document.write( "Question 568663: I'm not entirely sure this is the right subject, sorry if it's not...\r
\n" ); document.write( "\n" ); document.write( "A substance decomposes radioactively. Its half-life is 32 years. Find the constant \"k\" in the decay formula.
\n" ); document.write( "Use this formula instead: \"y=ne%5Ekt\"
\n" ); document.write( "y= final value
\n" ); document.write( "n= initial value
\n" ); document.write( "k= constant
\n" ); document.write( "t= time
\n" ); document.write( "e= antilogarithm
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Algebra.Com's Answer #367064 by solver91311(24713)\"\" \"About 
You can put this solution on YOUR website!
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\n" ); document.write( "\n" ); document.write( " is NOT the antilogarithm. is the base of the natural logs, that is: \r
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\n" ); document.write( "\n" ); document.write( "If one half-life has elapsed, then .\r
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\n" ); document.write( "\n" ); document.write( "Since the half-life of the substance in question is 32 years, then\r
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\n" ); document.write( "\n" ); document.write( "Take the natural log of both sides:\r
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\n" ); document.write( "\n" ); document.write( "Use\r
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\n" ); document.write( "\n" ); document.write( "Use \"the difference of the logs is the log of the quotient\",\r
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\n" ); document.write( "\n" ); document.write( "then use\r
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\n" ); document.write( "My calculator said it, I believe it, that settles it
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