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 March 26th, 2016, 11:31 PM #1 Newbie   Joined: Mar 2016 From: FL, USA Posts: 1 Thanks: 0 solving equation with log Let's say we have an equation: RES = log(PER + 1)/log(2) how can I solve this equation for PER with different values of RES, say 10 or 16? Thanks. Last edited by skipjack; March 27th, 2016 at 12:37 AM.
 March 27th, 2016, 12:08 AM #2 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,640 Thanks: 571 Math Focus: Yet to find out. There is an easy solution if your logs are base e. Are they?
 March 27th, 2016, 12:35 AM #3 Global Moderator   Joined: Dec 2006 Posts: 19,532 Thanks: 1750 ${\small\text{PER}} = 2^{\text{RES}} - 1$
 March 27th, 2016, 08:18 AM #4 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,126 Thanks: 914 Tattoo this on your wrist: if a^p = x then p = log(x) / log(a)
April 6th, 2016, 08:15 AM   #5
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Quote:
 Originally Posted by aliyesami Let's say we have an equation: RES = log(PER + 1)/log(2) how can I solve this equation for PER with different values of RES, say 10 or 16? Thanks.
I would start by multiplying both sides by log(2): RES log(2)= log(PER+ 1). Assuming the logarithms are base "a" (natural logarithms are base a= e, common logarithms are base a= 10).
$\displaystyle a^{RES log(2)}= (a^{log(2))^{RES}= a^{log(PER+ 1)}$
$\displaystyle 2^{RES}= PER+ 1$
$\displaystyle PER= 2^{RES}- 1$

Notice that this answer does not involve "a" so it does not matter what to what base the logarithm is.

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