1 MATH 16A LECTURE. OCTOBER 28, 2008. PROFESSOR: SO LET ME START WITH SOMETHING I'M SURE YOU ALL WANT TO HEAR ABOUT WHICH IS THE MIDTERM. THE NEXT MIDTERM. IT'S COMING UP, NOT THIS WEEK BUT THE NEXT WEEK. IT WILL COVER, SECTIONS, WELL ANYTHING IS FAIR GAME. IF I TOLD YOU HOW TO DIFFERENTIATE SOMETHING AT THE VERY BEGINNING THE COURSE I'LL ADD THAT. BUT IT'S BASICALLY GOING TO COVER STUFF THAT WE DID SINCE THE LAST MIDTERM. SECTIONS 1.7 TO 4.3. THERE ARE SAMPLE MIDTERM POSTED ALREADY. IN CLASS AS USUAL. SAME RULES. ALL LIKE YOU HAD LAST TIME. ANY QUESTIONS ABOUT THAT? STUDENT: DID YOU SAY WE MIGHT HAVE QUESTIONS FROM THE BEGINNING OF CLASS. PROFESSOR: IF I TAUGHT YOU SOMETHING, BASIC RULES OF ALGEBRA OR CALCULUS THIS SHOWED UP IN SECTIONS UP TO 1.6 I'M ALLOWED TO ASK YOU QUESTIONS USING THEM. BUT PARTICULAR QUESTIONS AS YOU'LL SEE FROM THE SAMPLE MIDTERM COME FROM THOSE SECTIONS. STUDENT: DID THE GRADE BREAKDOWN CHANGE FROM THE FIRST MIDTERM. PROFESSOR: THE TARGET IS STILL THE CURVE THAT I HAD ON WEB PAGES UNTIL WE FIND OUT HOW PEOPLE DO. I DON'T KNOW IF I NEED TO RECAP LATE. SO THE FIRST QUESTION, QUESTION FIVE I EXPERIMENTED WITH. MAYBE A LITTLE HARDER THAN LAST TIME. -- SO LET ME REVIEW A TINY BIT. SO I CAN KEEP GOING. WE LOOKED AT EXPONENTIAL FUNCTIONS. AND I THINK I TOLD YOU THAT IF YOU PLOT THEM THEY ALWAYS LOOK THE SAME. AS LONG AS THEY DON'T BOTHER TO LABEL THE AXIS, THIS IS X, THIS IS FUNCTION B-TO THE X-WILL BE SOME FUNCTION BIGGER THAN 2
ONE. AND I'M PARTICULARLY INTERESTED IN THE SLOPE OF THE TANGENT LINE RIGHT THERE BECAUSE THEY ALWAYS GO THROUGH.0 COMMA ONE. THAT'S IF YOU TAKE -- ALWAYS GO THROUGH THAT POINT. THIS WAS SORT OF GENERIC PICTURE OF WHAT ANY EXPONENTIAL FUNCTION LOOKED LIKE. LAST TIME I SHOWED THAT IF YOU TAKE THIS EXPONTENTIAL FUNCTION, AND YOU DIFFERENTIATE IT, THERE WAS ONE FORMULA, AND IT WAS VERY EASY, YOU GOT THE SAME FUNCTION BACK MULTIPLIED BY THE SLOPE OF THAT LINE. THAT WAS GEOMETRICALLY WHAT WE DID LAST TIME. WE DIFFERENTIATE. GET THE SAME FUNCTION BACK. POSSIBLY MULTIPLIED BY ONE NUMBER, WHICH IS THAT SLOPE. THEN WE CHOSE ONE MAGIC NUMBER, E, WHICH GOES ON FOREVER AT SOME IRRATIONAL NUMBER. TO MAKE THE SLOPE ONE. SO FOR A PARTICULAR VALUE OF E, WE GOT THAT D-D-X-TO THE E-X-WAS E-TO THE X-. IT CAME BACK BECAUSE THE SLOPE AT THIS POINT WAS EQUAL TO ONE. OKAY. AND THAT SUMMARIZES A