- NOW WE'LL LOOK AT THREE EXAMPLES
OF SOLVING LOG EQUATIONS.
IN GENERAL, TO SOLVE A LOG EQUATION,
WE WANT TO ISOLATE THE LOGARITHM
AND THEN WRITE THE LOG EQUATION
AS AN EXPONENTIAL EQUATION USING OUR DEFINITION BELOW.
SO FOR THIS FIRST EXAMPLE,
WE WANT TO ISOLATE LOG BASE 5 OF X,
SO WE'LL START BY ADDING 10 TO BOTH SIDES OF THE EQUATION.
SO NOW WE HAVE 2 x LOG BASE 5 OF X = -6 + 10 IS EQUAL TO 4,
AND NOW WE'LL DIVIDE BOTH SIDES OF THE EQUATION BY 2.
SO NOW WE HAVE LOG BASE 5 OF X IS EQUAL TO 4 DIVIDED BY 2
WHICH IS EQUAL TO 2.
AND NOW WE'RE GOING TO WRITE THE LOG EQUATION
AS AN EXPONENTIAL EQUATION TO SOLVE FOR X.
SO USING OUR DEFINITION,
WE WANT TO LOCATE THE BASE, THE EXPONENT AND THE NUMBER.
THE BASE IS 5, THE EXPONENT IS 2, AND THE NUMBER IS X.
AGAIN, A NICE WAY TO REMEMBER THIS IS TO WORK OUR WAY
AROUND THE EQUAL SIGN STARTING WITH THE BASE.
5 TO THE 2ND MUST EQUAL X.
SO IF 5 TO THE 2ND MUST EQUAL X,
AND 5 TO THE 2ND IS EQUAL TO 5 x 5 OR 25,
WE CAN SEE OUR SOLUTION IS X = 25.
LET'S TAKE A LOOK AT A SECOND EXAMPLE.
AGAIN, OUR GOAL HERE IS TO ISOLATE THE LOGARITHM
OR LOG BASE X OF 32.
SO FOR THE FIRST STEP,
WE'LL SUBTRACT 8 ON BOTH SIDES OF THE EQUATION.
THIS WILL GIVE US -3 LOG BASE X OF 32 = -7 - 8 IS -15,
AND NOW WE'LL DIVIDE BOTH SIDES OF THE EQUATION BY -3.
SO WE HAVE LOG BASE X OF 32 IS EQUAL TO -15 DIVIDED BY -3
IS +5.
NOW THAT WE'VE ISOLATE THE LOGARITHM,
WE'RE GONNA WRITE THIS AS AN EXPONENTIAL EQUATION.
AGAIN, WHERE X IS THE BASE, 5 IS THE EXPONENT,
AND 32 IS THE NUMBER.
WE'RE STARTING WITH THE BASE, WORKING AROUND THE EQUAL SIGN.
WE HAVE X RAISED TO THE 5TH MUST EQUAL 32.
SO IF X TO THE 5TH IS EQUAL TO 32,
WE COULD SOLVE FOR X BY TAKING THE 5TH ROOT
OF BOTH SIDES OF THE EQUATION,
BUT IF WE RECOGNIZE THAT THE PRIME FACTORIZATION OF 32
IS 2 TO THE 5TH,
WE BREAK DOWN 32 INTO ITS PRIME FACTORS,
WE HAVE 16 x 2, 4 x 4, AND 2 x 2.
NOTICE HOW WE HAVE 1, 2, 3, 4, 5 FACTORS OF 2
WHICH MEANS WE CAN WRITE 32 AS 2 TO THE 5TH.
SO IF X TO THE 5TH IS EQUAL TO 2 TO THE 5TH,
WELL, IF THESE ARE EQUAL AND THE EXPONENTS ARE THE SAME,
THEN THE BASES MUST BE EQUAL,
THEREFORE, OUR SOLUTION WOULD BE X = 2.
IF WE WEREN'T ABLE TO WRITE 32
AS SOMETHING RAISED TO THE 5TH POWER,
WE WOULD HAVE TO TAKE THE 5TH ROOT
OF BOTH SIDES OF THE EQUATION IN ORDER TO SOLVE FOR X.
NOW, LET'S TAKE A LOOK AT ONE MORE EXAMPLE.
THIS ONE'S A LITTLE BIT DIFFERENT,
BECAUSE NOTICE HOW THE VARIABLE X
IS NOT ACTUALLY IN THE LOGARITHM,
MEANING IT'S NOT THE BASE OR THE NUMBER OF THE LOGARITHM.
