- WELCOME TO A LESSON ON SOLVING LOGARITHMIC EQUATIONS.
THERE ARE REALLY TWO TYPES OF LOG EQUATIONS
WE'RE GOING TO TAKE A LOOK AT IN THIS VIDEO.
THE FIRST TYPE IS WHEN YOU HAVE AN EQUATION
THAT HAS ONLY TWO LOGS OF THE SAME BASE.
AND IF THAT'S THE CASE,
IF WE SET THE LOGS EQUAL TO EACH OTHER,
AS WE SEE HERE,
THEN IT FOLLOWS THAT THE ARGUMENT OF THE FIRST LOG
MUST EQUAL THE ARGUMENT OF THE SECOND LO.
SO WE'LL SET THOSE EQUAL TO EACH OTHER AND THEN SOLVE.
NOW, WE DO HAVE TO CHECK OUR ANSWERS THOUGH,
BECAUSE THE ARGUMENT OF THE LOG
HAS TO BE GREATER THAN ZERO.
SO IF WE OBTAIN A VALUE THAT MAKES THIS LESS THAN
OR EQUAL TO ZERO,
WE WOULD TO EXCLUDE IT.
LET'S GO AHEAD AND TRY A COUPLE OF THESE FIRST.
SO NOTICE WE HAVE TWO LOGS OF BASE 5 EQUAL TO EACH OTHER,
THEREFORE, IT FOLLOWS THAT 12 MUST EQUAL 2X - 3.
SO NOW WE CAN SOLVE THIS FOR X.
WE'LL ADD THREE TO BOTH SIDES.
THAT WOULD GIVE US 15 = 2X
AND DIVIDING BOTH SIDES BY 2, WE HAVE X = 15 HALVES,
OR WE COULD SAY 7.5.
NOW, LET'S GO AHEAD AND CHECK THIS,
AND THIS ONE IS PRETTY STRAIGHT FORWARD.
2 x 7.5, THAT WOULD BE 15 - 3, THAT'D BE 12.
SO WE HAVE LOG 12 ON THE RIGHT AND LOG 12 ON THE LEFT,
SO THAT SOLUTION CHECKS.
THIS NEXT ONE IS VERY SIMILAR,
EXCEPT WE HAVE TO SET THE TWO LOGS
EQUAL TO EACH OTHER FIRST.
SO IF WE ADD LOG BASE 7 OF 10 TO BOTH SIDES OF THE EQUATION,
WE WOULD HAVE LOG BASE 7 OF X SQUARED - 3X
= LOG BASE 7 OF 10.
AND IF THE LOGS ARE EQUAL TO EACH OTHER,
THAN X SQUARED - 3X MUST EQUAL 10.
WELL, NOW WE HAVE A QUADRATIC EQUATION.
WE'LL SET IT EQUAL TO ZERO AND SEE IF IT FACTORS,
SUBTRACTING 10 ON BOTH SIDES.
THERE ARE FACTORS OF -10 THAT ADD TO -3.
THAT WOULD BE X - 5 AND X + 2.
SO FROM THIS WE HAVE A SOLUTION OF X = 5 AND X = -2.
NOW, WE DO NEED TO CHECK THESE SOLUTIONS.
LET'S GO AHEAD AND CHECK THE EQUATION IN THIS FORM.
LET'S JUST MAKE SURE THIS IS EQUAL TO +10
FOR BOTH OF THESE X VALUES.
WHEN X = 5, WE'D HAVE 5 SQUARED - 3 x 5.
WE'D HAVE 25 - 15 WHICH = 10, AND THAT CHECKS.
AND NOT LET'S TRY X = -2.
-2 SQUARED IS 4 - 3 x -2, THAT WOULD BECOME 6.
4 + 6 DOES = 10, SO BOTH SOLUTIONS CHECK.
NOW, THERE ARE GUIDELINES FOR SOLVING
THE MORE GENERAL TYPE OF LOG EQUATION
AND HERE ARE THE STEPS.
IF THERE IS MORE THAN ONE LOG,
COMBINE THEM USING THE PROPERTIES OF LOGS.
THEN WE WANT TO ISOLATE THE SINGLE LOG,
AND THEN WE'RE GOING TO WRITE THE EQUATION
IS EXPONENTIAL FORM, SOLVE, AND CHECK.
THESE ARE THE PROPERTIES THAT WE DO HAVE TO BE AWARE OF.
THE PRODUCT, QUOTIENT, AND POWER PROPERTY,
AND SOMETIMES THE CASE OF BASE FORMULA.
LET'S TAKE A LOOK AT SOME OF THESE PROBLEMS.
IN ORDER TO ISOLATE THE LOG,
WE'LL FIRST DIVIDE BOTH SIDES OF THE EQUATION BY 2.
THAT'S GOING TO GIVE US LOG BASE 3 OF 3X = 3.
