document.write( "Question 1118991: Complete the table for the radioactive isotope. (Round your answer to two decimal places.) \r
\n" ); document.write( "\n" ); document.write( "isotope: RA^226
\n" ); document.write( "Half Life (Years): 1599
\n" ); document.write( "Initial Quantity: 10g
\n" ); document.write( "Amount after 1000 years: UNKNOWN
\n" ); document.write( "

Algebra.Com's Answer #734438 by greenestamps(13200)\"\" \"About 
You can put this solution on YOUR website!


\n" ); document.write( "The solution by tutor @josgarithmetic, using logs, ends up with the right answer but seems like a very inefficient way to solve the problem.

\n" ); document.write( "The solution by tutor @ikleyn is much simpler and also obtains the right answer. But I note that scientists like to use exponentials with negative exponents to indicate decay, so her initial equation is

\n" ); document.write( "\"M%28t%29+=+M%2A%282%5E%28-n%29%29\"

\n" ); document.write( "where n is the number of half lives.

\n" ); document.write( "For me, it is much more natural to write the equation in a way that clearly shows the amount remaining becomes half as much after each half life:

\n" ); document.write( "\"M%28t%29+=+M%2A%281%2F2%29%5En\"

\n" ); document.write( "So my path to the solution of the problem would be a single calculation:

\n" ); document.write( "\"10%281%2F2%29%5E%281000%2F1599%29+=+6.48\"

\n" ); document.write( "Use both methods as you practice other half life problems and see which works better for you.
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