Properties of Exponents

Return to The Properties

Return to The PropertiesSorry! Try Again!Time to head back and review the propertyReturn to the property

Quotient of PowersWhen dividing like bases, we have to SUBTRACT their exponents = xm-nExample: = x6

Now you choose the correct answer= ?x8x2x24

The PropertiesProduct of PowersPower of a PowerPower of a ProductQuotient of PowersZero ExponentNegative ExponentsPower of a QuotientThe Quiz

Notice if we were to break up the previous problem as the followingx5 x6 = ?x x x x x x x x x x x

Since x5 means x times itself five times and x6 means x times itself six times.

How many of the x times itself did we end up with? Power of a QuotientWhen a quotient is raised to a power, EVERYTHING in the quotient gets that power =

Example: =

Now you choose the correct answer = ?

Notice if we were to break up the previous problem as the followingx5 x6 = ?x x x x x x x x x x x

Since x5 means x times itself five times and x6 means x times itself six times.

How many of the x times itself did we end up with? Sorry! Try Again!Time to head back and review the propertyReturn to the property

Congratulations! You really know your exponent properties!

Show Mrs. Woudstra this screen so can award you full credit for completing this activity.Sorry! Try Again!Remember, if you have a quotient to a power, ALL terms receive that power

Now if we were to break up the previous problem as the following(x3)4 = ?(x x x)4

And continued to break these up using the ideas from the first property, we could get

(x x x) (x x x) (x x x) (x x x)

How many of the x times itself did we end up with? Now if we were to break up the previous problem as the following(xy)2 = ?(xy) (xy)

And thinking about what happens when we multiply like bases, what would the powers of each variable be?

Return to The PropertiesNow if we were to break up the previous problem as the following(xy)2 = ?(xy) (xy)

And thinking about what happens when we multiply like bases, what would the powers of each variable be?Now if we were to break up the previous problem as the following

Looking at the xs in the numerator and the denominator. If every x in the numerator was cancelled by one in the denominator, how many of the x times themselves would be left and where would they be?

Now if we were to break up the previous problem as the following

Looking at the xs in the numerator and the denominator. If every x in the numerator was cancelled by one in the denominator, how many of the x times themselves would be left and where would they be?

Now if we were to break up the previous problem as the following

Looking at the xs being multiplied in the numerator and the ys being multiplied in the denominator, how many of the x times themselves are in the numerator and how many of the y times themselves are in the denominator?

Now if we were to break up the previous problem as the following

Looking at the xs being multiplied in the numerator and the ys being multiplied in the denominator, how many of the x times themselves are in the numerator and how many of the y times themselves are in the denominator?

But Why?For a brief look at why anything to the power of zero is one, take a look at a few explanations here.x01Question #2:

(d6)3 = ?d63d9d18Question #3:

(ab)5 = ?a5b5ab5a5bQuestion #4:

= ?x2x4x32

Question #5:

= ?

Question #6:

(97rst)0 = ?97 1 0

Question #7:

= ?a6-a6a-6

Question #8:

(x2y3)4 = ?x6y7xy9x8y12Question #9:

(x4y5)2 (x3y2)3 = ?x17y16x72y60x36y42Question #10:

= ?