- WELCOME TO AN EXAMPLE 
OF HOW TO SOLVE A LOG EQUATION
THAT ALSO REQUIRES THE USE 
OF THE QUADRATIC FORMULA.
LOOKING AT THE GIVEN 
LOG EQUATION,
IF WE COMBINE THE 
TWO LOGARITHMS ON THE LEFT
WITH THESE TWO LOGS HERE,
NOTICE HOW WE HAVE A SUM 
OF TWO LOGS,
THEN WE CAN WRITE 
THE LOG EQUATION
AS AN EXPONENTIAL EQUATION 
TO SOLVE FOR X.
AND AGAIN BECAUSE WE HAVE 
A SUM OF TWO LOGS
WITH THE SAME BASE
WE CAN TAKE ADVANTAGE 
OF THE PRODUCT PROPERTY
OF LOGARITHMS GIVEN HERE
OR IF WE HAVE A SUM 
OF TWO LOGS
WE CAN WRITE THEM 
AS A SINGLE LOG
IF WE MULTIPLY THE NUMBER 
PARTS OF THE LOGARITHMS
WHICH MEANS THE LEFT SIDE 
OF THIS EQUATION
IS A GO TO LOG BASE 3 
OF THE QUANTITY X PLUS 2
TIMES A QUANTITY X MINUS 1.
AND THIS IS STILL EQUAL 
TO POSITIVE 2.
NOW WE HAVE A SINGLE LOGARITHM 
ON THE LEFT
WE CAN WRITE THIS LOG EQUATION 
AS AN EXPONENTIAL EQUATION
RECOGNIZING THAT THE BASE 
IS 3, THE EXPONENT IS 2,
AND THE NUMBER PART 
IS THIS PRODUCT HERE.
SO 3 RAISED TO THE SECOND 
POWER MUST EQUAL THIS PRODUCT.
THEREFORE 3 SQUARED EQUALS 
THE QUANTITY X PLUS 2
TIMES A QUANTITY X MINUS 1.
NOTICE HOW WE HAVE 
A QUADRATIC EQUATION
SO NOW WE'RE GOING 
TO MULTIPLY OUT THE RIGHT SIDE
AND THEN SET THIS EQUATION 
TO EQUAL TO ZERO.
WELL, 3 SQUARED IS 9
AND THEN HERE WE'LL HAVE 
FOUR PRODUCTS
ONE, TWO, THREE, AND FOUR.
SO X TIMES X IS X SQUARED.
X TIMES -1 IS -1 OR MINUS X, 
AND THEN PLUS 2X AND MINUS 2.
SO HERE WE HAVE 
TWO LIKE TERMS.
SO WE HAVE 9 EQUALS X SQUARED, 
NEGATIVE 1X PLUS 2X IS 1X
SO PLUS X MINUS 2.
LET SET THIS EQUAL TO ZERO 
AND SEE IF IT'S FACTORABLE.
SO IF WE SUBTRACT 9 
ON BOTH SIDE
AND WE GOT THE EQUATION 
ZERO EQUALS X SQUARED
PLUS X MINUS 11.
NOW, EARLIER WE HAD A PROBLEM 
JUST LIKE THIS
BUT THIS WAS FACTORABLE.
BUT NOTICE HOW IN THIS CASE 
THIS IS NOT FACTORABLE
BECAUSE THERE ARE NO FACTORS 
OF -11 THAT ADD TO POSITIVE 1.
WHICH MEANS TO SOLVE THIS,
WE'LL HAVE TO USE 
THE QUADRATIC FORMULA.
SO LET'S GO AHEAD AND FINISH 
THIS ON THE NEXT SLIDE.
SO AGAIN HERE'S THE EQUATION 
THAT WE WANT TO SOLVE.
NOTICE HOW I DID FLIP 
THE EQUATION
AROUND WHICH DOESN'T CHANGE 
ANYTHING.
