document.write( "Question 1198954: b) The half-life of a radioactive substance is 300 years. If the initial amount is q milligrams, then the quantity q(t) remaining after t years is given by q(t) = q2^(kt). Find k \n" ); document.write( "

Algebra.Com's Answer #832641 by josgarithmetic(39617) $\"\"$ $\"About$

You can put this solution on YOUR website!
Strange way of putting it.\r
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\n" ); document.write( "\n" ); document.write( " $\"q%28t%29=q2%5E%28kt%29\"$ \r
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\n" ); document.write( "\n" ); document.write( "If half life 300 years then
\n" ); document.write( " $\"1%2F2=1%2A2%5E%28300k%29\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"1%2F2=2%5E%28300k%29\"$ \r
\n" ); document.write( "\n" ); document.write( "Which base do you want? ...
\n" ); document.write( " $\"log%28%281%2F2%29%29=300k%2Alog%28%282%29%29\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"300k=log%28%281%2F2%29%29%2Flog%28%282%29%29\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"300k=-log%28%282%29%29%2Flog%28%282%29%29\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"300k=-1\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"highlight%28k=-1%2F300%29\"$ \r
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\n" ); document.write( "\n" ); document.write( "The function in that form is $\"highlight%28q%28t%29=q2%5E%28%28-1%2F300%29t%29%29\"$ . \n" ); document.write( "

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