- EXPANDING AND CONDENSING
LOGARITHMS.
ALL RIGHT. WE'RE GOING TO BE
USING THREE BASIC RULES.
THE FIRST RULE IS THE PROCESS
SUM RULE LOG BASE "A" OF XY,
X x Y = LOG BASE "A"
OF X + LOG BASE "A" OF Y.
SECOND RULE IS THE QUOTIENT
TWO DIFFERENCE RULE,
WHICH IS LOG BASE "A" OF X
DIVIDED BY Y = LOG BASE "A"
OF X - LOG BASE "A" OF Y.
THIRD RULE IS THE POWER RULE,
WHERE WE HAVE A BASE RAISED
TO A POWER, SAY POWER OF B
AND THAT B COMES DOWN
TO BE MULTIPLIED, LIKE SO.
SO WE'RE GOING TO BE
USING THESE THREE RULES,
FORWARDS AND BACKWARDS TO TRY
TO EITHER CONDENSE LOGARITHMS
INTO A SINGLE LOGARITHM
OR EXPAND A SINGLE LOGARITHM
INTO A BUNCH OF LOGARITHMS.
WE'RE GOING TO START
BY EXPANDING.
SO WE'RE GOING TO EXPAND THIS
AS FAR AS WE CAN GO.
OUR FIRST EXAMPLE IS GOING
TO BE COMMON LOG,
WHICH IS BASE 10 WHEN THERE'S
NOTHING WRITTEN THERE,
OF X SQUARED OVER Y SQUARED.
NOW, IN THIS CASE, AGAIN,
THE COMMON LOG IS MISSING,
THE BASE IS MISSING,
WHICH IS BASE 10,
BUT WE JUST CALL THEM
THE COMMON LOG.
I'M GOING TO USE THE SECOND
RULE, THE QUOTIENT RULE,
AND WRITE DOWN LOG OF X
SQUARED - LOG OF Y SQUARED.
NOW, I'M GOING TO APPLY
THE THIRD RULE
AND REWRITE AND BRING THAT
TWO DOWN, COMMON LOG OF X -
AND NOW, I'M GOING TO BRING
THE X ONE DOWN FROM THE Y,
I'M GOING TO GET 2,
AND THE COMMON LOG OF Y.
SO THAT WOULD BE
COMPLETELY EXPANDED,
AS EXPANDED AS WE CAN MAKE IT.
LET'S TRY ANOTHER ONE.
HERE WE HAVE LOG BASE 12
OF X SQUARED OVER (X + 3).
NOW, AGAIN, I HAVE
THE QUOTIENT RULE,
I'M DIVIDING IN THIS CASE,
WHICH IS RULE TWO FROM ABOVE.
IT'S GOING TO GIVE ME LOG BASE
12 OF X SQUARED - LOG BASE 12
OF (X + 3).
NOW, I'M GOING TO FURTHER USE
THIS ON THE FIRST PART HERE,
AND I'M GOING TO HAVE THIS 2
COME OUT IN FRONT,
USING RULE NUMBER THREE
UP ABOVE.
AND THIS NEXT ONE,
I CAN'T DO ANYTHING MORE.
THE LOG BASE 12 OF (X + 3) HAS
TO REMAIN EXACTLY AS IT IS.
I CAN'T DISTRIBUTE
THE LOG BASE 12 INTO BOTH
OR ANYTHING LIKE THAT.
RULE NUMBER ONE
DOES NOT APPLY.
IT JUST STAYS EXACTLY
LIKE IT IS.
AND THAT'S AS EXPANDED
AS WE CAN MAKE THAT ONE.
THE NEXT EXAMPLE
IS GOING TO BE
A LITTLE BIT MORE COMPLICATED.
WE'RE GOING TO HAVE LOG BASE 6
OF 4, X TO THE THIRD
OVER THE SQUARE ROOT OF Y.
AND WE HAVE FOUR BASES
IN THIS CASE.
WE'RE GOING TO EXPAND THEM
AS FAR AS WE CAN EXPAND THEM.
ON TOP, THE LOG BASE 6
OF 4 + LOG BASE 6 OF X CUBED,
SO I USE RULE NUMBER 1
FOR THE NUMERATOR.
THE DENOMINATOR BECOMES MINUS.
