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

Which of the following expressions does NOT describe the shown graph? [A] 𝑁 = 200 ⋅ (1/2)^(2𝑡) [B] 𝑁 = 200 ⋅ (1/4)^𝑡 [C] 𝑁 = 200 ⋅ 𝑒^(−1.39𝑡) [D] 𝑁 = 200 ⋅ (1/2)^(𝑡/2) [E] 𝑁 = 200 ⋅ (4)^(−𝑡).

06:47

### Video Transcript

Which of the following expressions does not describe the shown graph? Is it option A) 𝑁 equals 200 times a half to the power of two 𝑡, option B) 𝑁 equals 200 times a fourth to the power of 𝑡, option C) 𝑁 equals 200 times 𝑒 to the power of negative 1.39𝑡, option D) 𝑁 equals 200 times a half to the power of 𝑡 over two, or option E) 𝑁 equals 200 times four to the power of minus 𝑡?

There are five equations, four of which are the equations of the graph, just written in different forms, and one which is the odd one out. Before we try to find which equation is the odd one out, let’s talk about what all the equations have in common.

They are all of the form 𝑁 equals 200 times something which we’ll call 𝑎 to the power of a something else, we’ll call it 𝑏, times 𝑡. This means that 𝑁 is exponentially growing or decaying with 𝑡, starting with the initial value 200. We can show that this value 200 represents the initial value of 𝑁; that is, the value of 𝑁 when 𝑡 equals zero by substituting zero in.

Using the fact that anything to the power of zero is one, we therefore see that for all our options when 𝑡 is zero, 𝑁 is 200. This matches up nicely with what we see on our graph. The 𝑦-intercept, or should that be 𝑁-intercept, is 200. And of course the 𝑁-intercept is the value of 𝑁 when 𝑡 is zero. As all the options agree on this value of 200, we can’t use it to find the odd one out. We have to use the 𝑎 to the power of 𝑏𝑡 part instead.

The 200 gave us the initial value of 𝑁. This 𝑎 to the power of 𝑏𝑡 part determines how fast 𝑁 is exponentially growing or decaying. Looking at our graph, we can see that when 𝑡 is equal to one, 𝑁 is equal to 50. 50 is a fourth of 200. And so 𝑁 is a fourth of what it was originally.

And because we’re dealing with exponential things here, we know that this trend will continue. 𝑁 will always be a fourth of what it was when 𝑡 was one less. We can therefore write that 𝑁 is equal to 200 times a fourth to the power of 𝑡. We can check by substituting that when 𝑡 is equal to one, 𝑁 is 50.

Here we use the fact that a fourth to the power of one is just a fourth. So certainly the equation in B does describe the shown graph. Of course there are other ways to write a fourth to the power of 𝑡. We can write it as one over four to the power of 𝑡, and hence as four to the power of negative 𝑡 to get another expression of the 𝑁.

Therefore, the equation in option E also describes the shown graph. We’ve managed to write the equation with a base of four instead of a fourth. We can rewrite a fourth to the power of 𝑡 using pretty much any base. For any base 𝑎, a fourth is equal to 𝑎 to the power of log base 𝑎 of a fourth. And so a fourth to the power of 𝑡 is equal to 𝑎 to the power of log base 𝑎 of a fourth to the power of 𝑡.

And using one of our laws of exponents, we can write this as 𝑎 to the power of log base 𝑎 of a fourth times 𝑡. We substitute this into our expression for 𝑁. We see that the equation in option C has a base of 𝑒. We can take the equation that we know is true and rewrite it so that it has a base of 𝑒. And then we can compare that to the equation in option C to see if option C does accurately describe the shown graph.

We replace the arbitrary base 𝑎 by 𝑒. Log base 𝑒 is just the natural logarithm. And evaluating this logarithm using our calculator, we find that 𝑁 is equal to 200 times 𝑒 to the power of negative 1.38629 dot dot dot, this decimal is nonterminating, times 𝑡. If you compare this to the equation in option C, you’ll see that the number in the exponent has been rounded somewhat. But this is probably close enough.

We’re reading off the values from a graph. And so we can’t really expect it to be 100 percent accurate. We are left then with options A and D, both of which have a base of a half. And as they have different exponents, they can’t both be right. One of these options is going to be our odd one out, the one equation which does not describe the graph and so is not equivalent to all the other equations.

As we’re looking for a base of a half, we replace the arbitrary base 𝑎 by a half. If your calculator doesn’t allow you, you just type in log base a half of a quarter, then you’ll have to use one of the laws of logarithms to rewrite this logarithm in terms of logarithms to some other base.

I’ve chosen the natural logarithm. You should find out the value of this logarithm is two. And so 𝑁 is equal to 200 times a half to the power of two 𝑡. This is what we see in option A. Option A therefore does describe the shown graph. Our answer, the option which doesn’t describe the shown graph, is therefore option D: 𝑁 equals 200 times a half to the power of 𝑡 over two.

There is perhaps a quicker way to show that option A is equivalent to option B. We use the law of exponents shown, but in the opposite direction to normal, to write a half to the power of two 𝑡 as a half to the power of two to the power of 𝑡. And as a half to the power of two is a fourth, this gives us the equation we had in option B, which we knew was right.

And just as before, as options A and D have the same base and we know that option A does describe the shown graph, option D must not describe it. Let’s recap what we’ve seen in this question.

We’ve seen that the equation to describe an exponential decay situation is not unique. We can pick whatever base we want. But if you’ve chosen the base, you have no more choices to make. So if two equations have the same base but different something else, then they can’t possibly represent the same thing.