- WELCOME TO THE PRESENTATION ON LOGARITHMS.
GOALS OF THIS VIDEO ARE TO GRAPH A LOGARITHMIC FUNCTION
CONVERT BETWEEN LOGARITHMIC AND EXPONENTIAL EQUATIONS,
AND ALSO TO SOLVE EXPONENTIAL EQUATIONS USING LOGARITHMS.
NOW, THIS VIDEO DOES ASSUME THAT YOU HAVE BEEN EXPOSED
TO LOGARITHMS BEFORE,
AND IT DOES JUST SERVE AS A QUICK REVIEW.
A LOGARITHM IS DEFINED AS FOLLOWS.
Y IS THE POWER OF WHICH WE RAISE A TO GET X.
A IS CALLED A LOGARITHM BASE.
WE READ THIS AS THE LOG BASE A OF X IS = TO Y.
NOW, IT'S IMPORTANT THAT WE REALIZE
THE THREE MAJOR COMPONENTS OF A LOGARITHMIC EQUATION.
THIS LITTLE NUMBER DOWN HERE IS CALLED THE BASE.
SO WE'RE GOING TO REWRITE THIS IN EXPONENTIAL FORM.
A WILL BE THE BASE.
A LOGARITHM IS AN EXPONENT,
SO WHAT THE LOG IS EQUAL TO AND THAT BEING OUR EXPONENT.
THE VALUE OF X HERE IN THE LOG WILL BE THE NUMBER
OF WHAT ITS EQUAL TO IN EXPONENTIAL FORM.
LET'S GO AHEAD AND GIVE IT A TRY.
LET'S CONVERT THIS LOGARITHMIC EQUATION
TO AN EXPONENTIAL EQUATION.
WE HAVE TO BASICALLY ANSWER THESE THREE QUESTIONS.
WHAT'S THE BASE?
THE BASE IS 2.
SINCE A LOGARITHM IS AN EXPONENT, 3 IS OUR EXPONENT.
AND THEN NUMBER PART OF THE LOG IS = TO 8.
2 TO THE THIRD = 8 AND THAT'S CORRECT.
SO WE'RE GOING TO ASK THOSE SAME THREE QUESTIONS
OVER AND OVER AGAIN.
TAKING A LOOK AT THIS LOG EQUATION, THE BASE IS 3.
A LOG IS AN EXPONENT, SO OUR EXPONENT'S GOING TO BE 2,
AND IT'S = TO THE NUMBER PART, WHICH IS 1.
LET'S TRY IT IN REVERSE ORDER.
WE KNOW WE'RE GOING TO HAVE A LOG, SO LET'S WRITE THAT DOWN.
IT'S GOING TO BE EQUAL TO SOMETHING.
SO SAME THREE QUESTIONS, WHAT'S THE BASE.
THE BASE IS 6, ITS HERE.
THE EXPONENT IS 2, THEREFORE THE LOG MUST = 2.
AND THE NUMBER IS = TO 36.
LOG BASE 6 OF 36 IS = TO 2
BECAUSE 6 TO THE SECOND IS = TO 36.
LET'S TRY ONE MORE.
WE NEED A LOGARITHM.
NOW, IN THIS ONE THERE'S NO EXPONENT GIVEN,
SO I KNOW THIS IS E TO THE FIRST POWER.
SO OUR BASE IS E, THE EXPONENT IS = TO 1,
SO THE LOGARITHM IS = TO 1.
AND OUR NUMBER IS 2.718.
NOW, THIS WOULD BE APPROXIMATELY
BECAUSE WE DO HAVE AN APPROXIMATION OVER HERE.
NOW, ONE THING WE HAVE TO RECALL
IS THAT WE HAVE LOG BASE E, THAT'S CALLED NATURAL LOG,
AND WE DON'T USE THIS NOTATION,
WE USE LN FOR NATURAL LOG.
SO NATURAL LOG 2.718 IS APPROXIMATELY = TO 1.
BECAUSE E TO THE FIRST POWER IS APPROXIMATELY = TO 2.718.
OKAY, NOW LETS TAKE A LOOK AT THE GRAPH
OF A LOGARITHMIC FUNCTION.
THE FIRST THING WE HAVE TO DO TO GRAPH THIS
IS WRITE IT IN EXPONENTIAL FORM.
SO I'LL REWRITE THIS IN TERMS OF Y.
YOU HAVE Y = LOG BASE 2 OF X,
WHICH MEANS THAT 2 IS OUR BASE, Y WOULD BE OUR EXPONENT,
AND WOULD = X.
