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I need some help in solving a simple equation

first example: $n/2^i = 1$ then when we solve this we get : $2^i = n$ then $i=\log_2 n$ of base 2 . Okay so I understand this.

second problem: $n/3^i = 2*i$ okay so here I am stuck since i have $n=(3^i)2*i$ then what?? can you please help me so I tried to solve this but I could not so can somebody please provide with an answer with some explanation. Thank you

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  • $\begingroup$ I am almost sure that you won't be able to solve for $i$ in the second equation. Not every equation may be solved for a variable, at least in terms of elementary functions. $\endgroup$ – Cehhiro Oct 1 '15 at 4:56
  • $\begingroup$ In general that type of equation doesn't have a solution expressable in (finite) terms of normal fundamental functions. If you don't require that it is expressable in that way, then sure there's a solution - the question is just which form you want it expressed in? One form would be $f^{-1}(n)$, where $f^{-1}$ is the inverse of the function $f(i) = 3^i2i$. $\endgroup$ – skyking Oct 1 '15 at 5:23
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Hint:

Can be simplified to

$$ln(n) = iln(3) +ln(2) +ln(i)$$

Use the fact that $ln(i) \approx i-1$

$$i(ln(3) + 1) + ln(2)-1 = ln(n)$$

$$i = \dfrac{ln(\frac{n}{2}) +1}{ln(3) + 1}$$

This is an approximate solution.

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