### Video Transcript

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

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, letβs talk about what all the equations have in
common.

They are all of the form π equals
200 times something which weβll call π to the power of a something else, weβll call
it π, times π‘. This means that π is exponentially
growing or decaying with π‘, starting with the initial value 200. We can show that this value 200
represents the initial value of π; that is, the value of π when π‘ 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 π‘ is zero, π
is 200. This matches up nicely with what we
see on our graph. The π¦-intercept, or should that be
π-intercept, is 200. And of course the π-intercept is
the value of π when π‘ is zero. As all the options agree on this
value of 200, we canβt use it to find the odd one out. We have to use the π to the power
of ππ‘ part instead.

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

And because weβre dealing with
exponential things here, we know that this trend will continue. π will always be a fourth of what
it was when π‘ was one less. We can therefore write that π is
equal to 200 times a fourth to the power of π‘. We can check by substituting that
when π‘ is equal to one, π 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 π‘. We can write it as one over four to
the power of π‘, and hence as four to the power of negative π‘ to get another
expression of the π.

Therefore, the equation in option E
also describes the shown graph. Weβve managed to write the equation
with a base of four instead of a fourth. We can rewrite a fourth to the
power of π‘ using pretty much any base. For any base π, a fourth is equal
to π to the power of log base π of a fourth. And so a fourth to the power of π‘
is equal to π to the power of log base π of a fourth to the power of π‘.

And using one of our laws of
exponents, we can write this as π to the power of log base π of a fourth times
π‘. We substitute this into our
expression for π. We see that the equation in option
C has a base of π. We can take the equation that we
know is true and rewrite it so that it has a base of π. 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 π by
π. Log base π is just the natural
logarithm. And evaluating this logarithm using
our calculator, we find that π is equal to 200 times π to the power of negative
1.38629 dot dot dot, this decimal is nonterminating, times π‘. If you compare this to the equation
in option C, youβll see that the number in the exponent has been rounded
somewhat. But this is probably close
enough.

Weβre reading off the values from a
graph. And so we canβt 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 canβt 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 weβre looking for a base of a
half, we replace the arbitrary base π by a half. If your calculator doesnβt allow
you, you just type in log base a half of a quarter, then youβll have to use one of
the laws of logarithms to rewrite this logarithm in terms of logarithms to some
other base.

Iβve chosen the natural
logarithm. You should find out the value of
this logarithm is two. And so π is equal to 200 times a
half to the power of two π‘. This is what we see in option
A. Option A therefore does describe
the shown graph. Our answer, the option which
doesnβt describe the shown graph, is therefore option D: π equals 200 times a half
to the power of π‘ 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 π‘ as a
half to the power of two to the power of π‘. 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. Letβs recap what weβve seen in this
question.

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