document.write( "Question 1161308: Suppose that the population in a small city is 39,000 in the beginning of 2012 and the city council assumes that the population size t years later can be estimated by the equation\r
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document.write( "P=39,000e^0.09t\r
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document.write( "Approximately when will the city have a population of 60,000? \n" );
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Algebra.Com's Answer #784826 by Theo(13342)![]() ![]() You can put this solution on YOUR website! the equation is p = 39,000 * e^(.09*t) \n" ); document.write( "when the population is 60,000, the equation becomes: \n" ); document.write( "60,000 = 39,000 * e^(.09*t) \n" ); document.write( "divide both sides of the equation by 39,000 and simplify to get: \n" ); document.write( "60/39 = e^(.09*t) \n" ); document.write( "take the natural log of both sides of the equation to get: \n" ); document.write( "ln(60/39) = ln(e^(.09*t)) \n" ); document.write( "by the properties of logarithms, this is equivalent to: \n" ); document.write( "ln(60/39) = .09*t*ln(e) \n" ); document.write( "since ln(e) = 1, this becomes: \n" ); document.write( "ln(60/39) = .09*t \n" ); document.write( "divide both sides of this equation by .09 and solve for t to get: \n" ); document.write( "t = ln(60/39)/.09 = 4.786476845 \n" ); document.write( "confirm by replacing t in the original equation to get: \n" ); document.write( "f = 39,000 * e^(.09*4.786476845) = 60,000 \n" ); document.write( "this confirms the solution is correct. \n" ); document.write( "since 39,000 population is in the beginning of 2012, and the population reaches 60,000 4.786476845 years from then, the population reaches 60,000 sometime in the year 2016.\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |