It’s your turn on growth and decay problems

Hello. Welcome back. It’s your turn now. But before solving the exercises, let me remind you how logarithms are so powerful in order to solve equations in the exponent. For instance, if you wish to solve an equation of the form a to the x equals to b and the unknown is x, well, it is enough to take on both sides of the equality the logarithm to base anything you want, for instance, a, and you get the solution. For instance, if we apply the logarithm to base a on both sides of the equality, we get logarithm to base a of a to the x equals to logarithm to base a of b. But this is exactly x.

So we get immediately x equals to logarithm to base a of b. And this is the solution to our equation. Now, in Exercise 1, we’ve got a capital of 8 pounds that is invested with a compound interest of 5% per year. This means that after n years, we get the initial capital, 8000 pounds, multiplied by 1 plus 5 divided by 100 to the n. And we want this to be equal to 10000. So the equation is this quantity– find n in such a way that this quantity equals 10000. Well, this is equivalent to 1 plus 5 divided by 100 to the n equals 10 divided by 8, which is 5 over 4.

And so we get 1.05 to the n equals, well, 5/4. And then in order to get n, we apply a logarithm, any logarithm. But the most convenient is probably logarithm to base 1.05, because it gives just exactly n when we apply it to 1.05 to the n. So this is equivalent to logarithm to base 1.05 of 1.05 to the n equals to logarithm to base 1.05 of 5/4. Well, this is exactly n.

And so we get n equals to logarithm to base 1.05 of 5/4. Now, if you want to compute it, we can write that the logarithm to base 1.05 of 5/4 is expressed in terms of the natural logarithm as natural logarithm 5/4 divided by natural logarithm of 1.05. And this gives more or less 4.57 and so on. And this ends Exercise 1.

In Exercise 2, we are facing a problem concerning the mass of a radioactive substance, in particular the polonium. Actually, remind me of spy story. I don’t know why. Well, the mass is expressed as the initial mass, m0, times 2 to the minus t over 140. And they ask us how many days– this is expressed, t in days. So t is expressed in terms of days. And they ask us how many days does it take to decay, for the mass to decay to half of the initial mass. So we want to find t in such a way that m of t equals m0 divided by 2.

So the equation is m0, 2 to minus t to 140 equals to m0 divided by 2. We simplify and we get 2 times 2 minus t divided by 140 equals to 1, which is 2 to 1 minus t divided 140 equals to 1. 1 is also 2 to 0. And so this is equivalent since the 2 to the x is an injective function, this is– and it takes the value 1 just in 0, this is equivalent to 1 minus t over 140 equals to 0. That is, t equals 140 days.

And this answers Exercise 2.