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April 19th, 2016, 04:06 AM  #1 
Newbie Joined: Feb 2016 From: Cumbria Posts: 2 Thanks: 0  Quick Question: Make X the subject of Y = 2^X $\displaystyle y=2^x$ Super quick question, just really unsure how you would rearrange. Many thanks! 
April 19th, 2016, 04:16 AM  #2 
Senior Member Joined: Feb 2010 Posts: 706 Thanks: 140 
Use logarithms.

April 21st, 2016, 08:35 PM  #3 
Newbie Joined: Apr 2016 From: Europe Posts: 8 Thanks: 1 
$\displaystyle x = log2(Y)$ I don't know how to format that properly 
April 23rd, 2016, 05:37 AM  #4 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902  Use \log and an "underline", _, to get a subscript: \log_2(Y) gives $\displaystyle \log_2(Y)$
Last edited by skipjack; April 24th, 2016 at 02:00 PM. 
April 24th, 2016, 07:48 AM  #5 
Newbie Joined: Apr 2016 From: Europe Posts: 8 Thanks: 1  Thanks lol, well yeah like that.
Last edited by skipjack; April 24th, 2016 at 02:00 PM. 
April 24th, 2016, 01:49 PM  #6 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
Of course, all logarithms are equivalent: If $\displaystyle y= 2^x$ then $\displaystyle \log_2(y)= x$. If $\displaystyle y= 2^x$ then $\displaystyle \log(y)= x \log(2)$ so $\displaystyle x= \frac{\log(y)}{\log(2)}$ where "log" here is the "common logarithm" to base 10. If $\displaystyle y= 2^x$ then $\displaystyle \ln(y)= x \ln(2)$ so $\displaystyle x= \frac{\ln(y)}{\ln(2)}$ where "ln" here is the "natural logarithm" to base "e". In general, if $\displaystyle y= 2^x$ then $\displaystyle \log_b(y)= x \log_b(2)$ so $\displaystyle x= \frac{\log_b(y)}{\log_b(2)}$ where "b" is any positive number. $\displaystyle \log_2$ gives an especially simple answer because $\displaystyle \log_2(2)= 1$. Last edited by skipjack; April 24th, 2016 at 02:04 PM. 

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