### Video Transcript

If π₯ is equal to one and π¦ is
equal to root two, find π₯ to the power of negative two π¦ to the fourth power all
to the power of negative three in its simplest form.

Letβs begin by substituting the
values of π₯ and π¦ into the exponential expression. Doing so gives one to the power of
negative two multiplied by root two to the fourth power all raised to the power of
negative three.

We now need to simplify this
expression. First, we can recall that the value
one raised to any power is just one. So in particular, one to the power
of negative two is one. And the first term in the product
simplifies to one. Next, we can recall that raising a
base to a positive integer power means we multiply that number of bases
together. So root two to the fourth power
means four lots of root two multiplied together.

We can then recall that for
nonnegative values of π, the square root of π squared is equal to π. So, grouping the four root twos
into two pairs gives root two squared multiplied by root two squared. Applying the law weβve just
discussed, each of these factors simplifies to two. So the overall product simplifies
to two multiplied by two, which is four.

We can now substitute the values of
one to the power of negative two and root two to the fourth power back into the
expression. And it becomes one multiplied by
four all to the power of negative three. Of course, one multiplied by four
is just four. So the expression simplifies to
four to the power of negative three. Next, we recall that negative
exponents define reciprocals. For a nonzero base π, π to the
power of negative π is equal to one over π to the πth power. Four to the power of negative three
can therefore be written as one over four to the third power.

Finally, we evaluate four to the
third power, or four cubed, by recalling that this is equal to four multiplied by
four multiplied by four. We should recall from memory that
four cubed is equal to 64. Or we can work it out by first
multiplying four by four to give 16 and then multiplying this by four again to give
64. Substituting this value back into
the denominator gives our answer.

By applying laws of exponents,
weβve found that the value of the given expression, in its simplest form, is one
over 64.