LOG BASE 6 OF--AND I'M GOING
TO REWRITE IT AS--
WE JUST WANT THE NUMERATOR
BECAUSE WE HAVE THE MINUS
OUT IN FRONT.
I KNOW THAT 4, IS ACTUALLY 2
TO THE SECOND POWER.
SOMETIMES THIS IS NECESSARY,
SOMETIMES NOT SO MUCH.
LOG BASE 6 OF X TO THE THIRD.
NOW, HERE, I KNOW
THAT THE EXPONENT FOR Y
IS ACTUALLY Y
TO THE 1/2 POWER.
AND NOW, FOR ALL THREE
OF THESE,
I'M GOING TO APPLY
THE THIRD RULE UP ABOVE
WHERE I TAKE THE EXPONENT AND
THEN BRING IT DOWN IN FRONT.
THIS BECOMES 2 x LOG BASE 6
OF 2 + --NOW,
THE 3 IS GOING TO COME DOWN
FROM THE X, LOG BASE 3 OF X.
AND FOR THE Y, -1/2
IS NOW GOING TO BE IN FRONT,
LOG BASE 6 OF Y.
AND THAT'S AS EXPANDED
AS I CAN MAKE THAT ONE.
THIS NEXT ONE
IS QUITE COMPLICATED,
SO THIS ONE'S GOING TO BE A
LITTLE BIT LONGER. HERE WE GO.
IF YOU CAN DO THIS ONE,
YOU'RE DOING A FANTASTIC JOB.
I'M GOING TO USE
THE NATURAL LOG FOR THIS ONE
WHICH IS LOG BASE E,
AND WE'RE GOING TO START
OUT HERE
THAT THIS IS THE SEVENTH ROOT
OF THE WHOLE THING,
AND WE'RE GOING TO HAVE
A WHOLE LOT GOING ON HERE.
THIS IS X SQUARED TIMES
THE CUBE ROOT OF Y SQUARED,
OVER THE SQUARE ROOT
OF (Z + 5) CLOSED.
ALL RIGHT.
SO THIS IS QUITE A LARGE ONE.
SO WE'RE GOING TO HAVE TO GO
A LITTLE SLOW
AND TAKE OUR TIME WITH IT.
LET'S START WITH
THE BIG SEVENTH ROOT HERE,
AND WHAT WE'RE GOING TO DO IS,
I'M GOING TO REWRITE
THE NATURAL LOG,
AND THIS IS JUST GOING TO BE
X SQUARED x Y TO THE 2/3.
WE'LL JUST KEEP IT THERE
FOR NOW.
AND BELOW WE HAVE THE Z + 5
RAISED TO THE 1/2 POWER.
AND BECAUSE OF THE SEVENTH
ROOT OVER HERE,
THE ENTIRE THING IS GOING
TO BE RAISED TO THE 1/7 POWER,
ALL OF IT.
WHAT'S GOING TO HAPPEN NEXT,
IS I'M GOING TO HAVE TO TAKE
THAT 1/7
AND THAT'S GOING TO COME
ALL THE WAY DOWN IN FRONT.
I'M USING RULE NUMBER THREE
FROM UP ABOVE,
SO THAT 1/7 COMES DOWN.
I'M GOING TO PUT
THE NATURAL LOG
OF X SQUARED x Y TO THE 2/3,
AND BELOW WE HAVE (Z + 5)
RAISED TO THE 1/2 POWER.
ALL RIGHT. NOW, WE GOT TO
KEEP THIS 1/7 OUT HERE.
1/7, I'M JUST GOING TO PUT
A BIG BRACKET THERE.
AND NOW, I'M GOING TO USE RULE
NUMBER ONE FOR THE NUMERATOR,
SO NATURAL LOG OF X SQUARED
+ NATURAL LOG OF Y
TO THE 2/3 POWER.
AND DOWN BELOW,
I'M GOING TO USE MINUS,
WHICH IS RULE NUMBER TWO
UP ABOVE,
BECAUSE I'M DIVIDING BY THAT,
OF NATURAL LOG OF (Z + 5)
RAISED TO THE 1/2 POWER,
OH--BRACKET.
NOW, PLEASE NOTE, I STILL HAVE
THE 1/7 OUT IN FRONT.
