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April 19th, 2016, 04:06 AM   #1
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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!
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April 19th, 2016, 04:16 AM   #2
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Use logarithms.
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April 21st, 2016, 08:35 PM   #3
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$\displaystyle x = log2(Y)$

I don't know how to format that properly
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April 23rd, 2016, 05:37 AM   #4
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Quote:
Originally Posted by Schnibob View Post
$\displaystyle x = log2(Y)$

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.
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April 24th, 2016, 07:48 AM   #5
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Quote:
Originally Posted by Country Boy View Post
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.
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April 24th, 2016, 01:49 PM   #6
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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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