document.write( "Question 1162420: A certain radioactive isotope has a​ half-life of approximately 78 years. How many years would be required for a given amount of this isotope to decay to 60​% of that​ amount? ​(​Hint: first find k. Then use k to find​ t)
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Algebra.Com's Answer #786222 by ikleyn(52797)\"\" \"About 
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document.write( "We are given that the half-life period is 78 years;  therefore, we can write\r\n" );
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document.write( "    p(t) = \"%281%2F2%29%5E%28t%2F78%29\",      (1)\r\n" );
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document.write( "where p(t) is the remaining mass fraction. (It is the standard radioactive decay model in terms of half-life period).\r\n" );
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document.write( "The problem asks to determine the time \"t\" when p(t) = 0.6.\r\n" );
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document.write( "In this case, the equation (1) takes the form\r\n" );
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document.write( "    0.6 = \"%281%2F2%29%5E%28t%2F78%29\".\r\n" );
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document.write( "Take logarithm base 2 of both sides\r\n" );
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document.write( "    \"log%282%2C%280.6%29%29%29\" = \"%28t%2F78%29%2Alog%282%2C+%281%2F2%29%29\"\r\n" );
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document.write( "    t = \"%2878%2Alog%282%2C%280.6%29%29%29%2F%28%28-1%29%29\" = \"%28-78%29%2Alog%282%2C%280.6%29%29\" = 57.48 years.    ANSWER\r\n" );
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\n" ); document.write( "\n" ); document.write( "See the lesson\r
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\n" ); document.write( "\n" ); document.write( "Also,  you have this free of charge online textbook in ALGEBRA-I in this site\r
\n" ); document.write( "\n" ); document.write( "    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.\r
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\n" ); document.write( "\n" ); document.write( "The referred lessons are the part of this online textbook under the topic \"Logarithms\".\r
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\n" ); document.write( "\n" ); document.write( "Save the link to this online textbook together with its description\r
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\n" ); document.write( "\n" ); document.write( "Free of charge online textbook in ALGEBRA-I
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\n" ); document.write( "\n" ); document.write( "The lesson to learn from my post is THIS:\r
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document.write( "    If you are given input data in terms of half-life, you do not need to convert your data \r\n" );
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document.write( "    into ekt-model.  Such conversion is an excessive work and unnecessary calculations.\r\n" );
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document.write( "    You can complete all calculations in terms of the half-life model, working consistently with degrees of 2, \r\n" );
document.write( "    which is your base in this case..\r\n" );
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