In Exercises 107–114, simplify each exponential expression. Assum...

Question

In Exercises 107–114, simplify each exponential expression. Assume that variables represent nonzero real numbers. (2^−1x^−3y^−1)^−2(2x^−6y^4)^−2(9x^3y^−3)^0/(2x^−4y^−6)^2

Accepted Answer

Hello everyone in this video, we're gonna be looking at this exponential expression and you can see it's quite long and has many terms and we need to try to simplify it. So whenever we're approaching a complicated expression, it's always important to remember Pem does which sort of gives you a guideline into the steps that you should follow When approaching a complex problem such as this one. So remember that Kendall stands for parentheses, exponents, multiplication, division, addition and subtraction. So that's the order in which we're going to try to tackle these operations. First being parentheses Where we have four set of parentheses and you can see that the four set of parentheses have an exponent on the outside. So whenever you have an exponent on the outside you need to make sure to distribute that exponent to the inside exponents. So in this case we have negative three. You need to distribute that to the negative one, negative seven. And the negative four on the inside. And the same with the second set of parentheses, you need to make sure to distribute it. Yeah, we're also gonna do that with the third such where we have zero and whenever we're distributing a power of zero to any number, remember that any term to the power of zero is equal to one. So we're gonna keep that in mind when we're dealing with the third set of parentheses and then the last Is this three which needs to go to the eight and the two. So now I've sort of drawn the lines that tell me where I need to distribute the exponents. So I'm gonna go ahead and distribute them. So I have three to the negative one times negative three. Eight to the negative seven times negative three B to the negative four times negative three. And my second set of parentheses 3 to the one times negative three A. To the negative one times negative three. B to the fifth times negative three. Then I have my third set of parentheses or have to the first power times 08 to the 13th. Power time zero B to the negative 21st Power times zero. And all of those are being divided by A. To the eighth times three and B to the second times three. So now I finished distributing all the exponents on the outside of the parentheses. So I'm gonna go ahead and simplify. So I have three to the negative one times negative three. So negative one times negative three become positive three. I have A to the positive 21 B to the positive onto my second set of parentheses. I have three to the negative three. Aid to the positive three. B to the -15. And then I have my third set where I have 108 to the zero A. To the zero and B to the zero. However recall that any term to the zero is just one. So this parentheses is really one times one times one which is really just one. So in order to simplify this, I'm just gonna rewrite this set are this set of parentheses or the Parentheses? # three As 1. And that way my expression starts to look more simple And that's still being divided by a. To the 24 and B to the six. So now I have distributed the export outside of the parentheses. So I can sort of start dealing with the exponents on the inside of the parentheses and you can see that within each set of parentheses there's a sort of pattern where there's a number such as three to the third or three to the negative three. And then we have an A term A. To the 21 8 of the third. Eight to the 24 a B term B to the 12. B 29 to 15 and B to the sixth. So I'm gonna use that to sort of combine my different parentheses into those different parts. So the first one I have three to the third and that's being multiplied by three to the negative third. So when whenever we have multiplication with exponents that have the same base we can add their exponential values. So this becomes three to the third plus negative three. And there's no other number with a base of three. So I can Close that set of parentheses. So my next components are the a terms where I have AIDS of the 21 and that's being multiplied by eight to the third. So I'm gonna add The power of three. Then we also have this third a term eight to the 24. But that's on the denominator. So whenever we have exponents that are being divided we want to make sure to subtract the power that's on the denominator which in this case is 24. So I'm gonna say -24. And that deals with all of the A. Terms. So now I'm going to deal with the B terms the same way as I did with the A. Terms or have B To the 12th and that's being multiplied by B to the -15. And both of those are being divided by B to the 6th. So I need to subtract the 6th power. So now I've rewritten all the components. Now I can sort of simplify within these parentheses or I have three to the 3rd -3. So that becomes three to the zero. Then I have A to the 21 plus three. So I get to the 24 minus 24. So that becomes A to the zero and then I have B to the 12 -3. So negative 3 -6. So b to the -9. So now you can see that I have two terms that sort of turn into one because any term to the zero power is one. So I have one Times one and I'm just left with B to the -9. So I can rewrite B to the - as one over B to the ninth because remember that any exponents or any negative exponents can be represented In the denominator of a fraction as positive. So when I when I rewrite beats -9 as a fraction and I put B to the nine in the denominator it is positive, it loses the negative. So this is my final answer Yeah. For the simplification of the expression and you can see that this answer aligns with B. So hopefully this really helped and see you next time.

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