– WELCOME TO THE PRESENTATION ON THE PROPERTIES

OF EXPONENTS. THE GOAL OF THIS VIDEO IS TO USE

THE PROPERTIES OF EXPONENTS TO SIMPLIFY EXPRESSIONS. HERE IS A LIST

OF THE PROPERTIES OF EXPONENTS WE WILL BE DISCUSSING TODAY. YOU MAY WANT TO PAUSE THE VIDEO

NOW AND WRITE THESE DOWN. WE WILL CONSIDER THEM

ONE AT A TIME. THE FIRST PROPERTY

IS A PRODUCT PROPERTY. IT STATES IF YOU’RE MULTIPLYING

AND THE BASES ARE THE SAME, YOU ADD THE EXPONENTS. LET’S TAKE A LOOK

AT WHY THAT MAKES SENSE. IF WE WANT TO MULTIPLY 5

TO THE SECOND x 5 TO THE FOURTH, WE KNOW THAT

5 TO THE SECOND MEANS 5 x 5, AND WE KNOW THAT 5 TO THE FOURTH

WOULD BE 4 FACTORS OF 5. AND SO NOW WE CAN JUST COUNT

THE TOTAL FACTORS OF 5 AND WE CAN SEE WE’RE GOING

TO HAVE 6 FACTORS OF 5. WELL, IF WE TAKE ADVANTAGE OF

THE PRODUCT RULE THAT STATES THAT WE CAN JUST TAKE

THE EXPONENTS 2 AND 4 AND ADD THEM TO GET

5 TO THE SIXTH. THIS IS OBVIOUSLY

MUCH SHORTER, BUT IF YOU’RE NOT SURE

OF THE RULE, IT’S ALWAYS BEST TO EXPAND THE

EXPONENTS TO FIND THE PRODUCT. HERE WE HAVE X TO SIXTH x X. WELL, THE FIRST THING

WE HAVE TO RECOGNIZE IS THAT WHEN WE HAVE X

AND THERE’S NO EXPONENT LISTED THAT MEANS THE EXPONENT

WILL BE ONE. AND THAT’S ACTUALLY

ONE OF THE PROPERTIES LISTED ON THE PREVIOUS SCREEN. SO IF WE HAVE X TO THE SIXTH

x X TO THE FIRST, IF WE USED THE PRODUCT RULE

WE COULD JUST TAKE THE EXPONENTS AND ADD THEM TO GET

X TO THE SEVENTH. HOPEFULLY, THAT’S LOGICAL, BECAUSE WE KNOW

IF WE EXPAND X TO THE SIXTH WE’D HAVE 6 FACTORS OF X

x AN ADDITIONAL FACTOR OF X, WHICH OF COURSE WOULD GIVE US

A TOTAL OF 7 FACTORS OF X. THE QUOTIENT RULE STATES

THAT IF WE’RE DIVIDING AND THE BASES ARE THE SAME,

WE SUBTRACT THE EXPONENTS. SO FOR THIS FIRST PROBLEM, IF WE HAVE 3 TO THE SIXTH

DIVIDED BY 3 TO THE FOURTH, LET’S ASSUME THAT WE COULD NOT

REMEMBER THE QUOTIENT RULE. AGAIN, WE WOULD EXPAND

EACH OF THESE, SO WE’D HAVE

6 FACTORS OF 3 IN THE NUMERATOR, AND 4 FACTORS OF 3

IN THE DENOMINATOR. AND NOW WE CAN SIMPLIFY EVERY

3/3 WOULD SIMPLIFY TO 1, AND NOW WE CAN SEE,

WE HAVE 3 TO THE POWER OF 2 OR 3 SQUARED. WELL, USING THE QUOTIENT RULE,

WHAT IT’S STATING, IS WE COULD HAVE JUST RECOGNIZED

THAT THE BASES WERE THE SAME AND SIMPLIFIED IT TO

3 TO THE POWER OF 6 – 4, WHICH OF COURSE GIVES US

3 TO THE SECOND. BUT IF WE DON’T UNDERSTAND

WHY THIS RULE WORKS, IT’S BEST TO EXPAND IT

AND SIMPLIFY IT THE LONG WAY. SO FOR THIS NEXT PROBLEM, YOU CAN SEE

AS THE EXPONENTS GET LARGER, IT KIND OF GETS MORE TEDIOUS

TO WRITE OUT 15 FACTORS OF Y AND 8 FACTORS OF Y. SO WE WILL JUST

APPLY THE QUOTIENT RULE BY TAKING

THE EXPONENT AND THE NUMERATOR AND SUBTRACTING THE EXPONENT

FROM THE DENOMINATOR TO OBTAIN OUR SIMPLIFIED

EXPRESSION OF Y TO THE SEVENTH. AND THIS LAST ONE, WE KNOW THAT WHENEVER

WE HAVE SOMETHING OVER ITSELF IT IS GOING TO EQUAL ONE. IF WE APPLIED THE QUOTIENT RULE

TO THIS PROBLEM, WE WOULD HAVE

W TO THE 4 – 4 POWER OR W TO THE 0 POWER,

WHICH IS=TO 1. AND THIS IS ONE

OF THE PROPERTIES THAT WE SAW ON THE FIRST SCREEN. ANYTHING RAISED

