The problem describes exponential growth of a bacteria culture. The general formula for exponential growth is P(t) = P0 * e^(kt), where P0 is the initial population, k is the growth rate, and t is time.

(a) To find the expression for the number of bacteria after t hours, we first need to find the growth rate k. We know that the population increased to 420 after 1 hour, so we can set up the equation 420 = 100 * e^(k*1) and solve for k. This gives us k = ln(4.2). Therefore, the expression for the number of bacteria after t hours is P(t) = 100 * e^(ln(4.2)t).

(b) To find the number of bacteria after 4 hours, we substitute t = 4 into the equation from part (a). This gives us P(4) = 100 * e^(ln(4.2)*4). Evaluating this expression gives approximately 31121 bacteria.

(c) The rate of growth after 4 hours is given by the derivative of the population function, P'(t) = k * P(t). Substituting t = 4 and k = ln(4.2) gives P'(4) = ln(4.2) * 100 * e^(ln(4.2)*4). Evaluating this expression gives approximately 130482 bacteria per hour.

(d) To find when the population will reach 10,000, we set P(t) = 10000 and solve for t. This gives us the equation 10000 = 100 * e^(ln(4.2)t). Solving for t gives approximately 1.5 hours.

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The problem describes exponential growth of a bacteria culture. The general formula for exponential growth is P(t) = P0 * e^(kt), where P0 is the initial population, k is the growth rate, and t is time.

(a) To find the expression for the number of bacteria after t hours, we first need to find the growth rate k. We know that the population increased to 420 after 1 hour, so we can set up the equation 420 = 100 * e^(k*1) and solve for k. This gives us k = ln(4.2). Therefore, the expression for the number of bacteria after t hours is P(t) = 100 * e^(ln(4.2)t).

(b) To find the number of bacteria after 4 hours, we substitute t = 4 into the equation from part (a). This gives us P(4) = 100 * e^(ln(4.2)*4). Evaluating this expression gives approximately 31121 bacteria.

(c) The rate of growth after 4 hours is given by the derivative of the population function, P'(t) = k * P(t). Substituting t = 4 and k = ln(4.2) gives P'(4) = ln(4.2) * 100 * e^(ln(4.2)*4). Evaluating this expression gives approximately 130482 bacteria per hour.

(d) To find when the population will reach 10,000, we set P(t) = 10000 and solve for t. This gives us the equation 10000 = 100 * e^(ln(4.2)t). Solving for t gives approximately 1.5 hours.

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The problem describes exponential growth of a bacteria culture. The general formula for exponential growth is P(t) = P0 * e^(kt), where P0 is the initial population, k is the growth rate, and t is time.

(a) To find the expression for the number of bacteria after t hours, we first need to find the growth rate k. We know that the population increased to 420 after 1 hour, so we can set up the equation 420 = 100 * e^(k*1) and solve for k. This gives us k = ln(4.2). Therefore, the expression for the number of bacteria after t hours is P(t) = 100 * e^(ln(4.2)t).

(b) To find the number of bacteria after 4 hours, we substitute t = 4 into the equation from part (a). This gives us P(4) = 100 * e^(ln(4.2)*4). Evaluating this expression gives approximately 31121 bacteria.

(c) The rate of growth after 4 hours is given by the derivative of the population function, P'(t) = k * P(t). Substituting t = 4 and k = ln(4.2) gives P'(4) = ln(4.2) * 100 * e^(ln(4.2)*4). Evaluating this expression gives approximately 130482 bacteria per hour.

(d) To find when the population will reach 10,000, we set P(t) = 10000 and solve for t. This gives us the equation 10000 = 100 * e^(ln(4.2)t). Solving for t gives approximately 1.5 hours.

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