- WELCOME TO OUR LESSON ON 
THE PROPERTIES OF LOGARITHMS.
LET'S GO AHEAD 
AND GET STARTED.
THE FIRST PROPERTY IS LOG 
BASE A OF 1 IS EQUAL TO ZERO
AND THIS IS TRUE
BECAUSE A TO THE POWER OF ZERO 
WILL ALWAYS EQUAL 1.
AND IF WE WANT AN EXAMPLE 
OF THIS
WE COULD SAY LOG 1 
OF ANY BASE, LET'S JUST SAY 8,
WILL ALWAYS EQUAL ZERO
BECAUSE 8 TO THE ZERO 
IS EQUAL TO 1.
PROPERTY 2 SAYS LOG BASE A 
OF A IS EQUAL TO 1
SINCE A TO THE POWER OF 1 
WILL EQUAL A
SO IF THE BASE AND THE NUMBER 
ARE THE SAME
IT WILL ALWAYS EQUAL 1.
SO AN EXAMPLE OF THAT MIGHT BE 
LOG BASE 5 OF 5 WILL EQUAL 1.
AND THEN NO. 3, LOG BASE A OF 
A TO THE POWER OF A EQUALS X.
WELL, THAT'S TRUE
BECAUSE IF I TAKE A AND RAISE 
IT TO THE POWER OF X,
IT WILL EQUAL A TO THE X.
BUT THE PATTERN WE SEE HERE 
IS IF THE BASE AND THIS BASE
ARE THE SAME
IT WILL ALWAYS JUST EQUAL 
THE EXPONENT HERE.
SO AN EXAMPLE OF THIS MIGHT 
BE LOG BASE 2 OF 2 TO THE 7th
WOULD JUST EQUAL 7
BECAUSE 2 TO THE 7th 
IS EQUAL TO THE 7th.
OKAY, THE THREE MAIN 
PROPERTIES
WE'RE GOING TO LOOK AT 
IN THIS VIDEO
ARE THE PRODUCT, QUOTIENT
AND POWER PROPERTIES 
OF LOGARITHMS.
LET'S GO AHEAD AND TAKE A LOOK 
AT THOSE ONE AT A TIME.
THE PRODUCT PROPERTY 
OF LOGARITHMS
STATES THAT THE LOG 
OF A PRODUCT
EQUALS THE SUM OF THE LOGS.
SO IF WE HAVE 
LOG BASE A OF U TIMES V
IS EQUAL TO LOG BASE A OF U 
PLUS LOG BASE A OF V.
AND WE CAN ALSO STATE THE 
SAME RULE USING NATURAL LOGS.
NOW SINCE LOGARITHMS ARE 
EXPONENTS THERE'S A CONNECTION
BETWEEN THE PRODUCT PROPERTY 
FOR EXPONENTS
AND THE PRODUCT PROPERTY 
FOR LOGARITHMS.
NOTICE HERE 
WHEN YOU'RE MULTIPLYING
AND THE BASES ARE THE SAME 
YOU ADD YOUR EXPONENTS.
WELL, HERE WHEN YOU'RE 
MULTIPLYING YOUR NUMBERS
YOU ADD YOUR LOGARITHMS
BUT REMEMBER THE LOGARITHMS 
ARE EXPONENTS.
SO IF YOU'RE MULTIPLYING,
YOU ADD FOR BOTH THE PRODUCT 
PROPERTY OF LOGARITHMS
AND THE PRODUCT PROPERTY 
FOR EXPONENTS.
SO FOR EXAMPLE IF WE HAD 
LOG BASE 5 OF 6,
WE KNOW WE CAN WRITE 6 
AS 2 TIMES 3
WHICH MEANS IT CAN BE 
LIKE THIS AS LOG BASE 5 OF 2
PLUS LOG BASE 5 OF 3
USING THE PRODUCT PROPERTY 
OF LOGARITHMS.
NOW THE QUOTIENT PROPERTY 
OF LOGARITHMS
STATES THE LOG OF A QUOTIENT
EQUALS THE DIFFERENCE 
OF THE LOGS.
SO IF WE HAVE LOG BASE A OF U 
DIVIDED BY V
WE CAN REWRITE THIS 
AS LOG BASE A OF U
MINUS LOG BASE A OF V
AND AGAIN WE CAN WRITE THIS 
USING NATURAL LOGS AS WELL.