TO THE 0 POWER IS=TO 1. THE POWER RULE STATES

THAT WE HAVE A RAISED TO THE POWER OF M

ALL RAISED TO THE POWER OF N. THIS WILL=

A TO THE M x N POWER. AGAIN, BEFORE WE APPLY THIS, LET’S TAKE A LOOK AT IT

BY EXPANDING THIS. IF WE HAVE 3 CUBED

RAISED TO THE SECOND POWER, AS LONG AS WE UNDERSTAND

WHAT EXPONENTS MEAN WE CAN EXPAND ALL OF THIS AND WE

SHOULD GET THE SAME ANSWER THAT WE WOULD GET

FROM THE POWER RULE. SO IF WE KNOW

IF SOMETHING IS SQUARED THAT’S MULTIPLIED

BY ITSELF TWO TIMES, AND NOW WE ALSO KNOW THAT 2

CUBED MEANS 2 x ITSELF 3 TIMES. SO THERE’S

2 CUBED x ANOTHER 2 CUBED. NOW, WE CAN JUST

COUNT THE FACTORS AND WE HAVE

2 TO THE POWER OF 6. NOW, IF WE TRY

TO APPLY THE POWER RULE AND WE HAVE A POWER RATE,

SO POWER RATE MULTIPLY, OF COURSE, 2 TO THE POWER

OF 3 x 2, OF COURSE, GIVES US THE SAME ANSWER

OF 2 TO THE SIXTH. SO IF YOU EVER PANIC AND YOU

FORGET SOME OF THESE RULES, JUST EXPAND YOUR EXPONENTS AND YOU CAN STILL

DERIVE THEM THE LONG WAY. LOOKING AT OUR SECOND EXAMPLE, WE HAVE M TO THE FIFTH POWER

RAISED TO THE FOURTH POWER. WELL, AGAIN, THE SHORTCUT WOULD

BE TO RAISE M TO THE 5 x 4 POWER OR MULTIPLY OUR EXPONENTS, WHICH WOULD GIVE US

M TO THE TWENTIETH POWER. NEXT WE HAVE

THE PRODUCT OF POWER RULE AND ALSO

THE QUOTIENT TO POWER RULE. SO IF WE HAVE A,

B RAISED TO THE POWER OF N THAT WILL=A

TO THE N x B TO THE N. NOW, TO ME, I LOOK AT THIS

AS JUST AN EXPANSION OF THE PREVIOUS RULE WHERE YOU

HAVE A POWER TO A POWER, BECAUSE THIS IS A TO THE FIRST

AND THIS IS B TO THE FIRST. SO IF

WE MULTIPLY THESE EXPONENTS, THE RESULT WOULD BE THE SAME AND THE SAME THING

FOR THE QUOTIENT TO POWER RULE. THIS IS REALLY A TO THE FIRST

AND THIS IS B TO THE FIRST. SO I LOOK AT THIS AND SEE

WE HAVE POWERS RAISED TO POWERS, THEREFORE, WE CAN MULTIPLY

OUR EXPONENTS. OKAY. WE HAVE

SEVERAL EXAMPLES HERE. THE FIRST ONE

WE HAVE 7 X SQUARED, WELL, THIS IS ACTUALLY 7

TO THE FIRST AND X TO THE FIRST. SO APPLYING THIS

PRODUCT OF POWER RULE, YOU WOULD HAVE 7 SQUARED x

X SQUARED, WHICH=49 X SQUARED. A COMMON MISTAKE HERE IS DON’T FORGET TO SQUARE THE 7

AS WELL AS THE X. THE NEXT PROBLEM, I DON’T SEE

AN EXPONENT ON THE Z, SO REMEMBER,

THAT MEANS IT’S A 1. SO HERE WE’RE GOING TO HAVE

X TO THE 2 x 3 POWER THAT WOULD BE X TO THE SIXTH, Y TO THE 5 x 3 POWER

OR ITS FIFTEENTH POWER, AND THEN Z RAISED

TO THE 1 x 3 OR THIRD POWER. THE NEXT EXAMPLE, AGAIN,

WHENEVER I DON’T SEE AN EXPONENT I ALWAYS WRITE IT IN AS A ONE. SO I LOOK AT THIS AND I SEE

POWERS RAISED TO POWERS. SO I WOULD HAVE 2

TO THE FOURTH POWER DIVIDED BY 3

TO THE FOURTH POWER. AND SINCE WE CAN

MULTIPLY THIS OUT, WE SHOULD. TO THE FOURTH

WOULD GIVE US 16, AND 3 TO THE FOURTH WOULD BE 81. AND THE LAST EXAMPLE HERE, THIS WOULD BE

A TO THE 3 x 4 OR TWELFTH POWER, AND B TO THE 2 x 4

OR B TO THE EIGHTH POWER. NOW THAT WE’VE DISCUSSED THE

BASIC PROPERTIES OF EXPONENTS, LET’S TAKE A LOOK

AT A COUPLE OF PROBLEMS WHERE WE WILL SIMPLIFY

USING SEVERAL OF THE PROPERTIES. SO WE TAKE

A LOOK AT THIS PROBLEM, WE CANNOT

