### Video Transcript

Given that three π₯ minus π¦ equals 14, find the value of 27 to the π₯ power over three to the π¦ power.

At first, it might seem like we havenβt been given enough information to solve this problem. But we can solve it if we look closely and remember some exponent rules. We have three to the π¦ power in the denominator and 27 to the π₯ power in the numerator. The question is, can we rewrite 27 as a power with base three? Three cubed equals 27. And that means we can rewrite 27 to the π₯ power as three cubed to the π₯ power. 27 to the π₯ power is the same thing as three cubed to the π₯ power.

We now have three cubed to the π₯ power over three to the π¦ power. In our numerator, weβre taking a power to a power which we can simplify by multiplying these two exponents together. Three cubed to the π₯ power becomes three to the three π₯ power. Our denominator doesnβt change. Itβs three to the π¦ power. At this point, weβre starting to see some of the elements we were given at the start of the problem.

Because our numerator and our denominator have the same base of three, we can use the quotient rule, π to the π₯ power over π to the π¦ power is equal to π to the π₯ minus π¦ power. We start with the base of three. We then take the exponent from the numerator and subtract the exponent from the denominator. So, we have three to the three π₯ minus π¦ power.

We know what three π₯ minus π¦ is equal to. Three π₯ minus π¦ equals 14. And so, three to the three π₯ minus π¦ power equals three to the 14th power. So, 27 to the π₯ power over three to the π¦ power simplifies to three to the 14th power if we know that three π₯ minus π¦ equals 14. Now, if you didnβt remember the quotient rule, there is one other way we could solve this problem.

If we go back to when we had three to the three π₯ power over three to the π¦ power, we could use our fractional exponent rule, one over π to the π₯ power equals π to the negative π₯ power. The numerator stays the same, three to the three π₯ power, and it would be multiplied by three to the negative π¦ power.

And using one final rule, we could say that π to the π₯ power times π to the π¦ power is equal to π to the π₯ plus π¦ power. And so, weβll have three to the three π₯ minus π¦ power. Adding three π₯ and negative π¦ is three π₯ minus π¦. We then substitute 14 in for three π₯ minus π¦. Both methods give us three to the 14th power.