have an equation $\mathcal{p} = 3^n*I$ where $I\in\{0,1,2\}$ then can I find out $I$ using $\log$ ?. Assuming $n$ is unknown. And only $p$ is shared to you.

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closed as unclear what you're asking by user228113, drhab, yoknapatawpha, Alex Provost, user252450 Jan 23 '16 at 21:13

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  • $\begingroup$ I don't understand your question; the value of $I$ is irrelevant to the number of $3$'s that you multiply by, which is always $n$. $\endgroup$ – user296602 Jan 23 '16 at 17:21
  • $\begingroup$ If $I=0$, there is no way to evince $n$ from $3^n\times I$. $\endgroup$ – user228113 Jan 23 '16 at 17:24
  • $\begingroup$ $I$ is multiplied by $3^n$ once (1 time). whether n is known are not. I don't think that is what you meant to ask. But if n is known I don't know what you meant to ask. $\endgroup$ – fleablood Jan 23 '16 at 17:24
  • $\begingroup$ @G.Sassatelli If n is known then there is no need to evince n from 3^n*I. $\endgroup$ – fleablood Jan 23 '16 at 17:26
  • $\begingroup$ Let me make it simple. Let $n=2$ and $I=2$ then $p=18$. Now can we find out if $I=1$ or $I=2$ here if only $18$ is shared to you. by using logs only. $\endgroup$ – Bill Jan 23 '16 at 17:26
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Don't need log at all.

if $I = 0$ then $P = 3^n * I = 0$

if $I = 1$ then $P = 3^n * I = 3^n = $ odd.

if $I = 2$ then $P = 3^n * I = 2*3^n = $ even.

So as long as you have it on good authority that $P $ does $= 3^n *I$ for some legitimate natural number n, and I = 0, 1, 2, then you can find the value of I by seeing if P is 0, even, or odd.

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Which if you do want to use $\log$ can be expressed via

if $\log_3 P$ is undefined then $P \le 0$ which given our criteria means $I = 0$

if $\log_3 P <0$ then $P < 1$ which given our criteria is impossible.

if $\log_3 P = m$ a natural number, then $P = 3^m$ so $I = 1$ .

all others $\log_3 P = x$, then given our criteria $I = 2$ and $x = n* \log_3 2$. But there are several cases that are impossible with our criteria. $\log_3 2 = 0.63092975357145743709952711434276$ so if $x \ne k*0.63092975357145743709952711434276$ our criteria fail.

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  • $\begingroup$ wow that's logical. $\endgroup$ – Bill Jan 23 '16 at 17:44
  • $\begingroup$ If you get something like $p = 17$ or $p = 32$ or p not an integer, though, the guy who gave you the question was lying to you: $P \ne 3^n * I$ for I = 0,1,2 and $n \in \mathbb N$. $\endgroup$ – fleablood Jan 23 '16 at 17:55

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