- WE WANT TO SOLVE
THE EXPONENTIAL EQUATION 6
RAISED TO THE POWER OF X = 10.
NOTICE FOR THIS EQUATION
WE CANNOT GET A COMMON BASE
AND THEREFORE WE'LL HAVE
TO USE LOGARITHMS
TO SOLVE THIS EQUATION.
SO FOR AN EQUATION LIKE THIS
THE FIRST STEP IS TO ISOLATE
THE EXPONENTIAL PART
WHICH WOULD BE 6
RAISED TO THE POWER OF X.
SO NOTICE FOR THIS EQUATION
THE EXPONENTIAL PART
IS ALREADY ISOLATED
SO IN THIS FIRST CELL WE'LL
ENTER THE SAME EQUATION AGAIN.
6 RAISED TO THE POWER OF X
= 10.
WE CAN ENTER THIS TWO WAYS.
USING THE KEYBOARD
WE WOULD ENTER 6
AND THEN SHIFT 6 FOR THE CARAT
FOR THE EXPONENT
AND THEN X = 10
OR WE CAN CLICK IN
THE ANSWER CELL,
CLICK ON THE SMALL YELLOW
ARROW
AND USE THIS MATH PALETTE
USING THIS OPTION HERE
ENTERING THE EQUATION
AND THEN CLICK SAVE.
ONCE WE'VE ISOLATED
THE EXPONENTIAL PART
WE WANT TO WRITE
THE EXPONENTIAL EQUATION
AS A LOG EQUATION
USING OUR DEFINITION HERE
BY INDENTIFYING THE BASE,
THE EXPONENT, AND THE NUMBER.
NOTICE FOR OUR EXPONENTIAL
EQUATION THE BASE IS 6,
THE EXPONENT IS X,
AND THE NUMBER IS 10.
SO AS A LOG EQUATION,
WE KNOW WE'LL HAVE A LOG,
WE KNOW A LOGARITHM
IS AN EXPONENT
SO THE LOG IS GOING
TO EQUAL X, THE EXPONENT.
THE BASE IS 6
AND THE NUMBER IS 10.
NOTICE HOW IN THIS CASE
WHEN WE WROTE THE EXPONENTIAL
EQUATION IN LOG FORM
THE VARIABLE X
WAS ALREADY ISOLATED.
THIS ISN'T ALWAYS THE CASE
SO IF WE DIDN'T HAVE THIS X
HERE WE'D HAVE TO SOLVE FOR X
BUT BECAUSE OUR EQUATION
IS ALREADY SOLVED FOR X
THIS IS OUR EXACT SOLUTION SO
WE HAVE X = LOG BASE 6 OF 10.
AND AGAIN, THERE ARE TWO WAYS
TO ENTER THIS.
WE CAN ENTER X = LOG
UNDERSCORE 6 FOR THE BASE
AND THEN IN PARENTHESES 10
OR PROBABLY THE EASIER WAY
WOULD BE TO CLICK
IN THE ANSWER CELL,
CLICK ON THE YELLOW ARROW,
WHICH BRINGS UP
THIS MATH TOOL.
CLICK ON THE FUNCTIONS TAB
AND USE THIS OPTION HERE
AND CHANGE THE BASE TO 6,
ENTER THE EQUATION, CLICK SAVE
AND IT'S ALWAYS A GOOD IDEA
TO USE THE PREVIEW KEY
TO MAKE SURE YOU'VE ENTERED
EVERYTHING CORRECTLY.
NOW FOR THE LAST STEP WE WANT
TO GET A DECIMAL APPROXIMATION
FOR THE SOLUTION
OF 3 DECIMAL PLACES
SO WE'LL USE THE CHANGE
OF BASE FORMULA
ON THIS EQUATION HERE.
WHICH MEANS WE'LL TAKE
THE COMMON LOG OF THE NUMBER
AND DIVIDE BY THE COMMON LOG
OF THE BASE.
THIS WORKS FOR ANY BASE
BUT WE'LL USE EITHER
COMMON LOG OR NATURAL LOG
BECAUSE THESE ARE THE TWO
BASES ON THE CALCULATOR.
SO WE WOULD HAVE
X = COMMON LOG OF 10
DIVIDED BY COMMON LOG 6.
AND NOW, WE'LL GO
TO THE CALCULATOR.
SO WE HAVE COMMON LOG 10
DIVIDED BY COMMON LOG 6.
TO 3 DECIMAL PLACES, X WOULD
BE APPROXIMATELY 1.285.
EVEN THOUGH
THIS IS AN APPROXIMATION
FOR OUR HOMEWORK SYSTEM
WE WILL ENTER X = 1.285.
AND NOW, LET'S GO AHEAD AND
CHECK THIS SOLUTION TWO WAYS.
ONE WAY WOULD BE
TO SUB THIS VALUE HERE
BACK INTO THE ORIGINAL
EQUATION
TO MAKE SURE THE LEFT SIDE
IS APPROXIMATELY 10.
I SAY APPROXIMATELY
BECAUSE WE DID ROUND.
SO 6 RAISED TO THE POWER
OF 1.285
NOTICE HOW THIS IS
APPROXIMATELY 10.
IT'S A LITTLE BIT LESS
BECAUSE WE DID ROUND DOWN
TO GET THIS VALUE HERE.
THE LAST WAY TO CHECK IS WOULD
BE TO CHECK IT GRAPHICALLY
SO WE'LL ENTER Y1 = 6
TO THE POWER OF X, Y2 = 10,
AND IDENTIFY THE POINT
OF INTERSECTION.
SO WE'LL PRESS Y = AND THEN 6
RAISED TO THE POWER OF X
AND Y1 AND 10 IN Y2.
WE COULD TRY USING
THE STANDARD WINDOW
BUT WE'LL PROBABLY HAVE
TO ADJUST THAT.
LET'S TRY PRESSING ZOOM 6
FOR THE STANDARD WINDOW.
THERE'S THE EXPONENTIAL
AND NOTICE Y = 10
IS RIGHT ON THE EDGE.
SO LET'S INCREASE
THE Y MAXIMUM
AND WE CAN ALSO INCREASE
THE X MIN
AND DECREASE THE X MAX AS WELL
IF WE WANT.
LET'S PRESS WINDOW,
LETS CHANGE THE X VALUES
FROM -5 TO 5
AND LET'S INCREASE
THE Y MAXIMUM TO LET'S SAY 15.
AND NOW WE'LL PRESS GRAPH.
THE X COORDINATE OF THIS POINT
OF INTERSECTION
SHOULD BE APPROXIMATELY 1.285
IF OUR ANSWER IS CORRECT.
SO WE'LL PRESS 2nd TRACE
FOR THE CALCULATION MENU
AND WE WANT OPTION 5
FOR INTERSECTION.
SO PRESS 5.
BECAUSE WE ONLY HAVE 2 CURVES
OR 2 GRAPHS
ON THE COORDINATE PLANE WE
CAN JUST PRESS ENTER 3 TIMES.
SO ENTER, ENTER, ENTER,
AND NOTICE
HOW THIS DOES VERIFY
THAT OUR SOLUTION IS CORRECT.
X IS APPROXIMATELY 1.285.
I HOPE YOU FOUND THIS HELPFUL.
