- WE WANT TO MATCH THE GRAPHS
TO THE CORRECT EQUATIONS GIVEN HERE.
NOTICE THE FIRST AND LAST FUNCTIONS
ARE EXPONENTIAL FUNCTIONS
AND THE SECOND AND THIRD FUNCTIONS ARE LOG FUNCTIONS.
SO HOPEFULLY, BY NOW WE CAN RECOGNIZE
THAT THE BLUE AND THE RED FUNCTIONS
REPRESENT EXPONENTIAL GROWTH FUNCTIONS
AND THE BLACK AND GREEN FUNCTIONS
REPRESENT THE GRAPHS OF LOG FUNCTIONS.
SO THERE ARE SEVERAL WAYS TO IDENTIFY THE CORRECT EQUATIONS
FOR THE GRAPHS
BUT I THINK WHEN WORKING WITH LOG FUNCTIONS
IT'S OFTEN HELPFUL TO REWRITE THESE LOG FUNCTIONS
IN EXPONENTIAL FORM.
Y = LOG BASE 3 OF X
MEANS THAT 3 RAISED TO THE POWER OF Y MUST EQUAL X.
SO 3 TO THE POWER OF Y MUST EQUAL X
AND Y = LOG BASE 2 OF X
MEANS 2 RAISED TO THE POWER OF Y = X.
SO 2 TO THE POWER OF Y = X.
IN THIS FORM, WE SHOULD RECOGNIZE
THAT THIS FIRST EQUATION Y = 2 TO THE POWER OF X
AND THE THIRD EQUATION X = TO THE POWER OF Y,
ARE INVERSES OF ONE ANOTHER.
NOTICE HOW THE X AND Y VARIABLES
HAVE BEEN INTERCHANGED
AND THE SAME IS TRUE
FOR THE SECOND EQUATION AND THE FOURTH EQUATION.
NOTICE HOW THE X AND Y VARIABLES
HAVE BEEN INTERCHANGED.
SO KNOWING THE EXPONENTIAL FUNCTIONS
AND LOG FUNCTIONS ARE INVERSES OF ONE ANOTHER
IT SHOULD MAKE IT EASIER TO IDENTIFY OUR GRAPHS.
BUT LET'S START BY COMPARING THE TWO EXPONENTIAL FUNCTIONS.
WE HAVE Y = 2 RAISED TO THE POWER OF X
AND Y = 3 RAISED TO THE POWER OF X.
NOTICE HOW Y = 3 RAISED TO THE POWER OF X
HAS A BASE EQUAL TO 3
AND Y = 2 RAISED TO THE POWER OF X HAS A BASE OF 2.
SINCE 3 IS GREATER THAN 2, Y = 3 TO THE POWER OF X
REPRESENTS LARGER EXPONENTIAL GROWTH
AND THEREFORE THE GRAPH OF THIS EXPONENTIAL FUNCTION
SHOULD BE INCREASING FASTER
THAN THE GRAPH OF THIS FUNCTION HERE.
LOOKING AT THE RED AND BLUE FUNCTIONS
NOTICE HOW THE RED FUNCTION IS INCREASING FASTER
BECAUSE IT'S GOING UPHILL FASTER.
THEREFORE, Y = 3 RAISED TO THE POWER OF X
SHOULD BE THE RED GRAPH
AND Y = 2 RAISED TO THE POWER OF X SHOULD BE THE BLUE GRAPH.
OF COURSE, IF WE'RE STILL NOT CONVINCED
WE COULD FIND POINTS ON EACH FUNCTION,
SO FOR EXAMPLE, FOR Y = 2 RAISED TO THE POWER OF X
IF X = 1 NOTICE THAT Y = 2
AND FOR Y = 3 RAISED TO THE POWER OF X,
IF X = 1 ALL WOULD BE EQUAL TO 3.
NOTICE HOW THE RED FUNCTION DOES CONTAIN THE POINT (1, 3)
AND THE BLUE FUNCTION DOES CONTAIN THE POINT (1, 2).
SO WE HAVE IDENTIFIED THE EXPONENTIAL FUNCTIONS
CORRECTLY.
AND NOW IF WE RECOGNIZE THAT Y = LOG BASE 3 OF X
AND Y = 3 TO THE POWER OF X
ARE INVERSE FUNCTIONS OF ONE ANOTHER
IF THE EXPONENTIAL FUNCTION CONTAINS THE POINT (1, 3)
THEN THE INVERSE FUNCTION WOULD HAVE TO CONTAIN
THE POINT (3,1)
WHICH WOULD BE THIS POINT HERE ON THE BLACK FUNCTION.
NOTICE IN EXPONENTIAL FORM IF Y = 1, X WOULD BE 3.
AND SINCE Y = 2 TO THE X AND Y = LOG BASE 2 OF X ARE INVERSES
IF THE EXPONENTIAL CONTAINS THE POINT (1, 2)
THEN THE LOG FUNCTION MUST CONTAIN THE POINT (2,1)
WHICH WE CAN SEE HERE WOULD BE ON THE GREEN FUNCTION
AND THEREFORE, Y = LOG BASE 2 OF X WOULD BE THE GREEN GRAPH.
AGAIN NOTICE IF Y = 1, X WOULD EQUAL 2.
SO AS I MENTIONED IN THE BEGINNING
THERE ARE SEVERAL WAYS
TO IDENTIFY THESE GRAPHS CORRECTLY
BASED UPON THESE EQUATIONS
BUT IT IS HELPFUL TO RECOGNIZE
WHEN EXPONENTIALS AND LOGS ARE INVERSES OF ONE ANOTHER.
I HOPE YOU FOUND THIS HELPFUL.