WE JUST NEED TO KIND OF
KEEP THAT HANGING OUT THERE
FOR RIGHT NOW.
NEXT STEP, WE HAVE THE 1/7
AND INSIDE THE BRACKETS,
I NOTICE I HAVE NATURAL LOG
OF X SQUARED UP ABOVE,
THAT 2 IS GOING TO COME DOWN,
AND I'M GOING TO GET
2 x THE NATURAL LOG OF X.
WITH THE Y, WE HAVE IT--
IT'S RAISED TO THE 2/3
THAT IS ALSO GOING TO COME
DOWN, 2/3 NATURAL LOG OF Y.
AND FOR OUR MINUS HERE,
WE'RE GOING TO HAVE MINUS,
AND THE 1/2 POWER COMES
ALL THE WAY IN FRONT AGAIN,
USING RULE NUMBER THREE
FROM UP ABOVE.
NOW, AGAIN, THE Z + 5
JUST STAYS AS IT IS.
RULE NUMBER ONE
DOES NOT APPLY,
NOTHING CRAZY LIKE THAT.
SO I'M NOT GOING TO--
ANY DISTRIBUTING
OR ANYTHING LIKE THAT.
HOWEVER, NOW, WE ARE GOING
TO DISTRIBUTE THE 1/7
TO EACH OF THESE LOGS.
SO 1/7 x 2, GIVE ME 2/7--
1/7 x 2/3 IS GOING TO GIVE ME
+2/21 NATURAL LOG OF Y.
ALSO THE 1/7 IS GOING
TO STRETCH ALL THE WAY
OVER TO HERE.
I'M GOING TO GET -1/14
NATURAL LOG OF Z + 5.
AND THAT IS OUR FINAL ANSWER.
THAT IS A CRAZY ANSWER THERE,
CRAZY PROBLEM.
IF YOU DIDN'T FOLLOW THAT ONE,
THAT'S OKAY.
YOU CAN GO BACK AND REVIEW IT
A FEW TIMES.
BUT THIS IS DEFINITELY
THE MORE COMPLICATED ONE.
THE PREVIOUS THREE EXAMPLES,
YOU SHOULD BE ABLE TO FOLLOW
THOSE HOPEFULLY.
ALL RIGHT, SO NOW,
WHAT WE'RE GOING TO DO
IS WE'RE GOING TO CONDENSE
INTO A SINGLE LOG--LOGARITHM.
SO WE'RE GOING TO GO BACKWARDS
FROM WHAT WE JUST DID.
SO LET'S START
WITH A NEW EXAMPLE,
START WITH A NICE, SIMPLE ONE.
I'LL GO LOG BASE 2 OF 5 + LOG
BASE 2 OF X TO THE THIRD.
NOW, WE'RE GOING USE
RULE NUMBER TWO,
I'M SORRY, RULE NUMBER ONE
FROM BEFORE IS GOING TO APPLY.
NOTICE I HAVE A PLUS
IN-BETWEEN
AND SO I'M GOING TO BACKWARDS
AND CONDENSE IT
INTO A SINGLE LOGARITHM
BY MULTIPLYING LOG BASE 2
OF 5 x TO THE THIRD POWER.
SO NOW, IT'S BEEN REWRITTEN
AS A SINGLE LOGARITHM,
SO WE'RE GOOD TO GO.
NEXT EXAMPLE.
HERE I HAVE 2 LOG BASE 7
OF (Y + 6) LOG BASE 7 OF Z.
NOW, RULE NUMBER THREE
IS GOING TO APPLY HERE.
I NOTICE I HAVE THE 2
OUT IN FRONT.
THAT'S GOING TO MOVE HERE
AND BECOME THE EXPONENT,
SAME WITH THE 6.
SO LOG BASE 7 OF Y SQUARED
+ LOG BASE 7 OF (Z + 6).
NOW, AGAIN, RULE NUMBER ONE
IS GOING TO APPLY
BECAUSE I'M ADDING IN-BETWEEN.
SO THIS IS GOING TO GIVE ME
LOG BASE 7 OF Y SQUARED,
Z RAISED TO THE SIXTH POWER.
AND THERE WE'VE CONDENSED IT
AGAIN INTO A SINGLE LOGARITHM.
