- WELCOME TO A SECOND VIDEO ON 
SOLVING EXPONENTIAL EQUATIONS.
IN THIS VIDEO, WE'LL DISCUSS HOW 
TO SOLVE EXPONENTIAL EQUATIONS
THAT DO REQUIRE 
USING LOGARITHMS,
AND WE'LL ALSO SHOW HOW TO CHECK 
THEM GRAPHICALLY.
SO HERE ARE THE GENERAL STEPS 
THAT WE'LL TAKE
TO SOLVE THIS TYPE OF EQUATION.
WE'LL ISOLATE THE EXPONENTIAL 
PART OF THE EQUATION.
IF THERE ARE TWO 
EXPONENTIAL PARTS,
WE'LL PUT ONE ON 
EACH SIDE OF THE EQUATION,
AND THEN WE'LL TAKE 
EITHER THE COMMON LOG
OR THE NATURAL LOG OF BOTH SIDES 
OF THE EQUATION.
AND WE'LL USE 
EITHER OF THESE TWO LOGS,
BECAUSE THOSE ARE THE LOGS 
THAT ARE ON THE CALCULATOR.
AND THEN WE'LL SOLVE 
FOR THE VARIABLE
AND CHECK OUR SOLUTIONS.
SO FOR NUMBER ONE, WE NEED 
TO ISOLATE 7 TO THE POWER OF X.
SO WE'LL ADD 1 TO BOTH SIDES 
OF THE EQUATION,
SO IT'LL GIVE US 7 TO THE POWER 
OF X = 5.
WE OBVIOUSLY CANNOT GET A COMMON 
BASE HERE,
SO WHAT WE'LL DO NOW 
IS TAKE THE LOG
OF BOTH SIDES OF THE EQUATION.
I'M GOING TO TAKE THE NATURAL 
LOG.
SO IF WE TAKE THE NATURAL LOG 
OF 7 TO THE POWER OF X,
THAT MUST EQUAL 
THE NATURAL LOG OF 5.
NOW, WHAT WE CAN DO IS APPLY THE 
POWER PROPERTY TO THIS LOG,
SO WE CAN TAKE THIS EXPONENT 
AND MAKE IT A PRODUCT
OF X x NATURAL LOG 7.
SO X x NATURAL LOG 7 = NATURAL 
LOG 5.
AND NOW, WE CAN DIVIDE BOTH 
SIDES BY NATURAL LOG 7,
SO THIS QUOTIENT WILL GIVE US 
THE APPROXIMATE VALUE OF X.
LET'S GO AHEAD 
AND GET OUR CALCULATORS OUT.
SO WE HAVE NATURAL LOG 5 DIVIDED 
BY NATURAL LOG 7.
SO IF WE ROUND 
TO THE HUNDREDTHS,
IT'LL BE APPROXIMATELY .83.
NOW FOR THIS PROBLEM, 
I'M GOING TO SHOW
ONE ALTERNATIVE METHOD 
THAT WE COULD'VE USED.
IF YOU LOOK AT THE EQUATION
IN THIS FORM 7 TO THE POWER 
OF X = 5,
IF WE REWROTE THIS 
AS A LOGARITHMIC EQUATION,
WE WOULD HAVE LOG, 
THE BASE IS 7,
THE EXPONENT IS X, 
SO THE LOG WOULD EQUAL X,
AND 5 WOULD BE THE NUMBER 
OR THE ARGUMENT OF THE LOG.
IN THIS FORM, WE COULD APPLY 
THE CHANGE OF BASE FORMULA
SO THAT X WOULD BE EQUAL 
TO THE NATURAL LOG OF 5
DIVIDED BY THE NATURAL LOG 
OF THE BASE 7.
AND WHAT YOU'LL NOTICE IS THAT 
THIS QUOTIENT HERE
IS THE SAME QUOTIENT WE USED 
TO SOLVE FOR X ON THE LEFT.
THIS TECHNIQUE ONLY WORKS WHEN 
YOU HAVE ONE EXPONENTIAL PART,
SO WE DIDN'T GO AHEAD 
AND SHOW ALL THE PROBLEMS
USING THE STEPS OUTLINED 
HERE ON THE LEFT.
ALSO IF WE WANTED TO CHECK THIS 
ON THE GRAPHING CALCULATOR,
WHAT WE COULD DO IS TYPE 
THE LEFT SIDE INTO Y1
AND THE RIGHT SIDE INTO Y2
AND LOOK FOR THE X COORDINATE 
OF THE POINT OF INTERSECTION.
LET'S GO AHEAD AND TRY IT 
ON THIS ONE.