MULTIPLY THESE TWO TOGETHER BECAUSE 2 A SQUARED B

IS RAISED TO THE THIRD POWER. SO WE FIRST HAVE

TO ADDRESS THIS OUTER EXPONENT. THIS WOULD BE

2 TO THE THIRD POWER, A TO THE 2 x 3

OR A TO THE SIXTH POWER. AND THIS WOULD BE TO THE FIRST SO WE’D HAVE

B TO THE THIRD POWER. SINCE ALL OF THIS

IS BEING MULTIPLIED TOGETHER I’M GOING

TO REARRANGE THE ORDER. LET’S WRITE 3 x 2 CUBED X A

FACTOR OF A x A TO THE SIXTH x B

TO THE FOURTH x B CUBED. AGAIN, I’M ALLOWED TO DO THIS BECAUSE THIS IS ALL CONNECTED

BY MULTIPLICATION. SO NOW WE CAN

MULTIPLY THIS TOGETHER. 2 TO THE THIRD WOULD BE

8, 8 X 3 WOULD BE 24. A TO THE FIRST x A TO THE SIXTH WOULD GIVE US A TO THE SEVENTH SINCE WE ADD OUR EXPONENTS, AND B TO THE FOURTH x B

TO THE THIRD B TO THE SEVENTH. OKAY. THE LAST EXAMPLE, WE HAVE A FRACTION

INSIDE THESE PARENTHESIS BEING RAISED

TO THE THIRD POWER. THERE’S A COUPLE OF WAYS

TO DO THIS PROBLEM. WE COULD APPLY

THE POWER RULE HERE, BUT I DID NOTICE THAT

INSIDE THE PARENTHESIS THIS FRACTION CAN BE SIMPLIFIED. HERE WE HAVE X SQUARED, AND HERE WE HAVE

X TO THE FIRST. SO IF WE APPLIED

THE QUOTIENT RULE WE’D HAVE

X TO THE 2 – 1 POWER, WHICH WOULD GIVE US X

IN THE NUMERATOR, AND THIS IS STILL

RAISED TO THE THIRD POWER. SO IF I LOST YOU THERE,

WHAT HAPPENED IS WE HAD AN X TO THE SECOND

IN THE NUMERATOR AND WE HAD ONE FACTOR OF X

IN THE DENOMINATOR. SO WHEN WE SIMPLIFIED THIS WE WERE LEFT

WITH X IN THE NUMERATOR, WHICH IS THE SAME THING

YOU WOULD GET WHEN YOU APPLY THE QUOTIENT RULE WHERE YOU HAVE

X TO THE POWER OF 2 – 1 WHICH IS X TO THE FIRST. OKAY. NOW WE CAN

APPLY THE POWER RULE HERE, SO WE’D HAVE

X TO THE 1 x 3 POWER, X TO THE THIRD,

Y TO THE 4 X 3 POWER, WHICH WOULD BE

THE TWELFTH POWER. 2 TO THE 1 x 3

OR 2 TO THE THIRD POWER AND Z TO THE 1 x 3 WOULD GIVE US

Z TO THE THIRD. 2 TO THE THIRD

WOULD SIMPLIFY TO 8. SO LET’S REWRITE

THIS ONE MORE TIME, X CUBED, Y TO THE TWELFTH

DIVIDED BY 8 Z CUBED. OKAY. THE NEXT VIDEO WE’LL

DISCUSS NEGATIVE EXPONENTS. THERE WILL BE ADDITIONAL

EXAMPLES OF THESE PROPERTIES IN THAT VIDEO AS WELL. THANK YOU FOR WATCHING AND I HOPE YOU FOUND IT HELPFUL.

## 8 Comments

## Janet0696

Great video! hepled alot(:

what if you have …

r^2

—–

2r^3

???

it gets me confuse because the base arnt the same?

## Linsy Horn

The opposite of division is subtraction and the opposite of multiplication is addiction

## amerikani24

No offense, but this is why I hate math. I'm a very logical thinker so people tell me that math should be no problem because rules are rules. But there's always a "This is the rule to do this type of problem…….EXCEPT when…." I hate gray areas, to me logical means black & white while understanding there are things that are gray. But in math logical means black & white= gray.

## 800 Subscribers With no Videos

2017! are u still here?

## Kara Ann

You explained this better than my math teacher will ever know how to.

## ACeD

This is Gay

## Terra Dressler

Not helping with what i'm learning. Sorry 🙁

## Rachel June

Ah, and this is where I get stuck :/