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- WE'RE GIVEN THE LOGARITHMIC FUNCTION F OF X
= LOG BASE 5 OF X
AND ASKED A VARIETY OF QUESTIONS
ABOUT THE PROPERTIES AND CHARACTERISTICS
OF THE FUNCTION.
BEFORE WE ANSWER THE QUESTIONS, THOUGH,
LET'S GRAPH THIS FUNCTION
AND ALSO WRITE THE LOG EQUATION IN EXPONENTIAL FORM.
SO USING THE CHANGE OF BASE FORMULA GIVEN HERE
USING COMMON LOGS,
WE COULD WRITE F OF X AS F OF X = THE COMMON LOG OF X
DIVIDED BY THE COMMON LOG OF THE BASE 5.
WE'LL USE THIS TO GRAPH THE FUNCTION IN JUST A MOMENT.
TO WRITE THIS EQUATION IN EXPONENTIAL FORM,
LET'S REPLACE F OF X WITH Y
AND WRITE THIS AS Y = LOG BASE 5 OF X.
IN THIS FORM, WE SHOULD RECOGNIZE THE BASE IS 5,
THE EXPONENT IS Y, AND THE NUMBER IS X.
THEREFORE, THIS IS EQUIVALENT TO THE EXPONENTIAL EQUATION 5
RAISED TO THE POWER OF Y = X.
NOW, LET'S GO AHEAD AND GRAPH THIS LOG FUNCTION.
SO WE'LL PRESS Y=, TYPE IN COMMON LOG X,
CLOSE PARENTHESIS,
DIVIDED BY COMMON LOG 5, CLOSE PARENTHESIS.
I DID ADJUST THE WINDOW TO GET A BETTER VIEW.
THE X VALUES GO FROM -2 TO 10,
AND THEN Y VALUES GO FROM -10 TO 2.
LET'S GO AHEAD AND PRESS GRAPH.
HERE'S OUR LOGARITHMIC FUNCTION,
WHICH WE WILL USE TO ANSWER SOME OF THE QUESTIONS,
BUT I DID REPRODUCE THIS GRAPH ON SOME DIFFERENT SOFTWARE
SO WE CAN BETTER ANALYZE THE GRAPH.
SO I SHOULD BE USING THIS GRAPH HERE
TO ANALYZE THE LOG FUNCTION.
THE FIRST QUESTION ASKED FOR THE DOMAIN
OF THE LOGARITHMIC FUNCTION,
WHICH IS THE SET OF ALL POSSIBLE INPUTS
OR ALL POSSIBLE X VALUES.
TO DO THIS ANALYTICALLY, I THINK IT'S HELPFUL
TO LOOK AT THE EXPONENTIAL EQUATION HERE
WHILE THINKING ABOUT WHAT VALUES WE CAN GET FOR X
IF X = 5 RAISED TO THE POWER OF Y.
WHATEVER POWER WE RAISE 5 TO,
X IS NEVER GOING TO BE NEGATIVE.
IT'S ALSO NEVER GOING TO EQUAL 0,
AND THEREFORE, THE DOMAIN WOULD BE X > 0
OR THE OPEN INTERVAL FROM 0 TO 8 USING INTERVAL NOTATION.
IF WE LOOK AT THE GRAPH OF OUR FUNCTION,
NOTICE HOW IT NEVER = 0,
BECAUSE WE HAVE A VERTICAL ASYMPTOTE HERE.
NOTICE THE X VALUES ARE ONLY TO THE RIGHT OF X = 0,
VERIFYING THE DOMAIN IS THE OPEN INTERVAL FROM 0 TO 8.
NEXT, WE'RE ASKED ABOUT THE RANGE OF THE LOG FUNCTION.
AGAIN, LOOKING AT THE EXPONENTIAL FORM,
NOTICE HOW WE CAN RAISE 5 TO ANY REAL NUMBER,
AND THEREFORE, THE RANGE OF IT'D BE ALL REAL NUMBERS
OR THE OPEN INTERVAL FROM -8 TO +8.
GOING BACK TO THE GRAPH,
NOTICE HOW THE GRAPH GOES DOWN FOREVER,
AND EVEN THOUGH IT IS MOVING RIGHT VERY QUICKLY,
IT'S ALSO MOVING UP FOREVER WITH NO BREAKS,
VERIFYING OUR RANGE IS ALL REAL NUMBERS
OR THE INTERVAL FROM -8 TO +8.
AND NOW, TO FIND THE HORIZONTAL INTERCEPT,
WHICH IS THE SAME AS THE X INTERCEPT,
WE SET F OF X OR Y = 0 AND SOLVE FOR X.
TO DO THIS, WE'LL USE THE EXPONENTIAL FORM.
NOTICE HOW IF WE SET Y = 0,
WE WOULD HAVE THE EQUATION 5 TO THE 0 = X.
SO X = 5 TO THE 0, WHICH = 1.
SO IF THE X INTERCEPT OR HORIZONTAL INTERCEPT IS 1
OR THE POINT (1, 0),
WE CAN EASILY VERIFY THIS ON THE GRAPH.
HERE'S THE HORIZONTAL INTERCEPT,
WHICH IS THE POINT (1, 0).
