- WE WANT TO GRAPH THE 
EXPONENTIAL FUNCTION Y = 2
RAISED TO THE POWER OF X,
AND THE LOG FUNCTION Y = 
LOG BASE 2 OF X
ON THE SAME COORDINATE PLAIN.
THEN WE'LL DISCUSS 
THE RELATIONSHIP
BETWEEN THE TWO FUNCTIONS.
TO GRAPH THESE FUNCTIONS, 
WE'LL GRAPH THEM BY HAND
BY COMPLETING A TABLE 
OF VALUES.
SO FOR Y = 2 
RAISED TO THE POWER OF X,
NOTICE HOW I'VE ALREADY 
SELECTED X VALUES.
NOW WE'LL PERFORM SUBSTITUTION
TO DETERMINE 
THE CORRESPONDING Y VALUE.
SO WHEN X IS = TO 0, Y IS = 
TO 2 RAISED TO THE POWER OF 0.
ANYTHING RAISED TO THE POWER 
OF 0 IS = TO 1,
SO Y IS = TO 1.
WHEN X IS = TO 1, Y IS = TO 2 
TO THE FIRST OR 2.
WHEN X IS = TO 2,
Y IS = TO 2 TO THE SECOND 
POWER OF 2 SQUARE,
WHICH IS = TO 4.
WHEN IS 3, Y IS = TO 2 
TO THIRD, WHICH IS = TO 8.
AND NOTICE I ALSO SELECTED 
TWO NEGATIVE VALUES FOR X.
WHEN X IS = TO -1,
WE WOULD HAVE Y = 2 
RAISED TO THE POWER OF -1,
WHICH WOULD BE 1/2 
TO THE FIRST OR JUST 1/2.
WHEN X IS -2, 
Y IS = TO 2 TO THE -2,
WHICH IS = TO 1/2 SQUARED, 
WHICH WOULD BE 1/4.
LET'S GO AHEAD AND PLOT THESE 
POINTS ON A COORDINATE PLAIN
AND SKETCH OUR GRAPH.
AND WE'LL SKETCH THE 
EXPONENTIAL FUNCTION IN BLUE.
SO WE HAVE THE POINT (0,1), 
WHICH IS THE Y INTERCEPT.
WE HAVE THE POINT (1,2), POINT 
(2,4), POINT (3,8), (-1,1/2),
WHICH WOULD BE HERE,
AND (-2,1/4), 
WHICH WOULD BE HERE.
SO OUR FUNCTION LOOKS 
SOMETHING LIKE THIS.
NOTICE HOW WE HAVE 
EXPONENTIAL GROWTH
BECAUSE THE BASE IS GREATER 
THAN 1.
WE HAVE A HORIZONTAL ASYMPTOTE 
ALONG THE X AXIS OR Y = 0.
NOTICE THE DOMAIN, 
OR ALL POSSIBLE VALUES OF X,
WOULD BE ALL REAL NUMBERS.
BUT THE RANGE, 
OR THE Y VALUES,
WOULD BE Y IS ALWAYS GREATER 
THAN ZERO
BECAUSE THE FUNCTION 
IS ALWAYS ABOVE THE X AXIS.
NOW, TO SKETCH 
THE LOG FUNCTION,
WE'RE GOING TO REWRITE THIS 
IN EXPONENTIAL FORM.
REMEMBER Y = LOG BASE 2 OF X
IS EQUIVALENT TO 2 RAISED 
TO THE POWER OF Y MUST = X.
SO TO COMPLETE A TABLE 
OF VALUES FOR A LOG FUNCTION,
IT'S EASIER TO REWRITE IT 
AS AN EXPONENTIAL FUNCTION,
THEN INSTEAD OF SELECTING 
X VALUES WE SELECT Y VALUES,
AND THEN DETERMINE 
THE VALUE OF X.
