My Math Forum logarithms

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 May 20th, 2014, 05:58 PM #1 Banned Camp   Joined: May 2014 From: london Posts: 21 Thanks: 0 logarithms There are two primary equations used when dealing with standard radioactive decay. 1.) N = N0e^-kt 2.) K = ln2/r N represents the final amount remaining N0 represents the initial amount e is the exponential constant k is the constant exponent for that sample t represents the amount of time that passed r is the half-life of that sample ln stands for natural logarithm the half-life of radium-226 is 1600 years. Suppose you have a 22mg sample. After how long will only 18mg of the sample remain? Last edited by skipjack; May 21st, 2014 at 01:21 PM.
 May 20th, 2014, 11:34 PM #2 Senior Member     Joined: Nov 2013 From: Baku Posts: 502 Thanks: 56 Math Focus: Geometry If I understood it correctly $\displaystyle n=n_0 \cdot e^{-kt } \\ k= \frac{\ln 2}{r}$ Therefore $\displaystyle k= \frac{\ln 2}{1600} \\ 18=22 \cdot e^{-kt} \\ -kt =\ln {(9/11)}$ $\displaystyle t=-1600 \cdot \log_{2} (9/11)$
 May 21st, 2014, 01:25 PM #3 Global Moderator   Joined: Dec 2006 Posts: 19,270 Thanks: 1679 I can't be certain which is expected, but a value obtained by use of a calculator would often be acceptable for this type of question.
 May 21st, 2014, 05:52 PM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,341 Thanks: 2463 Math Focus: Mainly analysis and algebra I think you are supposed to understand how the answer is arrived at rather than simply typing some stuff into a calculator. It's possible to train monkeys to push buttons, after all. Thanks from tahirimanov
 May 22nd, 2014, 01:11 AM #5 Senior Member     Joined: Nov 2013 From: Baku Posts: 502 Thanks: 56 Math Focus: Geometry $\displaystyle \log_2 (9/11) \approx -0.289475$, therefor $\displaystyle t = -1600 \cdot (-0.289475) = 463.16$
May 22nd, 2014, 06:59 AM   #6
Senior Member

Joined: Nov 2013
From: Baku

Posts: 502
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Math Focus: Geometry
Quote:
 Originally Posted by ron246 oh no...i can use a calculator, i was just asking if that last line was the official answer or do i need to type that into the calcuator and press =
Do you have a calculator which calculates log base 2???

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