NOW, WHEN YOU'RE GRAPHING A LOGARITHMIC FUNCTION BY HAND,
NOTICE THAT Y IS THE EXPONENT,
SO IT'S GOING TO BE MORE CONVENIENT FOR US
TO CHOOSE VALUES OF Y
AND FIND THE CORRESPONDING VALUES OF X.
SO ESSENTIALLY WHAT WE'RE GOING TO DO
IS WE'RE GOING TO RAISE 2 TO THE POWER OF Y,
USE POWERS AND FIND THE VALUE OF X.
SO 2 TO THE FIRST POWER WOULD = 2.
2 TO THE SECOND POWER WOULD GIVE US AN X VALUE OF 4.
2 TO THE THIRD WOULD BE 8, 2 TO THE FOURTH WOULD BE 16,
2 TO THE POWER OF 0 WOULD GIVE US AN X VALUE OF 1.
REMEMBER 2 TO THE POWER OF -1
IS THE SAME AS 1/2 TO THE POWER OF +1,
WHICH OF COURSE IS = TO 1/2.
OKAY, LET'S PLOT THESE POINTS (2,1),(4,2),
REST OF THESE WILL BE OFF THE SCREEN,
BUT WE CAN PLOT THE POINT (1,0) AND THE POINT (1/2,-1),
AND OUR GRAPH WOULD LOOK SOMETHING LIKE THIS.
WE CAN SEE THAT THIS LOGARITHMIC FUNCTION
IS INCREASING OVER ITS DOMAIN.
ITS DOMAIN WOULD BE THE VALUES OF X GREATER THAN 0.
AND AT THIS POINT I THINK IT'S GOING TO BE HELPFUL FOR US
TO COMPARE THE GRAPH OF THIS LOGARITHMIC FUNCTION
WITH BASE 2 TO THE EXPONENTIAL FUNCTION OF BASE 2.
SO HERE WE HAVE IN BLUE OUR LOG FUNCTION,
AND IN BLACK WE HAVE OUR EXPONENTIAL FUNCTION.
AND I'VE ALSO GRAPHED THE LINE Y = X.
WHAT THIS SHOWS IS THEY'RE SYMMETRICAL
ACROSS THE LINE Y = X
AND THAT'S BECAUSE THEY'RE INVERSES OF ONE ANOTHER.
SO THERE'S A LOT GOING ON HERE,
BUT THE MAIN IDEA IS THEY'RE INVERSES OF ONE ANOTHER.
NOTICE HOW THIS HAS A Y INTERCEPT OF +1.
THE BLUE GRAPH HAS A X INTERCEPT OF +1,
AND THEY'RE SYMMETRICAL ACROSS THE LINE Y = X.
SO LOG FUNCTIONS AND EXPONENTIAL FUNCTIONS
ARE INVERSES OF ONE ANOTHER.
NOW, HERE'S A LIST OF A BUNCH OF PROPERTIES FOR LOGARITHMS.
I'M NOT GOING TO GO THROUGH EVERY SINGLE ONE OF THESE.
YOU MAY WANT TO PAUSE THE VIDEO
AND JUST REFRESH YOUR MEMORY.
THE MAJOR ONES HERE, WE HAVE THE PRODUCT PROPERTY,
THE QUOTIENT PROPERTY, AND THE POWER PROPERTY OF LOGS.
AND THE LAST ONE HERE WOULD BE OUR CHANGE OF BASE FORMULA.
LET'S GO AHEAD AND APPLY THESE
IN ORDER TO SOLVE EXPONENTIAL EQUATIONS.
IN GENERAL, THE BEST WAY TO SOLVE AN EXPONENTIAL EQUATION
IS ISOLATE THE EXPONENTIAL PART,
AND THEN TAKE THE LOG OF BOTH SIDES,
APPLY THE POWER PROPERTY, AND SOLVE FOR THE VARIABLE.
LET'S SEE HOW THIS WORKS.
IF WE WANT TO SOLVE THIS FOR T,
WE WANT TO REMOVE T FROM THE POSITION OF THE EXPONENT.
ONE WAY TO DO THAT WOULD BE TO TAKE THE NATURAL LOG
OF BOTH SIDES.
NOW, THE REASON I'M USING THE NATURAL LOG
IS BECAUSE THE NATURAL LOG IS ONE OF THE BASES
THAT'S ON THE GRAPHING CALCULATOR.
WE COULD ALSO USE COMMON LOG IF WE WANTED TO.
NOW, THE REASON WE TAKE THE LOG OF BOTH SIDES
IS NOW WE CAN APPLY THE POWER PROPERTY.
