document.write( "Question 1072026: A 15 g sample of radioactive iodine decays in such a way that the mass remaining after t
\n" ); document.write( "days is given by m(t) = 15e−0.087t, where m(t) is measured in grams. After how many days are there
\n" ); document.write( "only 5 g remaining?
\n" ); document.write( "

Algebra.Com's Answer #686938 by josgarithmetic(39617)\"\" \"About 
You can put this solution on YOUR website!
\"m%28t%29=cross%2815e-0.087t%29\"
\n" ); document.write( "\"m%28t%29=15e%5E%28-0.087t%29\"------THIS way.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"ln%28m%29=ln%2815%2Ae%5E%28-0.087t%29%29\"
\n" ); document.write( "\"ln%2815%29-0.087t=ln%28m%29\"
\n" ); document.write( "\"-0.087t=ln%28m%29-ln%2815%29\"
\n" ); document.write( "\"0.087t=ln%2815%29-ln%28m%29\"
\n" ); document.write( "\"highlight_green%28t=%28ln%2815%2Fm%29%29%2F0.087%29\"\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "m=5, evaluate t.
\n" ); document.write( "\"t=ln%2815%2F5%29%2F0.087\"
\n" ); document.write( "\"t=ln%283%29%2F0.087\"
\n" ); document.write( "\"highlight%28t=12.6%29\" or about 12 and a half day
\n" ); document.write( "
\n" );