- 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.