SO NOW THAT WE HAVE A SINGLE LOG,
WE'RE GOING TO REWRITE THIS IN EXPONENTIAL FORM.
SO 3 TO THE POWER OF 3 MUST EQUAL 3X,
AND NOW WE'RE GOING TO SOLVE THIS EQUATION
AND THEN CHECK OUR SOLUTION.
WELL, 3 TO THE 3RD WOULD BE 27.
SO NOW WE DIVIDE BOTH SIDES BY 3, AND WE HAVE X = 9.
LET'S GO AHEAD AND CHECK X = 9 IN THIS FORM OF THE EQUATION.
SO IF X = 9, WE'D HAVE THE LOG BASE 3 OF 27 = 3,
AND 3 TO THE 3RD DOES = 27, SO IT CHECKS.
LET'S TRY ANOTHER.
SO WE WANT TO COMBINE THESE TWO LOGS.
SINCE THE TWO LOGS ARE BEING ADDED TOGETHER,
WE CAN MULTIPLY THE NUMBER PART OF THE LOG.
SO WE CAN COMBINE THESE INTO LOG BASE 5
OF X - 2 x X + 2 = 1.
LET'S GO AHEAD AND MULTIPLY THIS OUT.
THIS IS GOING TO BE X SQUARED + 2x - 2X, THAT'S 0
AND THEN -4 = 1.
WE HAVE A SINGLE LOG,
SO NOW WE'LL REWRITE THIS IN EXPONENTIAL FORM.
5 TO THE 1ST POWER MUST EQUAL X SQUARED - 4.
LET'S GO AHEAD AND ADD 4 TO BOTH SIDES.
WE'LL HAVE X SQUARED = 9.
NOW WE CAN SQUARE ROOT BOTH SIDES OF THE EQUATION.
REMEMBER, WE'RE GOING TO HAVE TWO SOLUTIONS FROM THIS.
X = +/-3.
LET'S GO AHEAD AND CHECK THOSE SOLUTIONS.
WHEN X = 3, WE'RE GOING TO HAVE LOG BASE 5 OF 1,
THAT'S 0,
PLUS LOG BASE 5 OF 3 + 2, THAT'D BE 5.
LOG BASE 5 OF 5 = 1, THAT'S TRUE, SO THAT SOLUTION CHECKS.
X = -3 IS GOING TO BE A PROBLEM
BECAUSE WHEN WE SUB IN -3 HERE FOR X,
WE'RE GOING TO HAVE LOG BASE 5 OF -5,
WHICH IS NOT IN THE DOMAIN OF THIS LOGARITHM,
THEREFORE, IT'S NOT A SOLUTION.
WE ONLY HAVE ONE SOLUTION TO THIS EQUATION.
LET'S GO AHEAD AND TRY ONE MORE.
NOW, WE WANT TO COMBINE THESE TWO LOGS,
BUT IT'S A DIFFERENCE.
REMEMBER, IF IT'S A DIFFERENCE,
THEN THIS IS GOING TO END UP BEING THE LOG OF A QUOTIENT.
WE'RE GONNA HAVE LOG BASE 16 OF X DIVIDED BY X - 1 = 1/4.
WE HAVE A SINGLE LOG NOW,
SO WE'LL REWRITE THIS IN EXPONENTIAL FORM.
16 TO THE POWER OF 1/4 MUST EQUAL THIS QUOTIENT.
16 TO THE 1/4 POWER IS EQUAL TO 2.
WE CAN REWRITE 16 AS 2 TO THE 4TH
WHICH WOULD GIVE US JUST 2 TO THE 1ST.
THIS IS 2 MUST EQUAL X DIVIDED BY X - 1.
WHAT I WOULD PROBABLY DO HERE IS PUT THIS OVER 1.
WE HAVE A PROPORTION SO WE CAN CROSS MULTIPLY.
2 x X - 1 MUST EQUAL 1 x X WHICH IS X.
SO NOW WE CAN SOLVE THIS EQUATION.
2X - 2 = X.
LET'S GO AHEAD AND SUBTRACT 2X ON BOTH SIDES.
-2 = -X.
NOW WE CAN DIVIDE BY -1 IF WE WANT.
X = 2.
NOTICE WHEN X = 2
WE HAVE LOG BASE 16 OF 2 - LOG BASE 16 OF 1.
THIS WOULD BE 0.
16 TO THE 1/4 = 2 SO THIS DOES CHECK.
THE MAIN REASON FOR CHECKING THESE
IS THE ARGUMENT OF THE LOG HAS TO BE GREATER THAN ZERO,
AND IF IT'S NOT, IT WOULD NOT BE A SOLUTION.
OKAY. I HOPE YOU FOUND THIS VIDEO HELPFUL.
HAVE A NICE DAY.