SO Y1 WILL BE 7 TO THE POWER OF 
X - 1 AND THEN Y2 WILL BE 4.
PRESS GRAPH.
AND NOTICE HOW MY WINDOW 
IS NOT THE STANDARD WINDOW.
SO THE WAY TO GET IT BACK 
IS JUST TO PRESS ZOOM 6.
THERE'S THE EXPONENTIAL AND 
THERE'S THE HORIZONTAL LINE.
WE WANT THE X COORDINATE 
OF THIS POINT,
SO WE CAN PRESS 2nd TRACE, 
OPTION 5.
AND THEN WE CAN JUST PRESS ENTER 
THREE TIMES, ONE, TWO, THREE.
AND THIS VERIFIES OUR SOLUTION 
AS X IS APPROXIMATELY .83,
SO ALL THESE CAN BE CHECKED 
GRAPHICALLY AS WELL.
OKAY. ON NUMBER TWO, WE NEED 
TO ISOLATE THE EXPONENTIAL PART
OR 2 TO THE POWER OF X,
SO WE'LL FIRST ADD 2 
TO BOTH SIDES,
THAT WOULD GIVE US 3 x 2 
TO THE POWER OF X = 15.
AND NOW, 
WE'LL DIVIDE BOTH SIDES BY 3,
SO WE HAVE 2 TO THE POWER 
OF X = 5.
NOW THAT WE HAVE THE EXPONENTIAL 
PART ISOLATED,
WE'LL TAKE THE NATURAL LOG 
OF BOTH SIDES.
SO WE'LL HAVE THE NATURAL LOG
OF 2 TO THE POWER OF X = THE 
NATURAL LOG OF 5.
NOW, WE'LL APPLY THE POWER 
PROPERTY OF LOGS.
SO WE CAN TAKE THIS EXPONENT AND 
MULTIPLY IT BY NATURAL LOG 2,
SO WE HAVE X NATURAL LOG 
2 = NATURAL LOG 5,
AND NOW WE CAN DIVIDE 
BOTH SIDES BY NATURAL LOG 2.
AND SO WE'LL HAVE X 
IS APPROXIMATELY EQUAL
TO THIS QUOTIENT,
NATURAL LOG 5 
DIVIDED BY NATURAL LOG 2
IS APPROXIMATELY 2.32.
THIS DOES CHECK.
BUT IF WE WANTED TO, 
WE COULD CHECK IT GRAPHICALLY.
THE ONLY CONCERN I SEE HERE 
IS THAT Y2 WOULD EQUAL 13.
WE WOULD HAVE 
TO ADJUST THE WINDOW
TO SEE THAT HORIZONTAL LINE.
LET'S GO AHEAD 
AND DO ANOTHER EXAMPLE THOUGH.
SO ON THIS PROBLEM, WE DO HAVE 
TWO EXPONENTIAL PARTS,
SO WE HAVE ONE ON EACH SIDE.
SO FROM HERE, WE'LL TAKE 
THE NATURAL LOG OF BOTH SIDES.
THE NATURAL LOG 2/3 TO THE X = 
THE NATURAL LOG 5
TO THE POWER OF 3 - X.
NOW, WE'LL APPLY 
THE POWER PROPERTY
WHICH MEANS WE CAN TAKE THESE 
EXPONENTS,
MOVE THEM TO THE FRONT,
SO WE'LL HAVE X NATURAL LOG 2/3 
= 3 - X x NATURAL LOG 5.
NOW, WE'LL DISTRIBUTE 
THIS NATURAL LOG 5,
SO WE'LL HAVE 3 NATURAL LOG 
5 - X NATURAL LOG 5.
NOW, NOTICE WE HAVE TWO TERMS 
THAT INVOLVE X,
SO WE'RE GOING TO GO AHEAD 
AND ADD X NATURAL LOG 5
TO BOTH SIDES OF THE EQUATION.
SO IF WE DO THAT, 
WE'LL HAVE X NATURAL LOG 2/3,
WE MOVE THIS OVER, 
IT'S GOING TO BE + X
NATURAL LOG 5 = 3 NATURAL LOG 5.
NOW, WE CAN FACTOR OUT 
THE COMMON FACTOR OF X.
AND NOW, WE CAN DIVIDE 
BY THE SUM TO SOLVE FOR X.
SO TO DETERMINE OUR SOLUTION,
WE'LL HAVE TO EVALUATE THIS 
ON THE GRAPHING CALCULATOR.
WHEN WE ENTER THIS 
INTO THE CALCULATOR THOUGH,
WE'RE GOING TO HAVE A SET 
OF PARENTHESES
FOR THE NUMERATOR AND ANOTHER 
SET FOR THE DENOMINATOR
TO MAKE SURE WE EVALUATE 
THIS QUOTIENT CORRECTLY.
