### Video Transcript

Which of the following is equal to three-fifths to the negative sixth power times three-fifths to the negative three power over three-fifths to the eighth power? A) Three-fifths to the 11th power, B) Three-fifths to the negative one power, C) Three-fifths to the negative 11 power, D) Three-fifths to the negative 17 power, or E) Three-fifths to the negative 25th power.

First, let’s write down what we’re starting with. And before we do anything else, let’s try and remember some laws of exponents. We know the product rule which says that 𝑥 to the 𝑚 power times 𝑥 to the 𝑛 power is equal to the 𝑥 to the 𝑚 plus 𝑛 power. When we’re multiplying exponents with the same base, we add the two values together. The quotient rule when we’re dividing 𝑥 to the 𝑚 power by 𝑥 to the 𝑛 power, it will be equal to 𝑥 to the 𝑚 minus 𝑛 power. When we’re dividing two exponents with the same base, we subtract their values. And finally, negative exponents: If we have 𝑥 to the negative 𝑚 power, that’s the same thing as one over 𝑥 to the 𝑚 power.

So back to our problem. It’s really tempting to try and distribute this to the negative six power and to the negative three power, so that we get something like this. But in the end, that doesn’t help us simplify very much. We need a different strategy. Imagine that we had 𝑥 to the negative six power times 𝑥 to the negative three power divided by 𝑥 to the eighth power. First, we would use our product rule and add the two exponents in the numerator. The base stays the same 𝑥. And negative six plus negative three equals negative nine. Now, we have a statement that says 𝑥 to the negative nine over 𝑥 to the eighth power.

And so, we move on to the quotient rule. The base will stay the same. And we’ll subtract the exponent in the denominator from the exponent in the numerator. 𝑥 to the negative nine minus eight equals 𝑥 to the negative 17. We can say let 𝑥 equal three-fifths. And if 𝑥 equals three-fifths, we can plug that in to have three-fifths to the negative 17 power, which is option D. Using this method prevents us from making calculation mistakes. Let’s look at a longer method that doesn’t use variable substitution. We start at the same place. And we recognize that the two numerators have the same base. So we can add their exponents, negative six plus negative three. And then, we copy down our denominator. You could then distribute the exponent across the fractions.

Our numerator, three to the negative ninth power, over five to the negative ninth power is being divided by three to the eighth power over five to the eighth power. Division is the same thing as multiplying by the reciprocal. We have to multiply the numerator and multiply the denominator. But we switched the order because we can multiply in any order that we want. Which means we could say three to the negative ninth over three to the eighth times five to the eighth over five to the negative ninth. And then, we’ll have to use our quotient rule and use subtraction of exponents. Three to the negative ninth minus eight and five to the eighth minus negative nine gives us three to the negative 17 power times five to the 17th power.

If we move that three to the negative 17th into the denominator, it becomes three to the positive 17th, which is five-thirds to the 17th power. But that’s not an answer choice. If we change three to the 17th power into the numerator, it needs to be three to the negative 17th power. And five to the negative 17th power can be moved into the denominator. Which shows that three-fifths to the negative 17th power is equal to what we started with.

Taking so many steps to simplify could result very easily in us missing a negative sign or adjusting something incorrectly. At this stage, three over five to the negative ninth power over three-fifths to the eighth power. We could’ve simply said three over five to the negative nine minus eight power is equal to three over five to the negative 17th power. What matters here is recognizing that when we’re dealing with the same base, we can treat it as a single unit instead of trying to break it up into smaller pieces. And doing that saves us a lot of trouble. But both ways give us the final answer of D, three-fifths to the negative 17th power.