AGAIN, MAKING THE CONNECTION
TO THE QUOTIENT PROPERTY 
OF EXPONENTS
IF YOU'RE DIVIDING 
AND THE BASES ARE THE SAME
YOU SUBTRACT YOUR EXPONENTS.
SO WHEN YOU'RE DIVIDING 
THE NUMBERS IN THE LOG
YOU WOULD SUBTRACT 
YOUR LOGARITHMS.
REMEMBER LOGARITHMS 
ARE EXPONENTS.
SO FOR EXAMPLE, IF WE HAD LOG 
BASE 7 OF LET'S SAY 1.5,
WELL, 1.5 IS THE SAME 
AS THREE HALVES.
WE COULD REWRITE THIS 
AS LOG BASE 7 OF 3
MINUS LOG BASE 7 OF 2
USING THE QUOTIENT PROPERTY 
OF LOGARITHMS.
IF YOU HAVE A QUOTIENT
YOU CAN REWRITE THE LOGS 
AS A DIFFERENCE.
AND THE LAST PROPERTY IS THE 
POWER PROPERTY OF LOGARITHMS
AND IT STATES THAT THE LOG 
OF A POWER
EQUALS THE PRODUCT 
OF THE POWER AND THE LOG.
SO LOG BASE A OF U 
TO THE N POWER
IS EQUAL TO N 
TIMES LOG BASE A OF U
AND AGAIN WE CAN WRITE THIS 
USING NATURAL LOGS AS WELL.
THE CONNECTION YOU CAN MAKE
TO THE POWER PROPERTY 
OF EXPONENTS
IS EITHER POWER TO A POWER 
YOU MULTIPLY YOUR EXPONENTS.
WELL, HERE IF WE HAVE A LOG 
OF A POWER,
THEN WE MULTIPLY N TIMES 
THE LOG,
WELL, N IS AN EXPONENT AND 
THE LOG IS ALSO AN EXPONENT,
SO WE'RE MULTIPLYING 
OUR EXPONENTS.
AN EXAMPLE OF THIS ONE 
MIGHT BE
IF WE HAVE LOG BASE 12 OF 5 
TO THE 11th POWER.
THAT'S JUST EQUAL TO 11 
TIMES LOG BASE 12 OF 5.
THERE ARE USUALLY TWO TYPES 
OF PROBLEMS
THAT YOU'RE ASKED TO DO
TO ILLUSTRATE THE PROPERTIES 
OF LOGS.
THE FIRST TYPE IS TO EXPAND 
A LOG AS MUCH AS POSSIBLE
USING THE PROPERTIES OF LOGS.
SO ON THIS PROBLEM THE FIRST 
THING WE SHOULD NOTICE
IS THERE'S NO RULE HERE
THAT DEALS WITH SQUARE ROOTS 
OR RADICALS.
SO LET'S REWRITE THIS 
USING RATIONAL EXPONENTS.
SO THIS IS EQUAL TO LOG BASE 3 
OF XY CUBED OVER Z
TO THE POWER OF 1/2.
NOW, WE'LL TAKE THIS ONE STEP 
AT A TIME.
IF WE WANTED TO ELIMINATE 
THE FRACTION
WE COULD WRITE THIS 
AS A DIFFERENCE OF TWO LOGS.
IT WILL BE THE LOG 
OF THE NUMERATOR
MINUS THE LOG 
OF THE DENOMINATOR.
LET'S GO AHEAD AND DO THAT.
SO WE'D HAVE LOG BASE 3 
OF THE NUMERATOR XY CUBED,
MINUS LOB BASE 3 OF Z 
TO THE 1/2.
LET'S TAKE A LOOK 
AT THIS FIRST LOG.
NOW WE HAVE A PRODUCT 
SO WHAT WE CAN DO
IS REWRITE THE LOG 
OF THE PRODUCT
AS THE SUM OF TWO LOGS.
SO THIS WOULD BE EQUAL 
TO LOG BASE 3 OF X
PLUS LOG BASE 3 OF Y CUBED
AND THEN MINUS LOG BASE 3 OF Z 
TO THE 1/2.
NOW WE'RE GOING TO APPLY 
THE POWER PROPERTY OF LOGS
WHICH SAYS WE CAN TAKE THIS 
EXPONENT,
MOVE IT TO THE FRONT 
AS A PRODUCT
AND THE SAME THING HERE.
