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)![]() ![]() ![]() You can put this solution on YOUR website! 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( "-------- \n" ); document.write( "y = ab^x \n" ); document.write( "------- \n" ); document.write( "You have 2 points (0,900) and (40,500) \n" ); document.write( "--------------------- \n" ); document.write( "Using (0,900) 900 = ab^0 \n" ); document.write( "900 = a \n" ); document.write( "------- \n" ); document.write( "y = 900*b^x \n" ); document.write( "----- \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 \n" ); document.write( "---- \n" ); document.write( "(x/40) = log(1/3)/log(5/9) \n" ); document.write( "x/40 = 1.87 \n" ); document.write( "x = 75 minutes \n" ); document.write( "====================== \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( "================== \n" ); document.write( " |