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Video: Using Logarithms to Solve Geometric Sequence Problems

Tim Burnham

We explain how to use logarithms and your knowledge of geometric sequences to solve problems (e.g., which is the first term to exceed 12000 in a geometric sequence with a first term of 7 and common ratio of 2?).

12:46

Video Transcript

In this video, we’re gonna combine your knowledge of geometric sequences with your knowledge of logarithms to solve some problems.

First of all, remember that a geometric sequence is a sequence of numbers in which each term is multiplied by a common ratio, in order to obtain the next term. For example three, six, twelve, twenty-four, and so on. The first term, which we generally call π‘Ž one, in this case has the value three. And in each case, I’m multiplying one term by two to get the next term. So that common multiple, we call the common ratio π‘Ÿ, so in this case π‘Ÿ is equal to two. And we can come up with a formula for the 𝑛th term, π‘Ž 𝑛 is equal to π‘Ž one times π‘Ÿ to the power of 𝑛 minus one.

Now that works because our first term we called π‘Ž one, and the second term was the first term times the common ratio, so that’s π‘Ž one times π‘Ÿ to the power of one. And to get our third term, we multiply that second term by π‘Ÿ. So we’ve got π‘Ž one times π‘Ÿ times π‘Ÿ, or times π‘Ÿ squared. And to get our fourth term, we multiply that term by π‘Ÿ again. So we’ve got π‘Ž one times π‘Ÿ squared times π‘Ÿ, which is π‘Ž one times π‘Ÿ cubed. So the first term is π‘Ž one, not times π‘Ÿ, or you could say times one, which is π‘Ÿ to the power of zero. So the first term is π‘Ž one times π‘Ÿ to the power of zero, the second term is π‘Ž one times π‘Ÿ to the power of one, the third term is π‘Ž one times π‘Ÿ to the power of two, and the fourth term is π‘Ž one times π‘Ÿ to the power of three. So in each case, it’s π‘Ž one times π‘Ÿ to the power of one less than that position in the sequence, which is exactly what this formula is saying.

Now if you’re not told the value of π‘Ÿ, you can easily work it out. Because each term is simply π‘Ÿ times the previous term, we can work out the value of π‘Ÿ by doing one term divided by the previous term. So the second term divided by the first term will give you π‘Ÿ, the third term divided by the second term will also give you π‘Ÿ, and the fourth term divided by the third term will give you π‘Ÿ. Each term divided by the previous term will always give you the same answer, π‘Ÿ.

And also remember, about logarithms, if π‘Ž to the power of π‘₯ is equal to 𝑛 then log base π‘Ž of 𝑛 is equal to π‘₯. So log base π‘Ž of 𝑛 means, what power do I need to raise π‘Ž to in order to get 𝑛? Or, π‘Ž to the power of what equals 𝑛? So I can use these two things combined to solve some geometric sequence problems. Let’s take a look.

In the geometric sequence with first term seven and common ratio two, which is the first term to exceed twelve thousand?

So our first term is seven, π‘Ž one is equal to seven. And the common ratio is two, so π‘Ÿ is equal to two. Now remember, our general term for a geometric sequence is that the 𝑛th term π‘Ž 𝑛 is equal to the first term π‘Ž one times the common ratio π‘Ÿ to the power of 𝑛 minus one. So in this case, it means that our 𝑛th term is equal to seven times two to the power of 𝑛 minus one. Now remember, this is a sequence so- and it really makes sense for 𝑛 to have integer values, the first term, the second term, the third term, and so on. It doesn’t really make a lot of sense to talk about the twelve point six twoth term for example.

Now we could try putting π‘Ž 𝑛 equal to twelve thousand, so the 𝑛th term is twelve thousand, and see what value of 𝑛 we get. If it’s a whole number, then we found the term that equals twelve thousand. So if we add one to that, the next term will be the first one that exceeds it. And if 𝑛 is not a whole number, then the next integer up from that number will be the first term that exceeds twelve thousand. Let’s have a look.