NEXT, WE'RE ASKED TO FIND THE VERTICAL ASYMPTOTE,
AND WE BRIEFLY DISCUSSED THAT.
NOTICE HOW AS X APPROACHES 0 FROM THE RIGHT SIDE,
THE FUNCTION VALUES DECREASE WITHOUT BOUND AND APPROACH -8,
AND THEREFORE, WE HAVE A VERTICAL ASYMPTOTE AT X = 0.
NEXT, WE'RE ASKED TO FIND THE VERTICAL INTERCEPT,
WHICH WOULD BE THE SAME AS THE Y INTERCEPT.
SO WE SET X = 0 AND SOLVE FOR Y.
SO AGAIN, USING THE EXPONENTIAL FORM,
IF WE SET X = 0,
THIS WOULD GIVE US THE EQUATION 5
TO THE POWER OF Y = 0.
AND WE DISCUSSED EARLIER WE CAN'T RAISE 5 TO A POWER
AND GET A VALUE OF 0.
IT'S ALWAYS GOING TO BE POSITIVE,
AND THEREFORE, THIS DOES NOT HAVE A SOLUTION,
WHICH MEANS WE DO NOT HAVE A VERTICAL INTERCEPT
OR WE DO NOT HAVE A Y INTERCEPT.
SO FOR THE --, WE'LL TYPE IN DNE FOR DOES NOT EXIST.
NOW, WE'RE NOT DONE HERE. WE STILL HAVE MORE QUESTIONS.
I NOTICED THERE'S A TYPO HERE.
THIS SHOULD SAY, "WHAT IS THE LEFT-TO-RIGHT BEHAVIOR,
DESCRIBING IT AS INCREASING, DECREASING OR CONSTANT?"
SO GOING BACK TO THE GRAPH, NOTICE AS X INCREASES,
THE FUNCTION VALUES ALSO INCREASE.
SO THIS FUNCTION IS ALWAYS INCREASING,
BUT NOTICE AS X INCREASES, IT DOES NOT INCREASE AS FAST.
AND THEREFORE, WE CAN SAY THIS FUNCTION IS INCREASING,
BUT IS INCREASING AT A DECREASING RATE.
SO AGAIN, THIS ANSWER HERE IS IT'S INCREASING,
BUT IT'S INCREASING AT A DECREASING RATE.
NEXT, WE'RE ASKED TO DETERMINE THE VALUE
FOR WHICH THE FUNCTION IS > 0,
WHICH WOULD BE WHEN THE FUNCTION IS ABOVE THE X AXIS.
LOOKING AT OUR GRAPH, NOTICE HOW THE FUNCTION VALUES
OR Y VALUES ARE > 0
OR POSITIVE ABOVE THE X AXIS.
AND THEREFORE, WE CAN SAY THE FUNCTION IS POSITIVE
WHEN X > 1.
NOTICE HOW IT DOES ASK US TO USE INEQUALITY NOTATION,
NOT INTERVAL NOTATION.
SO WE'LL TYPE IN X > 1.
NEXT, WE'RE GOING TO FIND THE VALUES OF X
FOR WHICH F OF X < 0 OR NEGATIVE,
WHICH WOULD BE WHEN THE FUNCTION IS BELOW THE X AXIS.
WELL, IF WE JUST SAID THE FUNCTION WAS POSITIVE
OR ABOVE THE X AXIS ON THIS INTERVAL,
THAT MEANS IT WOULD BE NEGATIVE
OR < 0 WHEN IT'S BELOW THE X AXIS
OR OVER THIS INTERVAL HERE.
BUT REMEMBER, THE DOMAIN IS THE OPEN INTERVAL FROM 0 TO 8.
SO THIS WOULD BE WHEN X > 0 AND < 1.
SO AS A COMPOUND INEQUALITY, WE WOULD SAY X > 0 AND < 1.
NEXT, WE WANT TO FIND THE VALUE OF X
FOR WHICH F OF X = 0,
WHICH WOULD BE WHERE THE GRAPH CROSSES THE HORIZONTAL AXIS
OR X AXIS.
SO THE FUNCTION VALUE = 0
RIGHT HERE AT THE HORIZONTAL INTERCEPT OF X = +1.
AND NOW, FOR OUR LAST QUESTION,
WE WANT TO FIND THE VALUE OF X FOR WHICH F OF X = 1.
REMEMBER, F OF X IS THE SAME AS Y.
SO WE WANT TO FIND THE VALUE OF X WHEN Y = 1.
TO DO THIS, WE'LL USE THE EXPONENTIAL FORM
WHERE IF Y = LOG BASE 5 OF X,
THEN 5 TO THE POWER OF Y MUST = X.
SO 5 TO THE POWER OF Y = X.
WE CAN SUBSTITUTE 1 FOR Y.
5 TO THE 1 = X. SO X = 5.
WE CAN VERIFY THIS ON OUR GRAPH.
HERE'S WHERE X = 5 RIGHT ON THE EDGE,
AND NOTICE WHEN X = 5, THE FUNCTION VALUE IS +1.
SO THERE WAS A LOT OF GOOD INFORMATION
IN THIS ONE QUESTION.
I HOPE YOU FOUND THIS HELPFUL.