document.write( "Question 1198803: Radon is a naturally occurring radioactive gas which can collect in poorly ventilated structures. Radon gas is formed by the decomposition of radium-226 which has a half life of 1622 years. The half-life of radon is 3.82 days. Suppose a house basement contained 38 grams of radon gas when a family moved in. If the source of radium producing the radon gas is removed so that the radon gas eventually decays, how long will it take until there is only 6.8 grams of radon gas present? \n" ); document.write( "
Algebra.Com's Answer #832456 by math_tutor2020(3817)\"\" \"About 
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\n" ); document.write( "x = number of days
\n" ); document.write( "y = amount of radon leftover, in grams
\n" ); document.write( "a = starting amount of radon gas = 38 grams
\n" ); document.write( "H = half-life of radon gas = 3.82 days\r
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\n" ); document.write( "\n" ); document.write( "Half-life equation
\n" ); document.write( "y = a*(0.5)^(x/H)
\n" ); document.write( "y = 38*(0.5)^(x/3.82)\r
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\n" ); document.write( "\n" ); document.write( "We want to find out the value of x when there are y = 6.8 grams of radon gas leftover.\r
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\n" ); document.write( "\n" ); document.write( "Use logarithms to help isolate the exponent.
\n" ); document.write( "The relevant log rule used will be log(A^B) = B*log(A) to help pull down the exponent and isolate it.\r
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\n" ); document.write( "\n" ); document.write( "y = 38*(0.5)^(x/3.82)
\n" ); document.write( "6.8 = 38*(0.5)^(x/3.82)
\n" ); document.write( "6.8/38 = (0.5)^(x/3.82)
\n" ); document.write( "0.17894737 = (0.5)^(x/3.82)
\n" ); document.write( "Log(0.17894737) = Log( (0.5)^(x/3.82) )
\n" ); document.write( "Log(0.17894737) = (x/3.82)*Log(0.5)
\n" ); document.write( "3.82*Log(0.17894737) = x*Log(0.5)
\n" ); document.write( "x = 3.82*Log(0.17894737)/Log(0.5)
\n" ); document.write( "x = 9.48274032162061
\n" ); document.write( "x = 9.48\r
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\n" ); document.write( "\n" ); document.write( "It takes approximately 9.48 days for there to be 6.8 grams of radon gas remaining.
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