document.write( "Question 601067: given A(t)=Ao e^-0.00023t is a proper model for radioctive decay, find the halflife (years) \n" ); document.write( "
Algebra.Com's Answer #379595 by jim_thompson5910(35256)\"\" \"About 
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\"A%28t%29+=+A%5B0%5De%5E%28rt%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"50+=+100%2Ae%5E%28-0.00023t%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"50%2F100+=+e%5E%28-0.00023t%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"0.5+=+e%5E%28-0.00023t%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"ln%280.5%29+=+ln%28e%5E%28-0.00023t%29%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"ln%280.5%29+=+-0.00023t%2Aln%28e%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"ln%280.5%29+=+-0.00023t%2A%281%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"ln%280.5%29+=+-0.00023t\"\r
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\n" ); document.write( "\n" ); document.write( "\"ln%280.5%29%2F-0.00023+=+t\"\r
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\n" ); document.write( "\n" ); document.write( "\"-0.693147180559945%2F-0.00023+=+t\"\r
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\n" ); document.write( "\n" ); document.write( "\"3013.68339373889+=+t\"\r
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\n" ); document.write( "\n" ); document.write( "\"t+=+3013.68339373889\"\r
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\n" ); document.write( "\n" ); document.write( "So the half-life is roughly 3013.68339 years.
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