document.write( "Question 242015: the half life of radioactive potassium is 1.3 billion years. If ten grams are present now, how much will be present in 100 years? 1000 years? \n" ); document.write( "
Algebra.Com's Answer #177184 by nyc_function(2741)\"\" \"About 
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We use M(x) = c(.5^(x/h)), where c = initial mass, h = half-life and x = years.\r
\n" ); document.write( "\n" ); document.write( "NOTE: M(x) = the total amount of grams after certain years.\r
\n" ); document.write( "\n" ); document.write( "We first need to solve for h.\r
\n" ); document.write( "\n" ); document.write( "1.3 = 10(.5^(100/h))\r
\n" ); document.write( "\n" ); document.write( "1.3/10 = .5^(100/h)\r
\n" ); document.write( "\n" ); document.write( "0.13 = .5^(100/h)\r
\n" ); document.write( "\n" ); document.write( "Take log of both sides.\r
\n" ); document.write( "\n" ); document.write( "ln(0.13) = ln(.5^(100/h)\r
\n" ); document.write( "\n" ); document.write( "ln(0.13) = (100/h)(ln(.5))\r
\n" ); document.write( "\n" ); document.write( "Multiply both sides by h.\r
\n" ); document.write( "\n" ); document.write( "h(ln(0.13)) = 100(ln(.5))\r
\n" ); document.write( "\n" ); document.write( "Divide both sides by ln(0.13) to find h.\r
\n" ); document.write( "\n" ); document.write( "h ≈ 100(ln(.5))/ln(0.13)\r
\n" ); document.write( "\n" ); document.write( "h ≈ 33.97412529\r
\n" ); document.write( "\n" ); document.write( "We can round that off to the nearest year and the decimal number becomes 34 as the half-life.\r
\n" ); document.write( "\n" ); document.write( "I did enough. All you have to do now is replace x with 100 and simplify and then do the same but using x = 1000 in the formula and simplify.\r
\n" ); document.write( "\n" ); document.write( "Can you take it from here?\r
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