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after a special medicine is introduced into a petri dish full of bacteria the number of bacteria remaining in the decrease in the dish decreases rapidly the relationship between the elapsed time T in seconds and the number of bacteria n of T in the petri dish is modeled by the following function alright complete the following sentence about the half-life of the bacterial culture the number of bacteria is halved every blank seconds so T is being given to us in seconds so let's let's think about this a little bit let's think about a little table here I'll draw a little table here so if we have this is T and this is n of T and I'll start with a straightforward T at time equals zero right when we start this whole thing if this T is 0 then 1/2 to the 0 over 5 point 5 power that's just 1 after the zero that's all going to be 1 and you're just going to be left with one thousand one thousand bacteria in the petri dish now at what point do we get to multiply by 1/2 at what point do we get to say one thousand times 1/2 well in order to say one thousand times 1/2 the exponent here has to be one so at what time is the exponent here going to be one well the exponent here is going to be one this whole exponents going to be one when T is equal to five point five seconds so T is five point five seconds and likewise we wait another five point five seconds so if we go to eleven seconds then this is going to be one thousand times eleven divided by five point five is two so times 1/2 to the second power so times 1/2 times 1/2 so every five and a half seconds we will essentially have half of the bacteria that we had five and a half seconds ago so the number of bacteria is halved every five point five seconds and you see it in the formula and the in the the function definition right over there but it's nice to reason it through and really really digest why it makes sense let's do a few more of these the chemical the chemical element einsteinium 253 naturally loses its mass over time a sample of einsteinium 253 had an initial mass of 320 grams when we measured it the relationship between the elapsed time T in days in the mass M of T and grams left in the sample is modeled by the following function complete the following sentence about the rate of change in the mass sample the sample loses eighty seven point five percent of its mass every blank days so this one instead of saying how much we grew or shrunk by we're saying a percent change so if you lose eighty if you lose if you lose eighty seven point five percent that means that you are left with left with 12.5 percent 12.5 percent which is the same thing is that you have is the same way of saying that you have 0.125 percent of your mass so another way of thinking about it is the the sample has the sample is 0.125 of its mass of its original mass or how long does it take for the sample to be 0.125 of its mass and then we could do a similar a similar idea you see the 0.125 right over here and so I could draw a table if you like although I think you might guess where this is going but let me draw a little table here so T and M of T when T is zero M of T is three twenty and so at what point is T at what time is M of T going to be 320 times 0.125 because this going from this to this that is losing 87 point five percent of your mass losing losing 87 please so let me just write it this way so this is minus 87 point five percent of the mass you've lost 0.875 to get to 0.125 so this you could just use 0.125 to the first power so what what TD you have to make this exponent equal one well T has to be sixty one point four sixty one point four and we're T is in days sixty one point four days now you might be tempted always just pattern match so whatever is in the you know whatever's in the denominator here but I really encourage you to think about it because that's the whole point of these problems if you just our pattern matching these well I don't know how helpful that's going to be for you let's do one more of these Howard started studying how the number of branches on his tree grows over time the relationship between the elapsed time T in years since Howard started studying the tree and the number of its branches n of T is modeled by the following function complete the following sentence about the rate of change in the number of branches Howard's tree great gains four-fifths more branches every black years so gaining gaining 4/5 is equivalent to multiplying multiplying by you're gaining four-fifths of what you already are you're not just getting you know the number 4/5 you're getting 4/5 of what you already are so that's the equivalent of multiplying by one plus 4/5 or nine fifths so gaining 4/5 is the same thing as multiplying by nine fifths if I'm five years old and if I gain four fifths of my age I would gain five I've gained four years to get to be nine years old which means I've multiplied my age by nine fifths so Howard's tree you could say grows by a factor of nine fifths every how many years well you could see over here the common ratio is nine fifths and so you're going to grow by nine fifths every every time T is a multiple of seven point three or as you say every time T increases by seven point three you're going to then this exponent is going to increase by a whole and so you could view that as multiplying again by nine fifths so Howard's tree gains four fifths more branches every seven three years