document.write( "Question 320172: Please help me solve this word problem: A certain radioactive element decays at a rate of 4% per year. Find the half-life of the substance (i.e. the time it will take for one half of any given amount of the substance to decay). \n" ); document.write( "
Algebra.Com's Answer #229286 by Edwin McCravy(20062)\"\" \"About 
You can put this solution on YOUR website!
Please help me solve this word problem: A certain radioactive element decays at a rate of 4% per year. Find the half-life of the substance (i.e. the time it will take for one half of any given amount of the substance to decay).
\n" ); document.write( "
\r\n" );
document.write( "Let \"A\" be the original amount\r\n" );
document.write( "\r\n" );
document.write( "After 1 year only 96% will remain, which will be \"0.96A\"\r\n" );
document.write( "After 2 years only 96% of \"0.96A\" will remain, which is \"0.96%5E2%2AA\"\r\n" );
document.write( "After 3 years only 96% of \"0.96%5E2%2AA\" will remain, which is \"0.96%5E3%2AA\"\r\n" );
document.write( "...\r\n" );
document.write( "After n years only 96% of \"0.96%5E%28n-1%29A\" will remain, which is \"0.96%5En%2AA\"\r\n" );
document.write( "\r\n" );
document.write( "We want to know how many years before what will remain will be only \"1%2F2\"\"A\" or \"0.5A\" \r\n" );
document.write( "\r\n" );
document.write( "\"0.96%5En%2AA\"\"%22=%22\"\"0.5A\"\r\n" );
document.write( "\r\n" );
document.write( "Divide both sides by A\r\n" );
document.write( "\r\n" );
document.write( "\"0.96%5En\"\"%22=%22\"\"0.5\"\r\n" );
document.write( "\r\n" );
document.write( "Take logs of both sides:\r\n" );
document.write( "\r\n" );
document.write( "\"log%28%280.96%5En%29%29\"\"%22=%22\"\"log%28%280.5%29%29\"\r\n" );
document.write( "\r\n" );
document.write( "Use the rule of logs that says \"log%28B%2C%28A%5EC%29%29=C%2Alog%28B%2C%28A%29%29\" to bring\r\n" );
document.write( "the exponent n in front of the log\r\n" );
document.write( "\r\n" );
document.write( "\"n%2Alog%28%280.96%29%29\"\"%22=%22\"\"log%28%280.5%29%29\"\r\n" );
document.write( "\r\n" );
document.write( "Divide both sides by \"log%28%280.96%29%29\"\r\n" );
document.write( "\r\n" );
document.write( "\"n\"\"%22=%22\"\"log%28%280.5%29%29%2Flog%28%280.96%29%29\"\"%22=%22\"\"16.97974802\"\r\n" );
document.write( "\r\n" );
document.write( "or 17 years, rounded to the first whole year that the amount will be\r\n" );
document.write( "no more than half the original amount.\r\n" );
document.write( "\r\n" );
document.write( "Edwin
\n" ); document.write( "
\n" );