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March 26th, 2016, 11:31 PM   #1
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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.
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March 27th, 2016, 12:08 AM   #2
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There is an easy solution if your logs are base e. Are they?
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March 27th, 2016, 12:35 AM   #3
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${\small\text{PER}} = 2^{\text{RES}} - 1$
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March 27th, 2016, 08:18 AM   #4
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Tattoo this on your wrist:
if a^p = x
then p = log(x) / log(a)
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April 6th, 2016, 08:15 AM   #5
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Quote:
Originally Posted by aliyesami View Post
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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