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Advanced Math / Nonlinear functions Difficulty: Medium

The function $f (t) = 60,000 {(2)}^{\frac{t}{410}}$ gives the number of bacteria in a population $t$ minutes after an initial observation. How much time, in minutes, does it take for the number of bacteria in the population to double?

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Explanation

The correct answer is $410$ . It's given that $t$ minutes after an initial observation, the number of bacteria in a population is $60,000 left parenthesis 2 right parenthesis Superscript StartFraction t Over 410 EndFraction$ . This expression consists of the initial number of bacteria, $60,000$ , multiplied by the expression $2 Superscript StartFraction t Over 410 EndFraction$ . The time it takes for the number of bacteria to double is the increase in the value of $t$ that causes the expression $2 Superscript StartFraction t Over 410 EndFraction$ to double. Since the base of the expression $2 Superscript StartFraction t Over 410 EndFraction$ is $2$ , the expression $2 Superscript StartFraction t Over 410 EndFraction$ will double when the exponent increases by $1$ . Since the exponent of the expression $2 Superscript StartFraction t Over 410 EndFraction$ is $StartFraction t Over 410 EndFraction$ , the exponent will increase by $1$ when $t$ increases by $410$ . Therefore the time, in minutes, it takes for the number of bacteria in the population to double is $410$ .