document.write( "Question 924217: Can you please help me with this?\r
\n" ); document.write( "\n" ); document.write( "The amount of radioactive material, in grams, present after t days is modeled by:
\n" ); document.write( "A(t)=600e^-0.5t\r
\n" ); document.write( "\n" ); document.write( "a) Find the amount present after 12 days.
\n" ); document.write( "I think I got this one. \r
\n" ); document.write( "\n" ); document.write( "A(12)=600e^-0.5*12
\n" ); document.write( "A(12)=1.487g\r
\n" ); document.write( "\n" ); document.write( "b)find the half-life of the material
\n" ); document.write( "This one I had more trouble with. I got a negative answer when I tried it. \r
\n" ); document.write( "\n" ); document.write( "300=600e^-0.5t
\n" ); document.write( ".5=e^-0.5t
\n" ); document.write( "log base e (.5)=-0.5t
\n" ); document.write( "t=(log base e (.5))/-0.5
\n" ); document.write( "t=-.434yrs\r
\n" ); document.write( "\n" ); document.write( "but that doesn't feel right...
\n" ); document.write( "

Algebra.Com's Answer #560692 by MathLover1(20850)\"\" \"About 
You can put this solution on YOUR website!

\n" ); document.write( "a) Find the amount present after \"12\" days.
\n" ); document.write( "I think I got this one. \r
\n" ); document.write( "\n" ); document.write( "\"A%2812%29=600e%5E%28-0.5%2A12%29\"
\n" ); document.write( "\"A%2812%29=1.487g\"----you got this one\r
\n" ); document.write( "\n" ); document.write( "b)find the half-life of the material
\n" ); document.write( " \r
\n" ); document.write( "\n" ); document.write( "\"300=600e%5E%28-0.5t%29\"\r
\n" ); document.write( "\n" ); document.write( "\"300%2F600=e%5E%28-0.5t%29\"\r
\n" ); document.write( "\n" ); document.write( "\".5=e%5E%28-0.5t%29\" \r
\n" ); document.write( "\n" ); document.write( "\"ln%28.5%29=ln%28e%5E%28-0.5t%29%29\"\r
\n" ); document.write( "\n" ); document.write( "\"ln%28.5%29=%28-0.5t%29ln%28e%29\"......\"ln%28e%29=1\"\r
\n" ); document.write( "\n" ); document.write( "\"ln%28.5%29=-0.5t\"\r
\n" ); document.write( "\n" ); document.write( "\"-0.693147=-0.5t\"\r
\n" ); document.write( "\n" ); document.write( "\"-0.693147%2F-0.5=t\"\r
\n" ); document.write( "\n" ); document.write( "\"t=1.386294\"\r
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