document.write( "Question 236978: 2= 5e^-0.35h\r
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Algebra.Com's Answer #174334 by Theo(13342)\"\" \"About 
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2 = 5e^-0.35h\r
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\n" ); document.write( "\n" ); document.write( "2/5 = e^(-.35h)\r
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\n" ); document.write( "\n" ); document.write( "take natural log of both sides to get:\r
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\n" ); document.write( "\n" ); document.write( "ln(2/5) = ln(e^(-.35h)\r
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\n" ); document.write( "\n" ); document.write( "since log(b^x) = x*log(b), your equation becomes:\r
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\n" ); document.write( "\n" ); document.write( "ln(2/5) = -.35h * (ln(e))\r
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\n" ); document.write( "\n" ); document.write( "since ln(e) = 1, your equation becomes:\r
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\n" ); document.write( "\n" ); document.write( "ln(2/5) = -.35h\r
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\n" ); document.write( "\n" ); document.write( "divide both sides by -.35 to get:\r
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\n" ); document.write( "\n" ); document.write( "ln(2/5)/(-.35) = h\r
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\n" ); document.write( "\n" ); document.write( "since ln(2/5) = (-.916290732), your equation becomes:\r
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\n" ); document.write( "\n" ); document.write( "(-.916290732)/(-.35) = h\r
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\n" ); document.write( "\n" ); document.write( "simplify to get:\r
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\n" ); document.write( "\n" ); document.write( "h = 2.61797352\r
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\n" ); document.write( "\n" ); document.write( "to prove this is correct, replace h in the original equation to get:\r
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\n" ); document.write( "\n" ); document.write( "2/5 = e^(-.35h) becomes:\r
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\n" ); document.write( "\n" ); document.write( "2/5 = e^(-.35*2.61797352)\r
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\n" ); document.write( "\n" ); document.write( "this becomes:\r
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\n" ); document.write( "\n" ); document.write( "2/5 = e^(-.916290732)\r
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\n" ); document.write( "\n" ); document.write( "this becomes:\r
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\n" ); document.write( "\n" ); document.write( ".4 = .4 confirming the answer is correct.\r
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