Video: Constructing Equivalent Exponential Functions with Different Bases from a Graph

Which of the following expressions does NOT describe the shown graph? [A] 𝑁 = 80 β‹… (2^(ln 1.25/ln 2))^𝑑 [B] 𝑁 = 80 β‹… 2^((ln 1.25/ln 2)𝑑) [C] 𝑁 = 80 β‹… 2^((ln 2/ln 1.25)𝑑) [D] 𝑁 = 80 β‹… 𝑒^(ln 1.25 β‹… 𝑑) [E] 𝑁 = 80 β‹… 1.25^𝑑.

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Video Transcript

Which of the following expressions does not describe the shown graph? Is it A: 𝑁 equals 80 times two to the power of ln 1.25 over ln two, all to the power of 𝑑? Is it option B: 𝑁 equals 80 times two to the power of ln 1.25 over ln two times 𝑑? Is it option C: 𝑁 equals 80 times two to the power of ln two over ln 1.25 times 𝑑? Option D: 𝑁 equals 80 times 𝑒 to the power of ln 1.25 times 𝑑? Or is it option E: 𝑁 equals 80 times 1.25 to the power of 𝑑?

Notice that in each of these options, 𝑁 is an exponential function of 𝑑. So it’s written as 𝐢 some multiplicative constant times π‘Ž to the power of 𝑏𝑑. You can notice actually that each option has the same value of the multiplicative constant 𝐢; each value is 80. And this makes sense given what we see on our graph. The 𝑦-intercept or should we say 𝑁 intercept of our graph is 80. And in general, the value of the multiplicative constant 𝐢 is the value of the 𝑦-intercept or in this case 𝑁 intercept.

Remember that we’re looking for the option which does not describe the shown graph. That means that four of these options do describe the shown graph. And so four of those options were guaranteed to have the value of the multiplicative constant 𝐢 as 80. Unfortunately for us, the incorrect option also has a multiplicative constant of 80. And so we can’t use the fact that the multiplicative constant should be 80 to find out which option is wrong.

However, while these options all have the same multiplicative constant 𝐢, they all have different values of π‘Ž and 𝑏. You might think that this was a problem given that four of these options are supposed to describe the shown graph and therefore should all be equivalent. But actually, there are infinitely many ways to write any given exponential equation.

For example, let’s imagine we had a simpler equation 𝑁 equals five times eight to the power of one 𝑑. So 𝐢 is five, π‘Ž is eight, and 𝑏 is one. Of course, normally, we don’t bother to write in the one, so we would just write five times eight to the power of 𝑑. And of course eight is two to the power of three. So we can write this as five times two to the power of three to the power of 𝑑.

And we can use one of our exponent laws to write two to the power of three to the power of 𝑑 as two to the power of three 𝑑. So we have an equivalent exponential function, but with different values of π‘Ž and 𝑏. Now π‘Ž is two and 𝑏 is three, where before π‘Ž was eight and 𝑏 was one. If you’re given an exponential function, you can choose what you want π‘Ž to be and that will fix what 𝑏 is or you can choose what you want 𝑏 to be and that will fix what π‘Ž is.

Often, we choose that the base π‘Ž should be 𝑒, the base of the natural logarithm, and that fixes the value of 𝑏. Alternatively, we could choose that the number in the exponent 𝑏 should be one and that will fix the value of the base π‘Ž. So our plan is to write each option in the form: 𝑁 equals 𝐢 times π‘Ž to the power of 𝑑. So we’ve chosen that the value of 𝑏 is one.

There is a unique way of writing an exponential function in this form. And so once we’ve rewritten all the options into this form, we can see which four of them are equivalent and which one is the odd one out. That’s our plan. Let’s see if we can work out what the values of 𝐢 and π‘Ž should be from the graph.

We have already seen that the 𝑁 intercept is 80; in other words, when 𝑑 is zero, 𝑁 is 80. And of course, for all values of π‘Ž, π‘Ž to the power of zero is one and so 80 is equal to 𝐢. 𝐢 is equal to 80. Okay, so we know that 𝑁 is equal to 80 times π‘Ž to the power of 𝑑. And we just have to find out the value of π‘Ž.