I CAN TAKE THIS VALUE OF T AND MOVE IT TO THE FRONT,
SO WE WOULD HAVE T x NATURAL LOG E IS = TO NATURAL LOG 40.
NOW, NATURAL LOG E IS ACTUALLY = TO 1 BECAUSE THE BASE
AND THE NUMBER ARE THE SAME.
BUT JUST IN CASE WE DIDN'T REMEMBER THAT,
WE WOULD STILL DIVIDE BY NATURAL LOG E,
ISOLATE THE T, SIMPLIFY.
SO I WOULD HAVE T IS = TO NATURAL LOG 40
DIVIDED BY NATURAL LOG E,
BUT OF COURSE, AGAIN, THIS IS = TO 1.
IF WE EVALUATE THE NATURAL LOG OF 40
WE'LL GET APPROXIMATELY 3.6889.
AND IT'S PRETTY COMMON TO GO TO FOUR DECIMAL PLACES.
OF COURSE, WE COULD EASILY CHECK THIS.
WHAT WE'RE SAYING IS,
IF WE TAKE E, RAISE IT TO THE POWER OF 3.6889,
WE SHOULD GET APPROXIMATELY 40, AND WE DO.
MEAN IT'S A LITTLE BIT OFF, OF COURSE,
BECAUSE THIS IS ROUNDING.
NOW, THE SECOND METHOD TO SOLVE THIS
WOULD BE TO REWRITE THIS AS A LOG EQUATION,
MEANING WE'LL ASK THOSE SAME THREE QUESTIONS AGAIN.
OUR BASE IS E, OUR EXPONENT IS T, AND OUR NUMBER IS 40.
NOW, REMEMBER LOG BASE E IS NATURAL LOG = TO T.
IT'S A LITTLE BIT FASTER.
THIS METHOD IS A LITTLE BIT FASTER IN THIS CASE,
BUT SOMETIMES IT MAY NOT BE.
I PERSONALLY LIKE TO STICK TO THE STANDARD PROCEDURE
OF PICKING A LOG OF BOTH SIDES AND SOLVING FOR THE VARIABLE.
LET'S TAKE A LOOK AT ANOTHER.
FIRST STEP, WE NEED TO ISOLATE THE EXPONENTIAL PART,
SO WE NEED TO GET RID OF THIS -4 AND ALSO THIS 2 HERE.
SO STEP ONE WILL BE TO ADD 4 TO BOTH SIDES.
SO THIS WILL BE = TO 9.
NOW, NEED TO DIVIDE BY 2, SO WOULD BE = TO 4.5.
AND, AGAIN, WHAT I'M GOING TO DO IS BE CONSISTENT,
AND JUST TAKE THE NATURAL LOG OF BOTH SIDES,
APPLY THAT POWER PROPERTY, AND TAKE THIS EXPONENT
AND MOVE IT TO THE POSITION OF THE COEFFICIENT.
BUT WE KNOW NATURAL LOG E IS = TO 1,
SO NOW WE JUST NEED TO DIVIDE BY -0.05.
SO THE VALUE T WOULD BE APPROXIMATELY -30.0815.
LET'S TAKE A LOOK AT NUMBER THREE.
AGAIN, THE FIRST STEP
IS GOING TO BE TO ISOLATE THE EXPONENTIAL PART.
THAT'S THIS PART HERE.
SO WE'RE GOING TO ADD 19 TO BOTH SIDES.
341 + 19 WOULD BE 360.
NOW WE'RE GOING TO DIVIDE BY 2.
LET'S GO AHEAD AND TAKE THE NATURAL LOG OF BOTH SIDES.
AGAIN, I COULD CHECK THE COMMON LOG,
BUT I'M IN THE HABIT OF TAKING THE NATURAL LOG,
APPLY THE POWER PROPERTY.
SO WE'D HAVE X + 1 x NATURAL LOG 3 IS = TO NATURAL LOG 180.
NOW, TO SOLVE FOR X
I'D PROBABLY DIVIDE BY NATURAL LOG 3,
GET THE DECIMAL APPROXIMATION OF THIS,
WHICH FOR THE SAKE OF TIME IS 4.7268.
SUBTRACT 1 FROM BOTH SIDES, WE'D HAVE 3.7268.
AGAIN, IF WE WANT, WE CAN GO BACK UP HERE AND CHECK IT,
BUT I ALREADY HAVE, AND THAT IS THE CORRECT VALUE OF X.
OKAY. I HOPE THIS QUICK OVERVIEW OF LOGS WAS HELPFUL.
THANK YOU FOR WATCHING.