document.write( "Question 1118557: What is the half-life of a material that decays exponentially at the rate of 3.54% per year? Using exponential decay formula N(t)= Noe^(-kt) ? Thank you \n" ); document.write( "
Algebra.Com's Answer #733911 by josgarithmetic(39623)\"\" \"About 
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One year of time pass
\n" ); document.write( "\"1%2Ae%5E%28-k%2A1%29=1-0.0354\"\r
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\n" ); document.write( "\n" ); document.write( "\"e%5E%28-k%29=0.9646\"\r
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\n" ); document.write( "\n" ); document.write( "\"ln%28e%5E%28-k%29%29=ln%280.9646%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"-k=-0.03604\"\r
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\n" ); document.write( "\n" ); document.write( "\"k=0.03604\"\r
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\n" ); document.write( "\n" ); document.write( "The model equation: \"highlight_green%28N%28t%29=N%5Bo%5De%5E%28-0.03604t%29%29\"\r
\n" ); document.write( "\n" ); document.write( "FINDING HALF-LIFE\r
\n" ); document.write( "\n" ); document.write( "\"e%5E%28-0.03604t%29=0.5%2F1\"\r
\n" ); document.write( "\n" ); document.write( "\"e%5E%28-0.03604t%29=0.5\"\r
\n" ); document.write( "\n" ); document.write( "\"ln%28e%5E%28-0.03604t%29%29=ln%280.5%29\"\r
\n" ); document.write( "\n" ); document.write( "\"\"\r
\n" ); document.write( "\n" ); document.write( "\"t=ln%280.5%29%2F%28-0.03604%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"t=highlight%2819.23%29\" or about 19 years 3 months
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