SO WE'LL HAVE IN THE NUMERATOR 
3 NATURAL LOG 5 DIVIDED BY,
AND THEN OUR DENOMINATOR WILL BE 
NATURAL LOG 2/3 + NATURAL LOG 5,
AND THEN ANOTHER CLOSED 
PARENTHESIS FOR OUR DENOMINATOR,
SO IT'S APPROXIMATELY 4.01.
LET'S GO AHEAD AND CHECK THIS 
GRAPHICALLY AS WELL.
FOR THE SAKE OF TIME,
I'VE ALREADY ENTERED 
IN THE LEFT SIDE IN Y1
AND THE RIGHT SIDE IN Y2.
LET'S GO AHEAD AND PRESS ZOOM 
6 FOR THE STANDARD WINDOW.
WE'RE LOOKING 
FOR THE X COORDINATE
OF THE POINT OF INTERSECTION, 
AND THERE IT IS.
WHAT WE CAN DO NOW IS JUST PRESS 
2nd TRACE, OPTION 5
AND THEN PRESS ENTER 
THREE TIMES,
AND YOU CAN SEE THAT DOES VERIFY 
OUR SOLUTION.
OKAY, LET'S SEE IF WE HAVE TIME 
FOR ONE MORE.
WE HAVE TWO EXPONENTIAL PARTS, 
WE HAVE ONE ON EACH SIDE,
SO WE CAN GO AHEAD 
AND TAKE THE LOG OF BOTH SIDES.
I'M GOING TO GO AHEAD AND TAKE 
THE NATURAL LOG AGAIN.
NOW, WE'LL APPLY 
THE POWER PROPERTY,
SO WE CAN TAKE THESE EXPONENTS 
AND MAKE THIS INTO A PRODUCT.
SO WE'LL HAVE X - 3 x NATURAL 
LOG 5 MUST EQUAL
2X + 1 x NATURAL LOG 3.
NOW, WE'LL GO AHEAD AND 
DISTRIBUTE THE NATURAL LOGS.
WELL X NATURAL LOG 5 - 3 
NATURAL LOG 5 = 2X
NATURAL LOG 3 + NATURAL LOG 3.
WE HAVE TWO X TERMS,
SO LET'S GO AHEAD AND PUT THE X 
TERMS ON THE SAME SIDE,
AND WE'LL PUT THE CONSTANTS 
ON THE OTHER SIDE.
SO WE'LL MOVE THE VARIABLE TERMS 
TO THE LEFT SIDE,
SO WE'LL HAVE X NATURAL LOG 5.
IF WE MOVE THIS TERM OVER,
IT'S GOING TO BE - 2X NATURAL 
LOG 3 EQUALS,
AND THEN IF WE MOVE THIS TERM TO 
THE RIGHT SIDE,
IT'LL BECOME 3 NATURAL LOG 5 
+ THE NATURAL LOG 3
THAT'S ALREADY THERE.
NOW, WE'LL FACTOR OUT THE X,
AND OUR LAST STEP IS TO DIVIDE 
BY THIS DIFFERENCE.
NOTICE ON THE RIGHT SIDE, 
NOW, WE JUST HAVE X,
AND IT'LL BE APPROXIMATELY EQUAL 
TO AGAIN THIS QUOTIENT.
LET'S GO BACK TO OUR CALCULATOR 
AND SEE IF WE CAN EVALUATE THIS.
OUR NUMERATOR IS GOING TO BE 3 
NATURAL LOG 5 + NATURAL LOG 3
DIVIDED BY THIS DIFFERENCE,
SO WE'LL HAVE NATURAL LOG 5 - 2 
NATURAL LOG 3, ENTER,
APPROXIMATELY -10.1.
THIS ONE WAS A LOT OF WORK,
SO LET'S GO AHEAD AND CHECK THIS
JUST BY SUBSTITUTING X 
INTO THE ORIGINAL EQUATION.
SO ON THE RIGHT SIDE, WE'D HAVE 
5 TO THE POWER OF -10.1 - 3.
AND ON THE RIGHT SIDE, WE'D HAVE 
3 TO THE POWER OF 2 x -10.1 + 1.
AND IT'S A LITTLE BIT OFF, 
BECAUSE WE DID ROUND.
BUT THAT LOOKS VERY CLOSE, SO 
THAT WOULD VERIFY OUR SOLUTION.
OKAY, I HOPE YOU FOUND THESE 
EXPLANATIONS HELPFUL.
THANK YOU FOR WATCHING.
