### Video Transcript

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

Letβs begin by considering the
point zero, 200. This means that when π‘ is equal to
zero, π equals 200. If we substitute π‘ 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 π₯ to the power of
zero is equal to one. This means that for each of our
five options, when π‘ is equal to zero, π will be equal to 200. And as a result, the point zero,
200 satisfies all five equations.

Letβs now consider the point one,
50. In option (A), when π‘ is equal to
one, π is equal to 200 multiplied by four to the power of negative one. One of our laws of exponents or
indices states that π₯ to the power of negative π is equal to one over π₯ to the
power of π. This means that in option (A) π 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 π is equal to 200 multiplied by four to the power of negative
π‘.

Substituting π‘ equals one into
option (B), we have π 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 π‘ is equal to
one, π 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 π‘ is equal to
one, π 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 π‘ is equal to
one, π is equal to 200 multiplied by one-half to the power of one-half. Recalling that π₯ to the power of a
half is the same as the square root of π₯, we have π 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 π‘ is equal to
one, π is not equal to 50. And the expression for π does not
describe the shown graph. The correct answer is therefore
option (E). π is equal to 200 multiplied by
one-half to the power of π‘ over two does not describe the shown graph.

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

Next, we note that four is equal to
two squared. And since π₯ to the power of π all
raised to the power of π is equal to π₯ to the power of ππ, then two squared to
the power of negative π‘ can be rewritten as two to the power of negative two
π‘. And we can therefore conclude that
four to the power of negative π‘ is also equal to two to the power of negative two
π‘. The expressions for π 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).