- WE WANT TO EVALUATE
THE GIVEN EXPRESSION
IF NATURAL LOG X EQUALS 2,
NATURAL LOG Y EQUALS 3,
AND NATURAL LOG Z EQUALS 5.
WE HAVE TO BE CAREFUL
ABOUT THIS
BECAUSE NOTICE HOW WE HAVE
A QUOTIENT OF 2 LOGARITHMS
SO WE CANNOT WRITE THIS AS A
DIFFERENCE OF TWO LOGARITHMS
USING THE QUOTIENT PROPERTY
OF LOGS GIVEN HERE.
THIS TELLS US THE LOG OF X
DIVIDED BY Y
EQUALS LOG X MINUS LOG
BUT WE DON'T HAVE THE LOG
OF A QUOTIENT,
WE HAVE THE QUOTIENT
OF TWO LOGS.
SO, WE'LL HAVE TO EXPAND
THE NUMERATOR AND DENOMINATOR
SEPARATELY
AND THEN PERFORM SUBSTITUTION
FROM THE GIVEN INFORMATION.
BUT BEFORE WE START,
LET'S TAKE A LOOK
AT THE DENOMINATOR.
NOTICE HOW WE HAVE NATURAL
LOG X TIMES Z SQUARED.
SO WHILE WE COULD TAKE
THE EXPONENT OF 2
AND MOVE IT TO THE FRONT
OF THE NATURAL LOG
USING OUR POWER PROPERTY
OF LOGARITHMS HERE,
I THINK IT'LL BE EASIER
TO KEEP TRACK OF
IF WE USE OUR EXPONENT RULES
WHERE WE HAVE Y TO THE FIRST,
Z TO THE FIRST SQUARED,
AND THIS FORM,
MULTIPLY THE EXPONENTS.
SO, LET'S GO AHEAD AND WRITE
THIS AS NATURAL LOG
X TO THE NEGATIVE 1,
Y TO THE SECOND, DIVIDED BY
NATURAL LOG Y TO THE SECOND,
Z TO THE SECOND.
AND NOW IN THIS FORM,
NOTICE BOTH THE NUMERATOR
AND DENOMINATOR
CONTAIN NATURAL LOG
OF A PRODUCT.
SO, USING THE PRODUCT PROPERTY
OF LOGARITHMS HERE,
WE CAN REWRITE THE NUMERATOR
AND DENOMINATOR
AS A SUM OF TWO LOGARITHMS.
SO, FOR THE NUMERATOR,
WE'D HAVE NATURAL LOG X
TO THE NEGATIVE 1
PLUS NATURAL LOG Y
TO THE SECOND.
AGAIN, WE HAVE A SUM HERE
BECAUSE WE HAVE NATURAL LOG
OF X
TO THE NEGATIVE 1
TIMES Y SQUARED.
AND THEN FOR THE DENOMINATOR,
WE WOULD HAVE NATURAL LOG
Y SQUARED
PLUS NATURAL LOG Z SQUARED.
AND NOW, WE ARE GOING TO APPLY
THE POWER PROPERTY
OF LOGARITHMS,
WHICH MEANS WE CAN TAKE
THE EXPONENT, IN THIS CASE N,
AND WRITE IT AS N TIMES
THE LOGARITHM.
SO, WE CAN MOVE THE EXPONENT
OF NEGATIVE 1 TO THE FRONT,
MOVE THE EXPONENT OF 2
TO THE FRONT OF THE LOGARITHM,
AND HERE AND HERE AS WELL.
WHICH MEANS, NOW WE HAVE
NEGATIVE 1 NATURAL LOG X.
OF COURSE, THE 1 IS OPTIONAL.
WE COULD JUST WRITE
NEGATIVE NATURAL LOG X
AND THEN IT WOULD BE PLUS 2
NATURAL LOG Y,
DIVIDED BY THE QUANTITY 2
NATURAL LOG Y,
AND PLUS 2 NATURAL LOG Z.
AND NOW IN THIS FORM,
WE CAN PERFORM SUBSTITUTION
FOR NATURAL LOG X,
NATURAL LOG Y,
AND NATURAL LOG Z.
SO, WELL HAVE NEGATIVE 1
TIMES NATURAL LOG X,
WHICH IS EQUAL TO 2,
PLUS 2 TIMES NATURAL LOG Y,
WHICH IS EQUAL TO 3
AND THEN FOR THE DENOMINATOR,
WE'LL HAVE 2 TIMES NATURAL
LOG Y, WHICH AGAIN IS 3,
PLUS 2 TIMES NATURAL LOG Z,
WHICH IS 5.
SO, WE'LL HAVE NEGATIVE 2
PLUS 6 ALL OVER 6 PLUS 10,
WHICH IS EQUAL TO 4/16.
WELL 4/16 SIMPLIFIES TO 1/4
OR IF WE WANT, 0.25.
AND THAT'LL DO IT
FOR THIS EXAMPLE.
I HOPE YOU FOUND THIS HELPFUL.