Looking at the graph, we can see that when 𝑑 is equal to one, 𝑁 is equal to 100. Substituting those values in, we get that 100 is equal to 80 times π‘Ž to the power of one. We chose to look at when 𝑑 is equal to one because we know that π‘Ž to the power of one is π‘Ž. And so we get a nice, simple equation: 100 is equal to 80 times π‘Ž. And so π‘Ž is equal to 100 over 80 or five over four or 1.25. As a result, the equation of this graph in the form we’re looking for is 𝑁 equals 80 times 1.25 to the power of 𝑑.

And we can notice straight away that this is option E. And so option E certainly does describe the shown graph. So we know that option E is correct. Let’s see which other options are equivalent to option E.

So we can work our way up and try option D. We have 𝑁 equals 80 times 𝑒 to the power of ln 1.25 times 𝑑. And our plan is to write it in the form 𝑁 equals 𝐢 times π‘Ž to the power of 𝑑 and see if we get the same thing as in option E, which we know is correct.

We can use one of our power laws here to write this as 80 times 𝑒 to the power of ln 1.25 to the power of 𝑑. And this is because π‘Ž to the power of π‘š to the power of 𝑛 is equal to π‘Ž to the power of π‘š times 𝑛. And so we see we have it in this form basically: 𝐢 is 80 and π‘Ž is 𝑒 to the power of ln 1.25, but of course we can simplify this 𝑒 to the power of ln 1.25.

𝑒 to the power of ln 1.25 is just 1.25. This is because 𝑒 to the power of ln π‘₯; that is, 𝑒 to the power of the natural log of π‘₯, is equal to π‘₯. That is true for any value of π‘₯. And so it’s certainly true for π‘₯ equals 1.25. We can see now that this is the same as we have in E, which we know is correct; it does describe the graph shown. D is equivalent to E. And so D also will describe the shown graph.

So D and E describe the shown graph. How about option C? We have that 𝑁 is equal to 80 times two to the power of ln two divided by ln 1.25 times 𝑑. We can use the same trick, which is to write 𝑁 equals 80 times two to the power of ln two over ln 1.25 to the power of 𝑑.

So π‘Ž is two to the power of ln two over ln 1.25. Can we simplify this? We can use the fact that ln π‘₯ over ln 𝑦 is log base 𝑦 of π‘₯; in fact, this holds for any logarithm, not just the natural logarithm ln. And so the exponent of our base so to speak ln two over ln 1.25 is log base 1.25 of two. And so we have that 𝑁 is 80 times two to the power of log base 1.25 of two to the power of 𝑑.

Can we simplify our base further? Well, it’s not particularly obvious how to do this. If you enter this expression into a calculator, you get a number starting 8.611. Importantly, what you don’t get is 1.25. And so for option C, 𝑁 is not equal to 80 times 1.25 to the power of 𝑑, which is the equation which describes our graph.

Therefore, without looking at the other options, it looks like our answer is C. 𝑁 equals 80 times two to the power of ln two over ln 1.25 times 𝑑 is our odd one out. But of course, it makes sense to check the remaining options. So next, let’s check B.

The first thing we do is to use one of our exponent laws. So now we have 80 times something to the power of 𝑑. Let’s see if we can simplify that something. We can simplify this exponent by doing something similar to what we did in option C. ln 1.25 over ln two is a log base two of 1.25. Now, we have something similar to what we had in option C. We have 80 times two to the power of log base two of 1.25 to the power of 𝑑.

But unlike when we had two to the power of log base 1.25 of two, we can simplify what we have here further. For any base 𝑏, 𝑏 to the power of log base 𝑏 of π‘Ž is equal to π‘Ž. And so certainly, two to the power of log base two of 1.25 is 1.25. And so we get that 𝑁 is equal to 80 times 1.25 to the power of 𝑑. This is of course equivalent to options D and E. And so obviously, B is not our answer.

Finally, for option A, we can actually see that this is the second line of working when we were working out option B. And so this is obviously equivalent to option B and therefore equivalent to options D and E as well. So there we have it although they are written in different ways. Options A, B, D, and E are actually equivalent. And only option C does not describe the shown graph.

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