 My Math Forum Quick Question: Make X the subject of Y = 2^X
 User Name Remember Me? Password

 Algebra Pre-Algebra and Basic Algebra Math Forum

 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: 711 Thanks: 147 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

Quote:
 Originally Posted by Schnibob $\displaystyle x = log2(Y)$ I don't know how to format that properly
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

Quote:
 Originally Posted by Country Boy Use \log and an "underline", _, to get a subscript: \log_2(Y) gives $\displaystyle \log_2(Y)$
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. Tags make, question, quick, subject Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Tangeton Algebra 5 April 19th, 2016 02:09 PM Ailsa Algebra 2 May 30th, 2015 03:24 AM sander2798 Algebra 8 February 28th, 2015 05:23 AM mikeinitaly Algebra 6 February 26th, 2015 09:53 PM Battler. Abstract Algebra 4 February 19th, 2014 01:09 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2019 My Math Forum. All rights reserved.      