BUT BEFORE WE DO THIS,
ON THIS EQUATION HERE 
LOOK VERY SIMILAR
EXCEPT THE X AND Y VARIABLES 
HAVE BEEN INTERCHANGED.
SO BECAUSE Y IS NOW 
THE EXPONENT,
WE'LL SELECT Y VALUES 
AND DETERMINE X.
NOTICE HOW THESE Y VALUES HERE 
ARE THE SAME
AS THE X VALUES WE HAD HERE 
FOR THE EXPONENTIAL FUNCTION.
SO WHEN Y IS = TO 0 WE HAVE 
2 TO THE 0 = X, SO X IS 1.
WHEN Y IS 1, 2 
TO THE FIRST = X, WHICH IS 2.
WHEN Y IS 2 SQUARED, 
X WOULD BE 4.
WHEN Y IS 3, 
2 TO THE THIRD = X, OR 8 = X.
WHEN Y = -1, WE'D HAVE 2 
TO THE -1 = X, WHICH IS 1/2.
AND WHEN Y IS -2, WE'D HAVE 
Y TO THE -2 = X, WHICH IS 1/4.
NOW, LET'S GO AHEAD AND SKETCH 
THE LOG FUNCTION IN RED.
SO WE HAVE THE POINT (1,0),
OR THE X INTERCEPT, 
(2,1), (4,2), AND (8,3).
AS WELL AS (1/2,-1), 
AND (1/4,-2).
SO OUR LOG FUNCTION 
LOOKS SOMETHING LIKE THIS.
NOW, THERE ARE SEVERAL THINGS 
TO NOTICE ABOUT THIS FUNCTION.
THE LOG FUNCTION 
HAS A VERTICAL ASYMPTOTE,
WHICH WOULD THE Y AXIS 
WITH A VERTICAL LINE X = 0.
NOTICE THE DOMAIN 
OF THIS FUNCTION
WOULD BE X GREATER THAN ZERO,
AND THE RANGE OF THIS FUNCTION 
WOULD ACTUALLY BE ALL REALS
BECAUSE THIS FUNCTION 
GOES DOWN FOREVER,
AS WELL AS GOES UP FOREVER.
NOW, IF WE COMPARE 
THESE TWO FUNCTIONS,
NOTICE HOW LOOKING 
AT THE TABLE OF VALUES,
NOTICE HOW THE X AND Y 
COORDINATES
HAVE BEEN INTERCHANGED,
AS WELL AS WHEN BOTH EQUATIONS 
ARE WRITTEN
IN EXPONENTIAL FORM.
NOTICE HOW THE EXPONENTIAL 
FUNCTION
HAS A HORIZONTAL ASYMPTOTE 
OF Y = 0,
THE LOG FUNCTION HAS 
A VERTICAL ASYMPTOTE OF X = 0,
AND NOTICE HOW THE DOMAIN 
IN RANGE
HAVE ALSO BEEN INTERCHANGED 
BETWEEN THE TWO FUNCTIONS.
SO WE SHOULD RECOGNIZE 
THAT THESE TWO FUNCTIONS
ARE INVERSES OF ONE ANOTHER.
ANOTHER WAY TO SHOW THAT THESE 
TWO FUNCTIONS ARE INVERSES
IS TO GRAPH THE LINE Y = X
AND SHOW THESE TWO FUNCTIONS 
ARE SYMMETRICAL
ACROSS THIS LINE.
WELL, THIS LINE 
HAS A Y INTERCEPT OF 0
AND A SLOPE OF 1,
SO IT WOULD PASS THROUGH 
THESE POINTS HERE,
AND LOOK SOMETHING LIKE THIS.
SO NOTICE IF WE FOLDED THESE 
TWO FUNCTIONS ACROSS THIS LINE
THEY WOULD MATCH IT PERFECTLY,
AGAIN, SHOWING THEY ARE 
INVERSES OF ONE ANOTHER.
OKAY. I HOPE YOU FOUND 
THIS VIDEO HELPFUL.
