### Video Transcript

Express π₯ to the power of 16 over
three multiplied by π₯ to the power of 23 over two in the form the ~~πth~~
[πth] root of π to the power of π.

To solve this problem, we have to
write some rules of exponents thatβll help us out. So if weβre gonna express π₯ to the
power of 16 over three multiplied by π₯ to the power of 23 over two, the first rule
weβre gonna use is this one. Our first rule we are looking at is
that π to the power of π multiplied by π to the power of π is equal to π to the
power of π plus π.

So letβs apply this to our
question; thatβs gonna give us π₯ to the power of 16 over three plus 23 over
two. And to add these powers, weβre
gonna make the denominators equal. So weβre gonna have π₯ to the power
of- we get 23 over six. And we get that by multiplying the
numerator and the denominator both by two. And we do that because six is the
lowest common multiple of two and three, which are both for our denominators.

And then weβre gonna have plus 69
over six, again this time multiplying the numerators on it by three to give us that
denominator of six. And then when we add the powers
together, we get our final answer, which is π₯ to the power of 101 over six. But is this it? Is this the problem solved? Thatβs actually a no because having
a look at the original question, we can see that we need to leave it in this
particular form. As to be able to do this, what
weβre gonna do is actually use another rule of exponents. And this rule of exponents shows us
that π to the power of π over π is equal to the πth root of π to the power of
π.

Okay, great! Letβs apply this to the term we
have. Well first of all, we know that is
going to be sixth root because if we look at our rule of exponents we have that the
bottom of the denominator, so the π in this case, is equal to six in our term. And then our power of π₯ is gonna
be 101 because again looking at this, this is the numerator, so our π value if you
look at the rule of exponents. Itβs there we have it: π₯ to the
power of 16 over three multiplied by π₯ to the power of 23 over two is equal to the
sixth root of π₯ to the power of 101.