document.write( "Question 576524: Please help!!!\r
\n" ); document.write( "\n" ); document.write( "The half life of an element is 3.9 * 10^9 yr. How long does it take a sample of the element to decay to 2/5 of its original mass? Use A=A0(1/2)^(t/T) where AO is the initial amount, T is the half life, and t is the time. (express results in scientific notation, rounded to the nearest hundredth)
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Algebra.Com's Answer #369914 by KMST(5328)\"\" \"About 
You can put this solution on YOUR website!
\"A=A%5B0%5D%281%2F2%29%5E%28t%2FT%29\" is the equation given and you want the value of \"t\" that will make \"A%2FA%5B0%5D=2%2F5\"
\n" ); document.write( "(You know it has to be more than \"T=3.9+%2A+10%5E9years\" because at T you would have 1/2 left and 2/5<1/2).
\n" ); document.write( "We could divide both sides by \"A%5B0%5D\" to get
\n" ); document.write( "\"A%2FA%5B0%5D=%281%2F2%29%5E%28t%2FT%29\"
\n" ); document.write( "Taking logarithm of both sides
\n" ); document.write( "\"log%28%28A%2FA%5B0%5D%29%29=%28t%2FT%29%2Alog%28%281%2F2%29%29\" --> \"log%28%28A%2FA%5B0%5D%29%29=%28t%2FT%29%2A%28log%281%29-log%282%29%29\" --> \"log%28%28A%2FA%5B0%5D%29%29=%28t%2FT%29%2A%280-log%282%29%29\" --> \"log%28%28A%2FA%5B0%5D%29%29=%28t%2FT%29%2A%28-log%282%29%29\" --> \"log%28%28A%2FA%5B0%5D%29%29=-%28t%2FT%29%2Alog%282%29\"
\n" ); document.write( "Plugging in the values given,
\n" ); document.write( "\"log%28%282%2F5%29%29=-%28t%2F%283.9+%2A+10%5E9%29%29%2Alog%282%29\"
\n" ); document.write( "Using the approximate values for the logarithms
\n" ); document.write( "\"-0.39794=-0.30103%2At%2F%283.9+%2A+10%5E9%29\"
\n" ); document.write( "So, \"t=0.39794%2A%283.9%2A+10%5E9%29%2F0.30103\"
\n" ); document.write( "\"highlight%28t=5.16%2A10%5E9years%29\"
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