Question Video: Constructing Equivalent Exponential Functions with Different Bases from a Graph Mathematics • Second Year of Secondary School

Video Transcript

Which of the following expressions does not describe the shown graph? Is it (A) 𝑁 is equal to 200 multiplied by four to the power of negative 𝑡? (B) 𝑁 is equal to 200 multiplied by two to the power of negative two 𝑡. (C) 𝑁 is equal to 200 multiplied by a half to the power of two 𝑡. (D) 𝑁 is equal to 200 multiplied by one-quarter to the power of 𝑡. Or (E) 𝑁 is equal to 200 multiplied by a half to the power of 𝑡 over two.

There are lots of ways of solving this problem. One way would be to find coordinates from the graph and substitute these values into each of the equations. We would then be able to find which of the expressions does not describe the graph. Whilst we could choose any points on the graph, it is sensible to choose those with integer values. For example, we can see that the points zero, 200 and one, 50 both lie on the graph.

Let’s begin by considering the point zero, 200. This means that when 𝑡 is equal to zero, 𝑁 equals 200. If we substitute 𝑡 equals zero into each of our expressions, we will be left with 200 multiplied by a constant to the power of zero. We know that 𝑥 to the power of zero is equal to one. This means that for each of our five options, when 𝑡 is equal to zero, 𝑁 will be equal to 200. And as a result, the point zero, 200 satisfies all five equations.

Let’s now consider the point one, 50. In option (A), when 𝑡 is equal to one, 𝑁 is equal to 200 multiplied by four to the power of negative one. One of our laws of exponents or indices states that 𝑥 to the power of negative 𝑛 is equal to one over 𝑥 to the power of 𝑛. This means that in option (A) 𝑁 is equal to 200 multiplied by one over four to the power of one. This is the same as 200 multiplied by one-quarter, which is equal to 50. The point with coordinates one, 50 lies on the equation 𝑁 is equal to 200 multiplied by four to the power of negative 𝑡.

Substituting 𝑡 equals one into option (B), we have 𝑁 is equal to 200 multiplied by two to the power of negative two multiplied by one. This is equal to 200 multiplied by two to the power of negative two, which once again is equal to 200 multiplied by one-quarter. As this is equal to 50, the point with coordinates one, 50 also satisfies option (B). In option (C), when 𝑡 is equal to one, 𝑁 is equal to 200 multiplied by a half to the power of two multiplied by one. This is equal to 200 multiplied by a half squared, which is 200 multiplied by one-quarter. And once again, when 𝑡 is equal to one, 𝑁 is equal to 50. The same is true for option (D). This means that both of the points zero, 200 and one, 50 lie on equations (A) through (D).

In option (E), when 𝑡 is equal to one, 𝑁 is equal to 200 multiplied by one-half to the power of one-half. Recalling that 𝑥 to the power of a half is the same as the square root of 𝑥, we have 𝑁 is equal to 200 multiplied by the square root of one-half. This is equal to 200 multiplied by one over root two, which becomes 100 root two or 141.42 and so on. This means that when 𝑡 is equal to one, 𝑁 is not equal to 50. And the expression for 𝑁 does not describe the shown graph. The correct answer is therefore option (E). 𝑁 is equal to 200 multiplied by one-half to the power of 𝑡 over two does not describe the shown graph.

We could also have approached this question by recognizing that the expressions for 𝑁 in options (A) to (D) are equivalent. Since 𝑥 to the power of negative 𝑛 is equal to one over 𝑥 to the power of 𝑛, we know that four to the power of negative 𝑡 is equal to one-quarter to the power of 𝑡. This means that options (A) and (D) are clearly equivalent. Likewise, two to the power of negative two 𝑡 is equal to one-half to the power of two 𝑡. This means that the expressions for 𝑁 in options (B) and (C) are equivalent.

Next, we note that four is equal to two squared. And since 𝑥 to the power of 𝑎 all raised to the power of 𝑏 is equal to 𝑥 to the power of 𝑎𝑏, then two squared to the power of negative 𝑡 can be rewritten as two to the power of negative two 𝑡. And we can therefore conclude that four to the power of negative 𝑡 is also equal to two to the power of negative two 𝑡. The expressions for 𝑁 in options (A) through (D) are equivalent. And we therefore only need to compare one of these to the expression in option (E). Substituting in the coordinates one, 50, we would then once again have found that the expression that does not describe the shown graph is option (E).