Video Transcript
Which of these equations is
satisfied by 𝑥 equals two-thirds cubed squared and 𝑦 equals three times
three-halves to the fifth power minus three-halves to the sixth power? (A) 𝑥 equals 𝑦, (B) 𝑥 equals 𝑦
squared, (C) 𝑥 equals one over 𝑦, or (D) 𝑥 equals one over 𝑦 squared.
Let’s start with 𝑥. 𝑥 equals two-thirds cubed
squared. Using the power to a power rule, we
could rewrite this as two-thirds to the sixth power. And from there using another power
rule, we can simplify further to say two to the sixth power, three to the sixth
power. Now that this value is in its
simplest form, we don’t need to calculate yet. We can just leave it as two to the
sixth power over three to the sixth power.
𝑦 equals three times three-halves
to the fifth power minus three-halves to the sixth power. And at first, it’s not immediately
clear what can be done here. But one thing we can do is break up
this three-halves to the sixth power, using the principle that 𝑥 to the 𝑎 power
times 𝑥 to the 𝑏 power equals 𝑥 to the 𝑎 plus 𝑏 power. We know that six equals five plus
one. And that means three-halves to the
first power times three-halves to the fifth power is equal to three-halves to the
sixth power. And then we have three times
three-halves to the fifth power minus three-halves to the first power times
three-halves to the fifth power.
And looking at these fractions,
again, it might not seem clear what we should do. But imagine, if we replace this
three-halves with the variable 𝑥. It becomes a little bit easier to
see that both of these terms have a factor of three-halves to the fifth power. If you undistributed 𝑥 to the
fifth power, you’d have 𝑥 to the fifth power times three minus 𝑥 to the first
power. We want to follow this principle
using three-halves to the fifth power, which means we’ll have three-halves to the
fifth power times three minus three-halves to the first power.
Now, three-halves to the first
power is just three-halves. We wanna do the subtraction within
the brackets. And that means we’ll need to
rewrite three with the denominator of two. Three is equal to six over two. Six-halves minus three-halves
equals three-halves. And now, we have three over two to
the fifth power times three over two, which is three over two to the first
power. And we’re able to use this same
principle to make this three-halves to the sixth power, which we’ll rewrite as three
to the sixth power over two to the sixth power.
𝑦 equals three to the sixth power
over two to the sixth power and 𝑥 equals two to the sixth power over three to the
sixth power. What we’re seeing is that 𝑥 and 𝑦
are the inverse of one another. When we’re working with fractions,
the inverse switches the numerator and the denominator. And if 𝑥 and 𝑦 are the inverse of
one another, then 𝑥 equals one over 𝑦.
If you wanted to check that this
was true, we have a statement that says two to the sixth power over three to the
sixth power is equal to one over three to the sixth power over two to the sixth
power. That’s saying one divided by three
to the sixth power over two to the sixth power. We could rewrite that as one
divided by three to the sixth power over two to the sixth power. And we know dividing by a fraction
is the same as multiplying its reciprocal. So that would be one times two to
the sixth power over three to the sixth power, which is true. For these equations, 𝑥 is equal to
one over 𝑦.
