- WE WANT TO SOLVE 
THE GIVEN LOG EQUATIONS
BUT THESE TWO EQUATIONS 
ARE A LITTLE BIT DIFFERENT
THAN THE PREVIOUS EXAMPLES
BECAUSE NOTICE HOW EACH TERM 
IS A LOGARITHM.
SO NORMALLY WE COMBINE 
THE LOGARITHMS
AND THEN WRITE 
THE LOG EQUATIONS
AS AN EXPONENTIAL EQUATION.
BUT IN THIS CASE AFTER 
WE COMBINE THE LOGARITHMS
WE WILL NOT HAVE TO CONVERT 
THE LOG EQUATION
TO AN EXPONENTIAL EQUATION
BECAUSE IF WE CAN WRITE 
THE LOG EQUATION IN THE FORM
WHERE WE HAVE ONE LOGARITHM 
EQUAL TO ANOTHER LOGARITHM
WITH THE SAME BASE,
FOR EXAMPLE IN THIS FORM HERE,
THEN WE CAN CONCLUDE THAT THE 
NUMBER PARTS OF THE LOGARITHMS
MUST BE EQUAL
OR IN THIS CASE 
X MUST EQUAL Y.
SO LOOKING AT THIS 
FIRST EQUATION,
NOTICE HOW EACH LOGARITHM 
IS A COMMON LOG
WHICH MEANS IT'S LOG BASE 10.
SO THE FIRST STEP HERE WILL BE 
TO COMBINE THE LOGARITHMS
ON THE RIGHT
AND SINCE WE HAVE A SUM 
OF TWO LOGS WITH THE SAME BASE
WE USE THE PRODUCT PROPERTY 
OF LOGARITHMS HERE
TO COMBINE THE TWO LOGS,
WHICH MEANS 
BECAUSE WE HAVE A SUM,
WE ARE GOING TO MULTIPLY 
THE NUMBER PART
OF THE LOGARITHMS TOGETHER.
SO THE LEFT SIDE 
WILL BE THE SAME.
ON THE RIGHT SIDE, 
COMMON LOG X PLUS COMMON LOG 5
IS EQUAL TO THE COMMON LOG 
OF X TIMES 5 OR 5X.
AGAIN, NOW NOTICE HOW WE HAVE 
TWO LOGS WITH THE SAME BASE
EQUAL TO EACH OTHER
AND THEREFORE THE NUMBER 
OF PARTS OF THE LOGARITHMS
MUST BE EQUAL,
MEANING X PLUS 5 
MUST EQUAL 5X.
SO NOW TO SOLVE FOR X., WE'LL 
SUBTRACT X ON BOTH SIDES.
SO WE'LL HAVE 5 EQUALS 4X.
DIVIDE BOTH SIDES BY 4.
SO WE HAVE 5/4 EQUALS X, 
OR WE CAN SAY X EQUALS 5/4.
BUT BEFORE WE DECIDE THIS 
IS OUR SOLUTION,
WE SHOULD TAKE A LOOK BACK 
AT THE ORIGINAL EQUATION
TO DETERMINE 
THE POSSIBLE VALUES FOR X.
WHAT I MEAN BY THAT IS,
REMEMBER THE NUMBER PART 
OF A LOGARITHM
HAS TO BE GREATER THAN ZERO.
SO LOOKING 
AT THESE TWO LOGARITHMS,
NOTICE HOW THE DOMAIN 
OF COMMON LOG X
IS THE MOST RESTRICTIVE 
DOMAIN,
MEANING X MUST BE GREATER 
THAN ZERO.
AND SINCE 5/4 IS GREATER 
THAN ZERO,
WE CAN CONCLUDE 
THAT THIS IS OUR SOLUTION.
NOTICE HOW THE DOMAIN 
OF THIS LOGARITHM
WOULD BE X IS GREATER THAN -5,
BUT X BEING GREATER THAN ZERO 
IS MORE RESTRICTIVE.
SO THIS IS THE ONLY INTERVAL 
WE DO NEED TO CHECK.
NOW LET'S TAKE A LOOK 
AT OUR SECOND EXAMPLE.
NOTICE HOW EACH LOGARITHM 
IS NATURAL LOG OR LOG BASE E.
OUR FIRST STEP WILL BE 
TO COMBINE THE TWO LOGARITHMS
ON THE LEFT SIDE.
AGAIN NOTICE 
HOW WE HAVE A SUM,
SO WE CAN COMBINE THESE
BY MULTIPLYING THE NUMBER 
OF PARTS OF THE LOGARITHMS.
NATURAL LOG X PLUS NATURAL LOG 
OF THE QUANTITY X MINUS 4
IS EQUAL TO NATURAL LOG OF X 
TIMES X MINUS 4.
THIS IS STILL EQUAL 
TO NATURAL LOG 3X.
LET'S GO AHEAD 
AND MULTIPLY THIS OUT
BY DISTRIBUTING THE X.
SO WE HAVE NATURAL LOG 
OF X SQUARED MINUS 4X
EQUALS NATURAL LOG 3X.
AGAIN NOTICE HOW NOW WE HAVE 
TWO LOGARITHMS
WITH THE SAME BASE 
EQUAL TO EACH OTHER
AND THEREFORE THE NUMBER 
OF PARTS OF THE LOGS
MUST BE EQUAL,
MEANING X SQUARED MINUS 4X 
MUST EQUAL 3X.
SO NOW WE'LL SOLVE FOR X.
NOTICE HOW NOW WE HAVE 
A QUADRATIC EQUATION
SO WE'LL SET THIS 
EQUAL TO ZERO
AND SEE IF WE CAN FACTOR IT.
SO LET'S SUBTRACT 3X 
ON BOTH SIDES.
SO WE HAVE X SQUARED MINUS 7X 
EQUALS ZERO.
REMEMBER, THE FIRST STEP 
IN FACTORING
IS TO FACTOR OUT 
THE GREATEST COMMON FACTOR,
WHICH IN THIS CASE WOULD BE X.
SO IF X TIMES A QUANTITY X 
MINUS 7 EQUALS ZERO.
SO THIS QUADRATIC EQUATION 
HAS TWO SOLUTIONS,
X EQUALS ZERO OR X EQUALS 7.
BUT AGAIN WE CAN'T JUST ASSUME
THAT THIS LOG EQUATION 
HAS TWO SOLUTIONS.
WE NEED TO CONSIDER THE DOMAIN 
FOR EACH LOGARITHM.
AGAIN WHICH MEANS THE NUMBER 
PART OF THE LOGARITHMS
MUST ALWAYS BE GREATER 
THAN ZERO.
OR IN THIS CASE, 
THIS LOGARITHM HERE
HAS THE MOST RESTRICTIVE 
DOMAIN.
NOTICE HOW X HAS TO BE GREATER 
THAN 4.
FOR THESE TWO, 
X HAS TO BE GREATER THAN ZERO
BUT THIS IS THE MOST 
RESTRICTIVE DOMAIN,
SO OUR SOLUTION MUST BE 
IN THIS INTERVAL.
WELL, NOTICE HOW X EQUALS ZERO 
WILL NOT BE IN THIS INTERVAL
AND THEREFORE CANNOT BE 
ONE OF OUR SOLUTIONS.
BUT NOTICE HOW 7 IS GREATER 
THAN 4,
THEREFORE, OUR SOLUTION 
IS X EQUALS 7.
OKAY. HOPE YOU FOUND 
THIS HELPFUL.
