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
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.