ALL RIGHT. HERE, MOVING ALONG.
NEXT EXAMPLE IS GOING TO BE
A LITTLE BIT MORE LONG
AND THEN THE LAST EXAMPLE
WE'RE GOING TO BE DOING
IS GOING TO BE A LITTLE BIT
CRAZY LIKE BEFORE
WITH THE EXPANDING.
SO WE'RE GOING TO HAVE
LOG BASE 6 OF 18 + LOG BASE 6
OF 2 - LOG BASE 6 OF 9.
AND I'M GOING TO NOTICE,
I DON'T HAVE ANY EXPONENTS
TO BRING UP
SO I'M JUST GOING TO GO
STRAIGHT TO RULE NUMBER ONE
AND TWO.
WE'LL START UP
WITH RULE NUMBER ONE,
LOG BASE 6 OF 18 x 2
DIVIDED BY 9.
AND I NOTICED THESE ARE
ALL JUST NUMBERS,
WE CAN JUST SORT OF
CRUNCH THEM TOGETHER.
I HAVE 18 x 2 WHICH IS 36,
AND 36 DIVIDED BY 9 IS 4.
NOW, WE HAVE A CHOICE,
I COULD TAKE 4
AND BREAK IT INTO 2
RAISED TO THE SECOND POWER
AND BRING THE 2 DOWN.
BUT IN THIS CASE,
I'M JUST GOING TO STOP HERE.
PREVIOUSLY, WE DID BREAK IT
DOWN LIKE THAT.
I JUST WANT TO SHOW YOU THAT
IT DEPENDS ON THE PROBLEM.
IT DEPENDS WHAT YOUR TEACHER
IS ASKING FOR,
SOMETIMES YOU DO,
SOMETIMES YOU DON'T.
SO WE'LL JUST LEAVE IT
THIS TIME.
ALL RIGHT.
LAST EXAMPLE IS A LONG ONE.
3 LOG BASE 2 OF X + 1/2)
LOG BASE 2 OF (X - 2)
x LOG BASE 2 OF (X + 1).
SO I NOTICE I HAVE
SEVERAL PIECES
THAT ARE GOING TO TURN
INTO EXPONENTS HERE.
SO I'M GOING TO APPLY
RULE NUMBER THREE FIRST
AND BRING THOSE EXPONENTS UP.
SO THIS IS EQUAL TO LOG BASE 2
OF X TO THE THIRD
+ LOG BASE 2 OF X
TO THE 1/2 POWER -
LOG BASE 2 OF (X + 1)
TO THE SECOND POWER.
NOW, I HAVE ALL THE EXPONENTS
TAKEN CARE OF
WHICH WE'VE BEEN DOING FIRST,
WHICH IS NECESSARY.
NOW, I'M GOING TO APPLY
RULE NUMBER ONE AND TWO.
LET'S START
WITH RULE NUMBER ONE.
LOG BASE 2 OF--I HAVE X
TO THE THIRD POWER
x X TO THE HALF POWER
DIVIDED BY (X + 1)
RAISED TO THE SECOND POWER.
NOW, UP ABOVE
IN THE NUMERATOR,
I HAVE X TO THE THIRD x X
TO THE 1/2.
AND SO I'M GOING TO GO TO MY
SCRATCH PAPER OVER HERE,
X TO THE THIRD x X
TO THE 1/2 POWER.
AND IF YOU REMEMBER, THERE'S
AN EXPONENT RULE THAT SAYS
WHEN YOU MULTIPLY THE SAME
BASES, YOU ADD THE EXPONENTS.
(3 + 1/2) OR X RAISED
TO THE 6/2 + 1/2,
WHICH IS X TO THE 7/2.
COMING BACK TO OUR PROBLEM,
BACK HERE IN THE NUMERATOR,
LOG BASE 2.
IN THE NUMERATOR, THE X TO THE
THIRD x X TO THE 1/2 POWER,
IF I GO TO MY SCRATCH WORK,
THAT ALL JUST BECOMES X
RAISED TO THE 7/2 POWER.
DOWN BELOW, I HAVE (X + 1)
RAISED TO THE SECOND POWER.
AND THERE WE'VE CONDENSED IT
INTO A SINGLE LOGARITHM.