So this means that twelve thousand is equal to seven times two to the power of 𝑛 minus one. Well I could divide both sides by seven. And we’ve got twelve thousand over seven is equal to two to the power of 𝑛 minus one. Well we could do a lot of trial and error with different values of 𝑛 to see which one matches this equation. But a better idea is to take logarithms of both sides, and then we could use the rules of logarithms to help us solve the equation. So I’m gonna take logs base ten of both sides. So that log ten of twelve thousand over seven is equal to log ten of two to the power of 𝑛 minus one. Using the log power rule, log ten of two to the power of 𝑛 minus one is the same as 𝑛 minus one times log ten of two. And now I could divide both sides by log ten of two, so that I know what 𝑛 minus one is equal to. And then I could add one to both sides so that I’ve just got an equation for 𝑛.

And that’s a nice little job for my calculator, so when I type that in β€” Now as we said before, it only makes sense for 𝑛 to be an integer. We’re talking about positions in a sequence, so the first, second, third, and so on. So if 𝑛 was equal to eleven point seven, et cetera, et cetera, that would generate an exact answer of twelve thousand. So really our only options in terms of our sequence are, 𝑛 could be eleven, or 𝑛 could be twelve. Well if 𝑛 was eleven, then our term would be lower than twelve thousand. So if 𝑛 was twelve, then the value would be higher than twelve thousand. So our answer: The first term to exceed twelve thousand is the twelfth term.

So just quickly working out the value of the eleventh and the twelfth term, you don’t actually need to do this by the way but we’re gonna do it just to show you how it works, the eleventh term was seven thousand one hundred and sixty-eight and the twelfth term was fourteen thousand three hundred and thirty-six. So as you can see, the eleventh term is less than twelve thousand the twelfth term is the first term to be greater than twelve thousand. So our answer is: The twelfth term fourteen thousand three hundred and thirty-six is the first to exceed twelve thousand.

Let’s do one more example, this time one that involves depreciation. So it’s a decreasing geometric sequence.

Ella buys a car for thirty thousand dollars. It depreciates, or loses value, at a rate of six percent per year. How many years will it be before it’s worth less than two thousand five hundred dollars?

Now this question is different to the last one in two important ways. Firstly obviously, the numbers are going down rather than up because it’s depreciating geometric sequence. But secondly, we have to be really careful about how we label the terms. We’ve been given a starting value of thirty thousand dollars, but the first term in our sequence is how much it will be worth in one year’s time. Okay. So each year we’re losing six percent of the value of the car, which means we’re retaining ninety-four percent of the value of the car. Now to work out ninety-four percent of something, it’s ninety-four over a hundred times that number. And ninety-four over a hundred is nought point nine four. So it’s like if we multiply by nought point nine four, that will tell us what the car will be worth in a year’s time. So we could use a normal geometric sequence formula by saying that the first term is thirty thousand times nought point nine four, and that common ratio is nought point nine four. And then the 𝑛th term of the sequence is π‘Ž one times π‘Ÿ to the power of 𝑛 minus one. And if we do that when 𝑛 equals one, we’re finding the value in one year’s time. When 𝑛 equals two, we’re finding value in two years’ time. When 𝑛 equals three, we’re finding value in three years’ time, and so on.

But looking at that formula, we’re always multiplying thirty thousand by nought point nine for once, and then another 𝑛 minus one times. So in total, we’re multiplying thirty thousand by nought point four- nine four 𝑛 times. So to work out the residual value of the car in 𝑛 years’ time is just thirty thousand times nought point nine four to the power of 𝑛. Now before we move on, I’m just gonna look at another way to come to that same formula. In one year’s time, the car will be worth nought point nine four times thirty thousand. In two years’ time, we’ll take the value after one year and multiply that by nought point nine four, so it’s gonna be nought point nine four times nought point nine four times thirty thousand, or nought point nine four squared times thirty thousand. In three years’ time, we’re gonna take the value after two years and multiply that by nought point nine four. So that’s nought point nine four times nought point nine four squared times thirty thousand, which is nought point nine four cubed times thirty thousand. And in four years, we’ll take the value after three years and multiply that by nought point nine four again. So that will be nought point nine four times nought point nine four cubed times thirty thousand, or nought point nine four to the power of four times thirty thousand.

