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In this question, we are told that the culture of bacteria grows at a rate proportional to the current size of the culture.
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And we are also told that after 10 hours, there are 5 ,000 bacteria and after 12 hours, there are 6 ,000 bacteria.
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And we are asked to find the culture's initial population and to find the doubling time.
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First of all, since the population grows at a rate, which is dp over d t, proportional to its size, which means that it equals to k multiplied by p, this gives us a differential equation for finding p.
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If we divide both sides by p and multiply by d t, we are going to get that d p over p equals to k times d t.
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And after integrating both sides, we are going to get that ln of the absolute value of p equals to kt plus c.
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And after exponentiating both sides, we are going to get that p of t equals to c multiplied by e to the kt.
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And we need to find c and we need to find k.
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And that's why we need the initial conditions.
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Not initial, but the given conditions.
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On the one hand, p of 10 equals to 5 ,000.
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On the other hand, if we plug in t equals 10 in this formula, we are going to get c multiplied by e to the 10k.
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We are also told that p of 12 equals to 6 ,000.
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On the other hand, if we plug in t equals 12 in the formula, we are going to get c multiplied by e to the 12k power.
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What we are going to do next is divide y an equation by another.
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If we divide, let's divide p of 12 by p of 10.
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If we divide p of 12 by p of 10, on the one hand, we are going to get 6 over 5.
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On the other hand, we are going to get e to the 12k divide by e to the 10k because c canccccccccccccc equals to the 12k.
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And, you to the 12k over e to the 10k equals to e to the 2k.
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And, and this gives us an equation for finding k.
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E to 2k equals 6 over 5.
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Therefore, after applying the natural logarithm, we are going to get e to the ln of e to 2k equals to ln of 6 over 5.
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By using properties of logarithms, you can write the left -hand side as 2k multiplied by lne equals to ln of 6 over 5.
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And since lne equals to 1, we are going to get that k equals to 1 half times ln6 over 5, or ln6 over 2.
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And this means that we can rewrite the formula for p, this formula, we can replace k by the number we just found.
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We are going to get c times e to the power ln6 over 5 over 2 times t.
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Now we need to find the constant c.
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And to find c, we will use same equations, right? we still have two equations here.
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And if we plug in k we just found, we are going to get an equation for finding c.
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So let's plug in the value of k in the first equation.
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The equation is 5 ,000 equals to c times e to the 10k.
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And we just found that k equals to ln 6 over 5 divided by 2 this equals to c times e to the 5 ln 6 over 5 and therefore we are going to get that c equals to 5 000 divided by e now we can error by using the properties of logarithms we can write 5 ln 6 over 5 as ln of 6 over 5 to the 5th power.
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So what we did is we brought the 5 in front of the logarithm inside the logarithm as an exponent.
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The property we used here is that r multiplied by lna equals to ln of a to the r.
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This is a property we use here.
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R equals to 5 and a equals to 6 over 5.
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And then e and ln are inverse functions and they cancel each other and we are going to get 5 ,000 divided by 6 over 5 to the 5th power...