document.write( "Question 823004: Find the half-life of a radioactive substance if 240 grams of the substance decays to 180 grams in 9 years. Use Q(t)=e^kt, just not sure how to get the answer!
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Algebra.Com's Answer #495201 by Edwin McCravy(20060)\"\" \"About 
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Find the half-life of a radioactive substance if 240 grams of the substance decays to 180 grams in 9 years. Use Q(t)=e^(kt), just not sure how to get the answer!
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document.write( "You don't have the formula correct.  The formula must have \r\n" );
document.write( "Q(0) before the ekt.  The correct formula is\r\n" );
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document.write( "Q(t)=Q(0)ekt\r\n" );
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240 grams of the substance decays to 180 grams in 9 years.
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document.write( "That says Q(0)=240 and Q(9)=180 (that's when t=9).\r\n" );
document.write( "Plug those in to\r\n" );
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document.write( "Q(t) = Q(0)ekt\r\n" );
document.write( "Q(9) = 240ek·9\r\n" );
document.write( " 180 = 240e9k\r\n" );
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document.write( "Divide both sides by 240\r\n" );
document.write( " \"180%2F240\" = e9k\r\n" );
document.write( " 0.75 = e9k\r\n" );
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document.write( "Use the principle that exponential equation A=eB is equivalent \r\n" );
document.write( "to natural logarithm equation B=lnA:\r\n" );
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document.write( "9k = ln(0.75)\r\n" );
document.write( " k = ln(0.75)/9\r\n" );
document.write( "Use calculator:\r\n" );
document.write( " k = -0.0319646747\r\n" );
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document.write( "So the formula \r\n" );
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document.write( "Q(t) = Q(0)ekt\r\n" );
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document.write( "now becomes\r\n" );
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document.write( "Q(t) = Q(0)e-0.0319646747t\r\n" );
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Find the half-life...
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document.write( "Now to find the half life.  That will be when any original quantity\r\n" );
document.write( "reduces to one-half of its original quantity.  It's how long it takes \r\n" );
document.write( "240 grams to decay to 120 grams, or for 100 grams to decay to 50 grams \r\n" );
document.write( "or for any number of grams to decay to one-half of that number of grams.\r\n" );
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document.write( "In general, it's when the original quantity Q(0) decays to \"1%2F2\"Q(0).\r\n" );
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document.write( "So we substitute \"1%2F2\"Q(0) for Q(t) and solve for t\r\n" );
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document.write( "Q(t) = Q(0)e-0.0319646747t\r\n" );
document.write( "\"1%2F2\"Q(0) = Q(0)e-0.0319646747t\r\n" );
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document.write( "Divide both sides by Q(0)\r\n" );
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document.write( "\"1%2F2\" = e-0.0319646747t     \r\n" );
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document.write( "Use the principle that exponential equation A=eB is equivalent \r\n" );
document.write( "to natural logarithm equation B=lnA:\r\n" );
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document.write( "-0.0319646747t = \"ln%281%2F2%29\"\r\n" );
document.write( "-0.0319646747t = ln(0.5)\r\n" );
document.write( "Use calculator\r\n" );
document.write( "-0.0319646747t = -0.6931471806\r\n" );
document.write( "             t = \"%28-0.6931471806%29%2F%28-0.0319646747%29\"\r\n" );
document.write( "             t = 21.68478757 years\r\n" );
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document.write( "Edwin
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