document.write( "Question 1130481: Last one tonight and I believe this will help complete my study guide for this chapter. Thank you in advance.\r
\n" ); document.write( "\n" ); document.write( "The decay rate of a certain chemical is 9.5​% per year. What is its ​half-life? Use the exponential decay model ​P(t)=P0e Superscript -kt)
\n" ); document.write( "−kt where k is the decay​ rate, and P0 is the original amount of chemical.
\n" ); document.write( "The​ half-life of the chemical is ___years.
\n" ); document.write( "​(Round to the nearest​ integer.)
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Algebra.Com's Answer #747151 by josgarithmetic(39617)\"\" \"About 
You can put this solution on YOUR website!
Model is \"P%28t%29=P%5Bo%5De%5E%28-kt%29\", and half life is value of t when \"P%28t%29=%281%2F2%29%28P%5Bo%5D%29\".\r
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\n" ); document.write( "\n" ); document.write( "FIND k, first.
\n" ); document.write( "Some variables changes just for notational and keyboard conveniences.\r
\n" ); document.write( "\n" ); document.write( "y=p*e^(-kt) instead of \"P%28t%29\" and \"P%5Bo%5D\";\r
\n" ); document.write( "\n" ); document.write( "\"y=1%2Ae%5E%28-kt%29\"\r
\n" ); document.write( "\n" ); document.write( "\"1-0.095=1%2Ae%5E%28-k%2A1%29\", one year of decay time\r
\n" ); document.write( "\n" ); document.write( "\"0.905=e%5E%28-k%29\"\r
\n" ); document.write( "\n" ); document.write( "\"ln%280.905%29=ln%28e%5E%28-k%29%29\"\r
\n" ); document.write( "\n" ); document.write( "\"-k=ln%280.905%29\"\r
\n" ); document.write( "\n" ); document.write( "\"-k=-0.099820\"\r
\n" ); document.write( "\n" ); document.write( "\"k=0.099820\"\r
\n" ); document.write( "\n" ); document.write( "Model can be stated \"highlight%28P%28t%29=P%5Bo%5De%5E%28-0.09982t%29%29\".\r
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\n" ); document.write( "\n" ); document.write( "FIND HALF-LIFE\r
\n" ); document.write( "\n" ); document.write( "Using model as \"1%2F2=1%2Ae%5E%28-0.09982t%29\"
\n" ); document.write( "take natural logs of both sides, simplify, and solve for t.\r
\n" ); document.write( "\n" ); document.write( "result should be \"t%5B1%2F2%5D=ln%282%29%2F0.09982\".
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