- WE WANT TO SOLVE 
THE GIVEN LOG EQUATION,
AND WE'LL FIND 
THE EXACT SOLUTION
AND THE DECIMAL APPROXIMATION.
TO SOLVE THIS LOG EQUATION,
THE FIRST STEP IS TO COMBINE 
THE TWO LOGARITHMS
AND THEN WRITE 
THE LOG EQUATION
AS AN EXPONENTIAL EQUATION 
TO SOLVE FOR X.
NOTICE ON THE LEFT SIDE 
OF THIS EQUATION
WE HAVE A DIFFERENCE OF 
TWO LOGS WITH THE SAME BASE
WHICH MEANS WE CAN COMBINE 
THESE TWO LOGARITHMS
USING THE QUOTIENT PROPERTY 
OF LOGARITHMS HERE.
WHERE IF WE HAVE A DIFFERENCE 
OF TWO LOGARITHMS
WITH THE SAME BASE,
WE CAN COMBINE THEM 
BY DIVIDING THE NUMBER PARTS
OF THE LOGARITHMS.
SO LOG BASE 2 
OF X SQUARED - LOG BASE 2
OF THE QUANTITY X + 4
IS EQUAL TO LOG BASE 2 
OF X SQUARED
DIVIDED BY THE QUANTITY X + 4.
THIS IS STILL EQUAL TO +3.
NOW THAT WE HAVE A SINGLE LOG 
ON THE LEFT SIDE
OF THE EQUATION,
WE CAN WRITE THIS 
AS AN EXPONENTIAL EQUATION
BY RECOGNIZING 
THAT 2 IS THE BASE,
3 IS THE EXPONENT, AND 
OUR FRACTION IS THE NUMBER.
SO STARTING WITH THE BASE AND 
WORKING AROUND THE EQUAL SIGN,
WE HAVE 2 RAISED TO THE 3RD
MUST EQUAL X SQUARED 
DIVIDED BY X + 4.
AGAIN, 2 TO THE 3RD MUST EQUAL 
X SQUARED DIVIDED BY X + 4.
WELL, 2 TO THE 3RD 
IS EQUAL TO 8,
SO WE HAVE 8 = X SQUARED 
DIVIDED BY X + 4.
AND NOW TO CLEAR THE FRACTION 
HERE,
WE COULD PUT 8/1 
AND CROSS MULTIPLY,
OR MULTIPLY BOTH SIDES 
OF THE EQUATION BY X + 4.
NOTICE HOW THIS SIMPLIFIES 
TO 1,
SO WE'RE LEFT WITH 8 x THE 
QUANTITY X + 4 = X SQUARED.
LET'S GO AHEAD 
AND DISTRIBUTE HERE
SO WE HAVE 
8X + 32 = X SQUARED.
SINCE WE HAVE 
A QUADRATIC EQUATION,
WE WANT TO SET THIS EQUAL 
TO ZERO
AND THEN SEE 
IF IT'S FACTORABLE.
SO WE'LL SUBTRACT 8X 
ON BOTH SIDES,
AND SUBTRACT 32 ON BOTH SIDES
GIVING US I0 
= X SQUARED - 8X - 32.
WELL, UNFORTUNATELY, 
THERE ARE NO FACTORS OF -32
THAT ADD TO -8,
WHICH MEANS THIS IS NOT 
FACTORABLE,
SO WE'LL HAVE TO USE 
THE QUADRATIC FORMULA.
BUT BEFORE WE DO THIS, 
WE NEED TO RECOGNIZE
THAT A IS EQUAL 
TO THE COEFFICIENT
OF THE X SQUARED TERM OR 1,
B IS EQUAL TO THE COEFFICIENT 
OF THE X TERM OR -8,
AND C IS EQUAL 
TO THE CONSTANT TERM OF -32.
SO NOW WE'LL SUBSTITUTE 
THESE VALUES
INTO THE QUADRATIC FORMULA 
TO SOLVE FOR X.
SO WE'LL HAVE X EQUALS -B
OR -(-8) +/- THE SQUARE ROOT 
OF B SQUARED
WHICH IS -8 SQUARED - 4 x A 
WHICH IS 1 x C WHICH IS -32
AND THIS IS ALL DIVIDED 
BY 2 x A AND A IS 1.
SO NOW WE'LL BEGIN 
TO SIMPLIFY.
THIS WILL BE +8 +/- 
THE DISCRIMINANT
OR THE NUMBER UNDERNEATH 
THE SQUARE ROOT
WILL BE -8 SQUARED, THAT'S 64,
AND THE WILL BE +128.
SO WE'LL HAVE THE SQUARE ROOT 
OF 192
AND THIS IS ALL DIVIDED BY 2.