SO WE CAN ACTUALLY SOLVE THIS TWO WAYS,
BUT FOR CONSISTENCY, LET'S FIRST SOLVE IT
BY ISOLATING THE LOGARITHM,
AND THEN WRITING THE LOG EQUATION
AS AN EXPONENTIAL EQUATION.
IF WE WANT TO ISOLATE LOG BASE 3 OF 81,
WE'LL ADD 12 TO BOTH SIDES.
SO THIS WOULD GIVE US 4 LOG BASE 3 OF 81 EQUALS --
THIS WOULD BE 4X + 12.
NOW WE'LL DIVIDE BOTH SIDES OF THE EQUATION BY 4.
SO WE HAVE LOG BASE 3 OF 81 IS EQUAL TO THE QUANTITY 4X + 12
DIVIDED BY 4.
LET'S GO AHEAD AND DIVIDE EACH TERM BY 4.
THAT WOULD GIVE US 4X DIVIDED BY 4 + 12 DIVIDED BY 4.
SO WE HAVE LOG BASE 3 OF 81 IS EQUAL TO THIS WOULD BE X + 3.
NOW WE'LL WRITE THE LOG EQUATION
AS AN EXPONENTIAL EQUATION.
LET'S GO AHEAD AND PUT THE QUANTITY (X + 3)
IN PARENTHESES,
SO WE'D HAVE 3 RAISED TO THE POWER OF X + 3 MUST EQUAL 81.
SO 3 TO THE POWER OF X + 3 MUST EQUAL 81,
AND NOW IF WE CAN WRITE 81 USING BASE 3,
WE CAN THEN SOLVE FOR X.
WELL, 81 IS EQUAL TO 9 x 9, 9 IS EQUAL TO 3 x 3,
WHICH MEANS 81 IS EQUAL TO 3 RAISED TO THE 4TH POWER.
SO WE HAVE 3 TO THE POWER OF X + 3 = 3 TO THE 4TH.
NOW, IF THESE ARE EQUAL AND THE BASES ARE THE SAME,
THE EXPONENTS MUST BE EQUAL,
WHICH MEANS X + 3 MUST EQUAL 4 AND IF X + 3 IS EQUAL TO 4,
OF COURSE WE WOULD SUBTRACT 3 ON BOTH SIDES.
X IS EQUAL TO +1.
SO THIS IS ONE WAY THAT WE COULD SOLVE THIS LOG EQUATION.
BUT I ALSO WANT TO SHOW ANOTHER WAY
THAT WE CAN SOLVE THIS LOG EQUATION.
AS I MENTIONED IN THE BEGINNING,
NOTICE HOW THE VARIABLE IS NOT PART OF THE LOGARITHM,
SO WHAT WE COULD DO IS EVALUATE THIS LOGARITHM HERE
TO LOG BASE THREE OF 81 AND THEN SOLVE FOR X.
SO IF WE WANT TO EVALUATE LOG BASE 3 OF 81,
WE WANT TO KNOW WHAT EXPONENT WE WOULD RAISE 3 TO
IN ORDER TO OBTAIN 81.
MEANING 3 RAISED TO SOME POWER MUST EQUAL 81.
3 RAISED TO WHAT POWER MUST EQUAL 81?
WELL, WE JUST SAW THAT 3 TO THE 4TH IS EQUAL TO 81
WHICH MEANS LOG BASE THREE OF 81 IS EQUAL TO 4,
SO WE CAN REPLACE THE LOGARITHM WITH 4,
THAT WOULD GIVE US 4 x 4 - 12 = 4X.
WELL, 4 x 4 IS EQUAL TO 16.
WE HAVE 16 - 12 = 4X, 16 - 12 IS EQUAL TO 4, SO 4 = 4X.
FINALLY, TO SOLVE FOR X WE WOULD DIVIDE BOTH SIDES BY 4.
YOU CAN SEE WE GET THE SAME SOLUTION, X = +1.
SO BECAUSE THE VARIABLE IS NOT IN THE LOGARITHM,
WE CAN ACTUALLY EVALUATE THIS LOGARITHM FIRST.
IF THIS LOGARITHM DIDN'T COME OUT TO A NICE VALUE,
WE COULD ALWAYS USE THE CHANGE OF BASE FORMULA
TO EVALUATE THIS LOGARITHM.
I HOPE THESE EXAMPLES WERE HELPFUL.