My Math Forum A * 2^n = B, solving for n

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 February 5th, 2010, 04:25 AM #1 Member   Joined: Feb 2008 Posts: 39 Thanks: 1 A * 2^n = B, solving for n I have a problem which seemed easy enough, but I which found difficult to solve. I have two variables, A => B. I need to know how many times (n) you need to double A to get B, that is I need to find n for: A * 2^n = B Since I don't know what kind of math I need to solve this, I didn't know in which forum to post this thread. Sorry if this is the wrong place. Thanks /Göran
 February 5th, 2010, 06:19 AM #2 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: A * 2^n = B, solving for n You need logarithms: $A2^n=B\\2^n=\frac{B}{A}\\n=\log_2\left(\frac{B}{A} \right)\\n=\frac{\ln B - \ln A}{\ln 2}$ Have a look at: http://www.purplemath.com/modules/logs.htm http://www.math.utah.edu/~pa/math/log.html
 February 5th, 2010, 07:55 AM #3 Member   Joined: Feb 2008 Posts: 39 Thanks: 1 Re: A * 2^n = B, solving for n Thanks! Works perfectly, will read up on the subject.

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