document.write( "Question 753000: Under the onslaught of the college algebra second period class, a pile of homework problems decreased exponentially. It decreased form 900 to 500 problems in only 40 minutes. How long would it take until only 300 problems remained? \n" ); document.write( "
Algebra.Com's Answer #458170 by stanbon(75887)\"\" \"About 
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Under the onslaught of the college algebra second period class, a pile of homework problems decreased exponentially. It decreased form 900 to 500 problems in only 40 minutes. How long would it take until only 300 problems remained?
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\n" ); document.write( "y = ab^x
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\n" ); document.write( "You have 2 points (0,900) and (40,500)
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\n" ); document.write( "Using (0,900) 900 = ab^0
\n" ); document.write( "900 = a
\n" ); document.write( "-------
\n" ); document.write( "y = 900*b^x
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\n" ); document.write( "Using (40,500) solve for \"b\":
\n" ); document.write( "500 = 900*b^40
\n" ); document.write( "b^40 = 5/9
\n" ); document.write( "------
\n" ); document.write( "Equation:
\n" ); document.write( "f(x) = 900(5/9)^(x/40)
\n" ); document.write( "-------------------
\n" ); document.write( "How long would it take until only 300 problems remained?
\n" ); document.write( "Solve: 300 = 900(5/9)^(x/40)
\n" ); document.write( "(5/9)^(x/40) = 1/3
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\n" ); document.write( "(x/40) = log(1/3)/log(5/9)
\n" ); document.write( "x/40 = 1.87
\n" ); document.write( "x = 75 minutes
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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