SO THIS IS EXPANDED AS MUCH AS 
POSSIBLE AS LOG BASE 3 OF X
PLUS 3 TIMES LOG BASE 3 OF Y 
MINUS 1/2 LOG BASE 3 OF Z.
AND THIS IS CONSIDERED 
AS EXPANDED AS POSSIBLE.
SO WE'RE NOT REALLY 
SOLVING HERE.
WE'RE JUST DEMONSTRATING
THAT WE UNDERSTAND 
THESE THREE PROPERTIES.
LET'S GO AHEAD AND TAKE A LOOK 
AT THE SECOND TYPE OF PROBLEM
WHERE WE'RE GIVEN A SUM 
OR DIFFERENCE OF LOGS
AND WE WANT TO ADD IT 
AS A SINGLE LOG.
IN ORDER TO UTILIZE 
THE PRODUCT
AND QUOTIENT PROPERTY 
OF LOGARITHMS,
THE COEFFICIENTS OF THE LOGS 
HAVE TO BE 1.
SO WE FIRST HAVE TO UTILIZE 
THE POWER PROPERTY OF LOGS
IN THE OPPOSITE DIRECTION 
THAT WE JUST DID.
WE'RE GOING TO TAKE 
THE COEFFICIENT OF THE LOG
AND MOVE IT TO THE POSITION 
OF THE EXPONENT.
SO LET'S GO AHEAD 
AND DO THAT FIRST.
WE'RE GOING TO MOVE THIS 2 SO 
IT'S NATURAL LOG OF X SQUARED,
PLUS--NOW THE NUMBER HERE IS X 
PLUS 3 SO WE'LL TAKE THIS 1/3
AND MOVE IT TO THE POSITION 
OF THE EXPONENTS.
WE HAVE NATURAL LOG, 
X PLUS 3 TO THE 1/3 POWER
MINUS--TAKE THIS 4, MOVE IT TO 
THE POSITION OF THE EXPONENT.
SO WE HAVE NATURAL LOG 
OF 2X TO THE 4th.
NOW LET'S GO AHEAD AND TAKE 
THIS ONE STEP AT A TIME.
WE CAN COMBINE THESE TWO, 
SINCE IT'S A SUM OF TWO LOGS,
WE CAN MULTIPLY THE NUMBER 
OF PARTS TOGETHER.
SO THIS WOULD BE NATURAL LOG 
X SQUARED
TIMES X PLUS 3 
TO THE POWER OF 1/3
MINUS NATURAL LOG 2X 
TO THE 4th.
NOW WE HAVE A DIFFERENCE 
OF TWO LOGS
SO WE CAN COMBINE THOSE 
BY WRITING IT AS A QUOTIENT.
SO WE'LL HAVE THE NATURAL LOG
AND THEN IN THE NUMERATOR 
WE'RE GOING TO HAVE X SQUARED,
X PLUS 3 TO THE 1/3 POWER
AND OUR DENOMINATOR WILL BE 2X 
RAISED TO THE POWER OF 4.
NOW, THERE IS ONE MORE THING
THAT THEY MIGHT TRY TO DO 
TO US HERE.
2X TO THE POWER OF 4 
IS ACTUALLY 16X TO THE 4th.
LET'S TAKE A LOOK AT THIS 
EXPRESSION INSIDE THE LOG.
WE HAVE X TO THE 2ND, 
X PLUS 3 TO THE POWER OF 1/3,
AND AGAIN OUR DENOMINATOR 
WILL BE 2X TO THE POWER OF 4.
THAT'D BE 16X TO THE 4th.
WE COULD SIMPLIFY THE X 
SQUARED AND THE X TO THE 4th.
THIS WOULD SIMPLIFY OUT
AND THIS WOULD BECOME X 
TO THE 2nd.
SO LET'S GO AHEAD AND REWRITE 
THIS ONE MORE TIME.
3 TO THE NATURAL LOG,
X PLUS 3 TO THE 1/3 POWER 
IN OUR NUMERATOR
AND OUR DENOMINATOR 
WOULD JUST BE 16X SQUARED.
SO AFTER COMBINING YOUR LOGS,
IF YOU CAN SIMPLIFY 
THIS EXPRESSION,
YOU SHOULD IF POSSIBLE.
OKAY. THAT'LL DO IT 
FOR THIS VIDEO.
I HOPE YOU FOUND IT HELPFUL.
THANK YOU.
