document.write( "Question 79517: The amount of a radioactive tracer remaining after t days is given by A=Aoe^-0.058t, where Ao is the starting amount at the beginning of the time period. How many days will it take for one half of the original amount to decay? \n" ); document.write( "
Algebra.Com's Answer #57127 by ankor@dixie-net.com(22740)\"\" \"About 
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The amount of a radioactive tracer remaining after t days is given by A=Aoe^-0.058t, where Ao is the starting amount at the beginning of the time period. How many days will it take for one half of the original amount to decay?
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\n" ); document.write( "Let Ao = 2; (the original amt)
\n" ); document.write( "Let A = 1; (half the original amt)
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\n" ); document.write( "Ao*(e^-,058t) = A
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\n" ); document.write( "2*(e^-.058t) = 1
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\n" ); document.write( "e^-.058t = 1/2: divided both sides by 2
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\n" ); document.write( "ln(e^-.058t) = ln(1/2); nat log of both sides
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\n" ); document.write( "-.058t*ln(e) = ln(1/2); log equiv of exponents
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\n" ); document.write( "-.058t = -.693147; remember ln(e) = 1
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\n" ); document.write( "t = -.693147/-.058
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\n" ); document.write( "t = +11.95 days for half of the original amt to decay.
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\n" ); document.write( "Check on a good calc; enter: ln(e^(-.058*11.95)) = .5000\r
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