document.write( "Question 447946: The population of a city is growing exponentially. Initially the population was 43,000 while 7 years later the population was 79,700. Find\r
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document.write( "the population 4 years later (11 years from the beginning)\r
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document.write( "and the doubling time. \n" );
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Algebra.Com's Answer #308299 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! The population of a city is growing exponentially. Initially the population was 43,000 while 7 years later the population was 79,700. \n" ); document.write( "------------------------------- \n" ); document.write( "Find the population 4 years later (11 years from the beginning) \n" ); document.write( "and the doubling time. \n" ); document.write( "---------------------- \n" ); document.write( "p = ab^t \n" ); document.write( "--- \n" ); document.write( "Solve for a and b: \n" ); document.write( "--- \n" ); document.write( "Initial conditions (0,43000) so a = 43000 \n" ); document.write( "---- \n" ); document.write( "p(7) = 43000*b^7 = 79,700 \n" ); document.write( "--- \n" ); document.write( "Divide by 43000 and solve for \"b\": \n" ); document.write( "b^7 = 1.8535 \n" ); document.write( "--- \n" ); document.write( "b = 1.0922 \n" ); document.write( "--------------------- \n" ); document.write( "p(t) = 43000*(1.0922)^t \n" ); document.write( "---------------------- \n" ); document.write( "Find the population 4 years later (11 years from the beginning) \n" ); document.write( "p(11) = 43000*(1.0922)^11 = 113,395.39 \n" ); document.write( "--------- \n" ); document.write( "and the doubling time. \n" ); document.write( "Solve 1.0922^t = 2 \n" ); document.write( "--- \n" ); document.write( "t*log(1.0922) = log(2) \n" ); document.write( "--- \n" ); document.write( "t = 7.86 years \n" ); document.write( "===================== \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " \n" ); document.write( " |