document.write( "Question 1104707: An isotope has a half-life of 92 years. How much of a 27-gram sample is left after 150 years? \n" ); document.write( "

Algebra.Com's Answer #719461 by greenestamps(13200) $\"\"$ $\"About$

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\n" ); document.write( "If you are going to use the $\"Ae%5E%28-kt%29\"$ formula, and you want a high degree of accuracy in your answer, then you need to keep several digits of the logarithm for your calculations.

\n" ); document.write( "One of the tutors who answered your question got an answer of 8.77 grams; a more accurate answer is 8.72 grams. The better accuracy requires a more accurate value for the logarithm involved in the calculation.

\n" ); document.write( "But you can avoid the logarithm problem completely by using a purely mathematical calculation, instead of using the $\"Ae%5E%28-kt%29\"$ formula.

\n" ); document.write( "The amount of the original sample after n half-lives is the original amount, multiplied by one-half raised to the power n.

\n" ); document.write( "In this problem, the number of half-lives is 150/92. So the amount remaining after 150 years is
\n" ); document.write( " $\"27%28.5%29%5E%28150%2F92%29+=+8.72075\"$ to 5 decimal places.

\n" ); document.write( "There is no need with this calculation to worry about rounding errors.... \n" ); document.write( "

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