OG详解-OG7 数学1 Q38

Choice B is correct. It’s given that t minutes after an initial observation, the  t   number of bacteria in a population is 40000(2)t​790​ . This expression consists of the initial number of bacteria, 40,000, multiplied by the expression ​2t790​ . The time, in minutes, it takes for the number of bacteria to double is the increase in the value of t that causes the expression 2t790​to double. Since the base is 2, the expression 2t790​ will double when the exponent increases by 1. Since the exponent of this expression is ​t790​ , the exponent will increase by 1 when t increases by 790. Alternate approach: The initial number of bacteria in the population can be found 0 by substituting 0 for t in the given function. This yields f(0) = 40000(2)0​790​ , or f(0) = 40,000. Therefore, the initial number of bacteria present in the population is 40,000, so the bacteria population will have doubled when f(t) = 80,000. Substituting 80,000 for f(t) in the given function yields 80,000 = 40000(2)t​790​. Dividing both sides of this equation by 40,000 yields 2 =2​t790​ , or 2¹ = 2​t790​ . It follows that 1 = ​t790​  . Multiplying both sides of this equation by 790 yields 790 = t. Therefore, the time, in minutes, it takes for the number of bacteria in the population to double is 790.
Choice A is incorrect. This is the base of the exponent, not the time it takes for the number of bacteria in the population to double. Choice C is incorrect. This is the number of minutes it takes for the population to double twice. Choice D is incorrect. This is the number of bacteria that are initially observed, not the time it takes for the number of bacteria in the population to double.