Properties of Exponents
Articles,  Blog

Properties of Exponents


– 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

Leave a Reply

Your email address will not be published. Required fields are marked *