SO THE FIRST THING 
TO RECOGNIZE
IS THAT A, THE LEADING 
COEFFICIENT, IS EQUAL TO 1,
B THE COEFFICIENT OF X 
IS EQUAL TO 1
AND C IS EQUAL TO -11.
SO A EQUALS 1, B EQUALS 1 
AND C EQUALS 011.
SO NOW WE'LL SUB THESE VALUES 
INTO THE QUADRATIC FORMULA.
SO WE'LL HAVE X EQUALS 
NEGATIVE B OR -1
PLUS OR MINUS THE SQUARE ROOT 
OF B SQUARED
OR 1 SQUARED MINUS 4 TIMES A,
WHICH IS 1 TIMES C 
WHICH IS EQUAL TO -11.
THIS IS ALL DIVIDED BY TWO 
TIMES A
OR IN THIS CASE 2 TIMES 1.
SO WE HAVE X EQUALS NEGATIVE 1 
PLUS OR MINUS THE SQUARE ROOT.
AND WE NEED TO BE CAREFUL HERE 
WITH THE DISCRIMINANT,
1 SQUARED IS 1
AND THIS IS GOING TO BE MINUS 
-44 PLUS 44 DIVIDED BY 2.
SO WE HAVE X EQUALS -1 PLUS 
OR MINUS THE SQUARE ROOT OF 45
DIVIDED BY 2.
NOW LET'S SEE 
IF IT CAN SIMPLIFY
THE SQUARE ROOT OF 45.
WELL, 45 IS EQUAL TO 9 TIMES 5
AND THE SQUARE ROOT OF 9 
IS EQUAL TO 3.
SO THIS SIMPLIFIES NICELY 
TO 3 SQUARE ROOT 5.
THEREFOR WE HAVE X EQUALS 
NEGATIVE 1 PLUS OR MINUS 3
SQUARE ROOT 5 DIVIDED BY 2.
NOW REMEMBER THIS REPRESENTS 
TWO SOLUTIONS.
ONE SOLUTION IS X EQUALS -1 
PLUS 3 SQUARE ROOT 5
DIVIDED BY 2
AND THE OTHER SOLUTION 
IS X EQUALS -1
MINUS 3 SQUARE ROOT 5 
DIVIDED BY 2.
NOW BEFORE WE CHECK 
OUR SOLUTIONS,
LET'S GET THE DECIMAL 
APPROXIMATIONS
FOR THESE VALUES.
THIS FIRST SOLUTION 
IS APPROXIMATELY 2.8541
AND THE SECOND SOLUTION 
IS APPROXIMATELY -3.8541.
WHEN I CHECK OUR SOLUTIONS, 
THE MAIN THING TO REMEMBER
IS THAT THE NUMBER PART--
THAT THE LOGARITHMS
OR THE ARGUMENTS 
OF THE LOGARITHMS
WHICH WOULD BE X PLUS 2 
AND X MINUS 1
MUST ALWAYS BE POSITIVE.
SO -- IF WE TAKE A LOOK 
AT THE SOLUTION
THAT'S APPROXIMATELY -3.8541,
BOTH X PLUS 2 AND X MINUS 1 
WOULD BE NEGATIVE
AND THEREFORE THIS IS NOT A 
SOLUTION TO THE LOG EQUATION.
AND THEREFORE WE MUST EXCLUDE 
THIS AS A SOLUTION.
BUT NOTICE HOW WHEN X 
IS APPROXIMATELY 2.8541,
BOTH X PLUS 2 AND X MINUS 1 
WOULD BE POSITIVE
AND THEREFORE THIS DOES CHECK 
AS OUR SOLUTION
WHEN WE HAVE ONE SOLUTION 
TO OUR LOG EQUATION
WHICH WOULD BE THIS VALUE 
HERE.
AGAIN WE HAVE THE EXACT VALUE
AND -- MORE APPROXIMATION 
FOR OUR SOLUTION.
THAT'S GOING TO DO IT 
FOR THIS EXAMPLE.
I HOPE YOU FOUND THIS HELPFUL.
