The Properties of Logarithms
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The Properties of Logarithms


– 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.  

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