Now looking at this first formula, nought point nine four, that’s the same as nought point nine four to the power of one. So what you’ll notice is that the power of nought point nine four is always equal to the year number that we’re looking at. So nought point nine four cubed is the third year, nought point nine four to the power of four is the fourth year, and so on. So the value after 𝑛 years, let’s call it π‘Ž 𝑛, is equal to nought point nine four to the power of 𝑛 times thirty thousand. And we want to know the value of 𝑛 when this whole expression becomes worth less than two thousand five hundred dollars. So that’s the calculation we’re gonna try to do. Well, except, we’re gonna do the same thing that we did last time. We’re actually gonna find- try and find the value of 𝑛 that makes this equal to two thousand five hundred, and then work out which is the integer value of 𝑛 which takes this value to below two thousand five hundred. So the car is worth two thousand five hundred dollars when nought point nine four to the power of 𝑛 is times thirty thousand is equal to two thousand five hundred. Now if I divide both sides by thirty thousand, I get nought point nine four to the power of 𝑛 is it equal to two thousand five hundred over thirty thousand, and that cancels to one-twelfth.

So nought point nine four to the power of 𝑛 is equal to a twelfth. Now I’m gonna take log base ten of both sides. So log base ten of nought point nine four to the power of 𝑛 is equal to log base ten of a twelfth and now I’m gonna use the log power rule to rearrange the left-hand side. And that enables me to say that 𝑛 times log ten of nought point nine four is equal to log ten of a twelfth. So now I can divide both sides by log ten of nought point nine four and that leaves me with an expression for 𝑛. And again, I can just tap this into my calculator and then I find that 𝑛 would be equal to forty point one five nine four five four four and so on.

So this means that theoretically, if this was a continual thing, when 𝑛 is just over forty, so in just over forty years’ time, the car will be worth exactly two thousand five hundred dollars. So that means in forty years’ time, it will still be worth slightly more than two thousand five hundred dollars. But in forty-one years’ time, the value will have dipped to below two thousand five hundred dollars. So just quickly checking those two values, when 𝑛 is forty, the value of the car is two thousand five hundred and twenty-four dollars eighty-five cents, just rounding to two decimal places obviously because it’s money. And when 𝑛 is forty-one, plugging that number into the equation, I get a value of two thousand three hundred and seventy-three dollars and thirty-six cents. So our answer is: The car will be worth less than two thousand five hundred dollars in forty-one years’ time. Of course, it will also be worth less than two thousand five hundred dollars in forty-two, forty-three, and so on. So the first time it becomes worth less than two thousand five hundred dollars, is in forty-one years.

So just a couple of tips and pointers then for things to watch out for, in that sort of question. First, we were depreciating by six percent per year, now that’s the same as working out what is ninety-four percent of the number. So rather than trying to take away six percent, we work out the residual value of the ninety-four percent that’s gonna remain. The other thing is, in these sorts of questions where you’re told an initial value when something is happening every year, you’ve gotta be very careful of what you mean when you talk about 𝑛, what’s your definition of the first term, what’s your definition of the value of 𝑛. The standard formula for geometric sequences is π‘Ž 𝑛 equals π‘Ž one times π‘Ÿ to the power of 𝑛 minus one. But because of our definition of 𝑛, and π‘Ÿ, and π‘Ž one in this case, we’ve got a formula that looks more like the 𝑛th term of this sequence, is: Thirty thousand, initial value of the car, times nought point nine four to the power of 𝑛. So you just need to be careful about that discrepancy between 𝑛 and 𝑛 minus one.