- WE WANT TO SOLVE THE GIVEN LOG EQUATIONS.
BEFORE WE SOLVE THESE,
LET'S REVIEW THE GUIDELINES FOR SOLVING LOG EQUATIONS.
STEP 1, IF THERE'S MORE THAN 1 LOG,
WE WANT TO COMBINE THEM USING THE PROPERTIES OF LOGARITHMS.
STEP 2, WE WANT TO ISOLATE THE LOG ON ONE SIDE OF THE EQUATION.
STEP 3, WRITE THE LOG EQUATION AS AN EXPONENTIAL EQUATION.
STEP 4, SOLVE FOR THE VARIABLE.
AND STEP 5, CHECK THE ANSWER.
SO LOOKING AT OUR FIRST EXAMPLE,
WE HAVE NATURAL LOG OF THE QUANTITY 2X - 1 = 5.
WE HAVE A SINGLE LOG, AND IT'S ALREADY ISOLATED.
SO WE'LL GO AHEAD AND WRITE THIS IN EXPONENTIAL FORM.
WE NEED TO RECOGNIZE THAT SINCE IT'S NATURAL LOG, THE BASE IS E.
THEREFORE IN EXPONENTIAL FORM, AGAIN E IS THE BASE,
5 IS THE EXPONENT, AND THE NUMBER WOULD BE 2X - 1.
SO E TO THE 5th MUST EQUAL THE QUANTITY 2X - 1.
AND NOW, WE'LL SOLVE THIS FOR X.
SO WE'LL ADD 1 TO BOTH SIDES OF THE EQUATION.
WE'D HAVE E TO THE 5th + 1 = 2X, DIVIDE BOTH SIDES BY 2.
THIS SIMPLIFIES TO X,
SO WE CAN WRITE THIS AS X = E TO THE 5th + 1 DIVIDED BY 2.
LET'S ALSO GET A DECIMAL APPROXIMATION,
SO WE'LL USE THE CALCULATOR.
THE NUMERATOR IS E TO THE 5th + 1.
SO (2nd NATURAL LOG BRINGS UP E TO THE POWER 5 + 1,
SO THERE'S THE NUMERATOR DIVIDED BY 2.
SO X IS APPROXIMATELY 74.7066.
TO CHECK THIS, WE PERFORM SUBSTITUTION
BACK INTO THE ORIGINAL EQUATION.
SO IF THIS IS CORRECT,
THEN THE NATURAL LOG OF THE QUANTITY 2 x 74.7066 - 1
WOULD BE APPROXIMATELY EQUAL TO 5 SINCE THIS IS ROUNDED.
LET'S CHECK THAT.
WE DID ROUND UP
AND THAT'S WHY IT'S A LITTLE BIT LARGER THAN 5,
BUT THIS DOES VERIFY OUR SOLUTION.
AND LOOKING AT OUR SECOND EQUATION,
WE DO HAVE A SINGLE LOG,
BUT WE DO NEED TO ISOLATE THE LOG
BEFORE WE WRITE THIS AS AN EXPONENTIAL EQUATION.
SO WE NEED TO UNDO THIS + 3
AND UNDO THIS MULTIPLICATION HERE WITH THE 2.
SO WE'LL START BY SUBTRACTING 3 ON BOTH SIDES OF THE EQUATION,
SO IT'D BE 0.
SO WE HAVE 2 x THE COMMON LOG OF X - 2 = 4,
AND THEN UNDO THIS MULTIPLICATION.
WE DIVIDE BOTH SIDES BY 2.
SO THIS SIMPLIFIES TO THE COMMON LOG OF X - 2 = 2.
NOW, WE WANT TO WRITE THIS AS AN EXPONENTIAL EQUATION.
IT'S COMMON LOG, SO THE BASE IS 10,
THE EXPONENT IS 2, AND THE NUMBER IS X - 2.
SO 10 TO THE SECOND POWER = X - 2.
10 SQUARED IS A HUNDRED,
ADD 2 TO BOTH SIDES OF THE EQUATION,
WE HAVE 102 = X,
OR IF WE WANT, WE CAN WRITE X = 102.
LET'S GO AHEAD AND CHECK THIS.
WE'LL SUBSTITUTE 102 INTO THE ORIGINAL EQUATION.
SO WE'D HAVE 2 COMMON LOG, 102 - 2 WOULD BE 100 + 3 = 7.
WELL, THE COMMON LOG OF 100 IS EQUAL TO 2,
SINCE 10 TO THE 2nd IS EQUAL TO 100.
SO THIS WOULD BE 2 x 2 + 3 = 7, AND THAT DOES EQUAL 7.
AGAIN IF WE WEREN'T SURE ABOUT LOG 100,
WE COULD USE THE CALCULATOR
BUT WE SHOULD KNOW THAT THAT'S EQUAL TO 2.
JUST TO CHECK IT, LOG 100 IS EQUAL TO 2,
SO OUR ANSWER IS CORRECT.
OKAY, WE'LL TAKE A LOOK AT TWO MORE EXAMPLES OF LOG EQUATIONS
IN THE NEXT VIDEO.