NOW, IF OUR GOAL 
IS TO ONLY FIND
THE DECIMAL APPROXIMATIONS 
FOR X,
WE COULD STOP HERE 
AND USE THE CALCULATOR,
BUT I'M GOING TO GO AHEAD 
AND CONTINUE SIMPLIFYING
TO FIND THE EXACT VALUES OF X 
FOR THE SOLUTION.
LET'S GO AHEAD 
AND DO THIS ON THE NEXT SLIDE.
SO THE NEXT STEP 
IS TO SIMPLIFY
TO SQUARE ROOT OF 192.
TO DO THIS, IT'S BEST TO LOOK 
AT THE PRIME FACTORIZATION
TO MAKE SURE WE IDENTIFY ALL 
THE PERFECT SQUARE FACTORS.
TO SAVE SOME TIME, 
I'VE ALREADY DONE THIS.
THE PRIME FACTORIZATION OF 192 
IS EQUAL TO 2 TO THE 6TH x 3.
SO WE WOULD HAVE X = 8 +/- THE 
SQUARE ROOT OF 2 TO THE 2ND
IS A PERFECT SQUARE.
THERE'S TWO FACTORS OF 2.
2 TO THE 2ND, 
THAT'S FOUR FACTORS OF 2.
2 TO THE 2ND, 
THAT'S 6 FACTORS OF 2,
AND THEN WE HAVE A FACTOR 
OF THREE.
SO THESE ARE PERFECT SQUARE 
FACTORS,
THEREFORE, THIS WILL SIMPLIFY, 
THIS WILL SIMPLIFY,
AND THIS WILL SIMPLIFY.
SO WE HAVE X = 8 +/- WE'LL 
HAVE THREE FACTORS OF 2
OUTSIDE THE SQUARE ROOT,
AND 2 TO THE 4D IS EQUAL TO 8,
AND THEN WE HAVE THE SQUARE 
ROOT OF 3 ALL OVER 2.
NOW, TO SIMPLIFY THIS SO WE 
DON'T MAKE A CARELESS MISTAKE,
I'D BREAK THIS UP 
INTO TWO FRACTIONS.
WE'D HAVE 8/2 +/- 8 
SQUARE ROOT 3 ALL OVER 2.
SO WE HAVE X = 4 +/- 4 
SQUARE ROOT 3.
NOW, REMEMBER, 
THIS REPRESENTS TWO SOLUTIONS.
ONE SOLUTION 
IS X = 4 + 4 SQUARE ROOT 3,
OR X = 4 - 4 SQUARE ROOT 3.
NOW, THESE ARE THE SOLUTIONS 
TO THE QUADRATIC EQUATION,
BUT WE CAN'T ASSUME 
THESE ARE THE SOLUTIONS
TO THE LOG EQUATION
BECAUSE REMEMBER LOOKING 
AT THE LOG EQUATION HERE,
THE NUMBER PART 
OF THE LOGARITHMS
HAS TO BE GREATER THAN 0.
SO LOOKING AT THIS LOG HERE,
SINCE WE HAVE X SQUARED, X CAN 
BE BOTH POSITIVE OR NEGATIVE,
BUT LOOKING AT THIS LOG HERE,
IF X + 4 HAS TO BE GREATER 
THAN 0,
THEN X MUST BE GREATER 
THAN -4.
SO WE NEED TO MAKE SURE 
THAT BOTH OF THESE SOLUTIONS
ARE GREATER THAN -4.
IT LOOKS LIKE THEY BOTH ARE,
BUT JUST TO BE SURE, 
LET'S GO AHEAD
AND GET THE DECIMAL 
APPROXIMATIONS FOR BOTH
BEFORE WE DECIDE 
OUR FINAL SOLUTIONS.
SO WE'LL CONVERT THIS 
TO A DECIMAL,
AND WE'LL CONVERT THIS 
TO A DECIMAL
AND MAKE SURE THAT X 
IS GREATER THAN -4.
IF THEY BOTH ARE, 
THEN WE HAVE TWO SOLUTIONS.
SO WE HAVE 
4 + 4 SQUARE ROOT 3.
WELL, OF COURSE, 
THIS IS LARGER THAN -4.
THIS IS APPROXIMATELY 10.9282.
AND THEN WE HAVE 4 - 4 
SQUARE ROOT 3.
SO I'M GOING TO PRESS 
SECOND ENTER
AND THEN JUST CHANGE THIS 
PLUS SIGN TO A MINUS SIGN.
AND WE HAVE X IS APPROXIMATELY 
-2.9282.
WELL, THIS VALUE IS GREATER 
THAN -4
BECAUSE IT'S TO THE RIGHT 
OF -4,
WHICH MEANS WE DO HAVE 
TWO SOLUTIONS TO LOG EQUATION.
HERE ARE THE TWO EXACT 
SOLUTIONS TO THE LOG EQUATION,
AND HERE ARE THE TWO DECIMAL 
APPROXIMATIONS
FOR THE SOLUTION.
I HOPE YOU FOUND THIS HELPFUL.