WHOLE LOT OF WHAT WE DID LAST TIME. THAT'S THE PICTURE FOR B-TO THE X-WHEN B-IS GREATER THAN ONE. IF YOU KEEP MULTIPLIED BY B IT KEEPS GET BEING BIGGER. WE ALSO DREW THIS PICTURE. AND THIS IS B-TO THE X-WHEN B-WAS LESS THAN ONE. BECAUSE IF YOU KEEP MULTIPLYING NUMBER LESS THAN ONE BY ITSELF IT KEEPS GETTING SMALLER. THIS IS THE SAME PICTURE BACKWARDS. IT STILL GOES THROUGH ZERO COMMA ONE. THAT WAS A SUMMARY. AND NOW I WANT TO SPEND MORE TIME IF WE BELIEVE ALL THIS TO HOW TO DIFFERENTIATE FUNCTIONS THAT LOOK LIKE E-TO THE X-THAT ARE VARIATIONS ON E-TO THE X. USE ONE MAGIC NUMBER 2.718... SO LET'S GO ON TO SECTION 4.3. WHICH WE STARTED QUICKLY LAST TIME. SO RULES, ALL BASED ON 3 THE CHAIN RULE FOR DIFFERENTIATING FUNCTIONS THAT LOOK LIKE E-TO THE X. SO LET'S START BY USING THE CHAIN RULE TO DIFFERENTIATE
E-TO THE K-TIMES X-WHERE K-IS GOING TO BE A CONSTANT. I THINK I DID THIS LAST TIME. BUT LET ME DO IT AGAIN. TO USE THE CHAIN RULE I'M GOING TO WRITE THIS AS F-OF G-OF X-WHERE F-OF X, THE FIRST THING I DO IS TAKE X-AND MULTIPLY IT BY K. THAT'S THE FUNCTION G AND THEN I TAKE THAT AND I EXPONENTIATE IT. SO IF I DO F-OF G-OF X-I GET E-TO THE K-X. IF I WANT TO USE THE CHAIN RULE, D-D-X-E-TO THE K-X- I KNOW THE CHAIN RULE TELLS ME THIS. (ON BOARD). I BETTER FIGURE OUT WHAT G-PRIME AND F-PRIME ARE. F-PRIME IS MEANT TO BE PARTICULARLY EASY. THAT FUNCTION DOESN'T CHANGE WHEN DIFFERENTIATE IT, THAT'S WHY WE CHOSE E-THE WAY WE DID. G-PRIME IS JUST A CONSTANT TIMES X. SO IF I PLUG THAT ALL IN I GET THAT THIS IS JUST E-TO THE K-X-TIMES K. (ON BOARD). THERE'S K. AND THERE'S E-TO THE K-X. SO DIFFERENTIATING A FUNCTION WHERE YOU TOOK A CONSTANT UP THERE, YOU BRING IT DOWN TO THE FRONT. TO SUMMARIZE, (ON BOARD). JUST BRING THE CONSTANTS DOWN TO THE FRONT. AWFULLY EASY FUNCTION TO DIFFERENTIATE. AND SO HERE, LET ME DO ONE EXAMPLE WE DID LAST TIME. D-D-X-TO E-TO THE MINUS X-THAT APPLIES HERE BECAUSE I HAVE TO THINK THAT MINUS X-IS, OF COURSE, NEGATIVE ONE TIMES X. AND APPLY THE RULE AND YOU GET NEGATIVE ONE TIMES E-TO THE MINUS X. SO THERE'S THE EXAMPLE. AND IT'S NOT MUCH FARTHER TO WRITE DOWN THE GENERAL CHAIN RULE FOR TAKING D-D-X-THE DERIVATIVE OF E-TO THE ANY OLD FUNCTION G-OF X-. WE'RE GOING TO BE DOING THAT. 4 THAT TELLS US IT'S THE SAME RULE AS UP THERE. F-PRIME OF G-OF X. THERE'S THE F-PRIME OF G-OF X-TIMES G-PRIME. SO DIFFERENTIATE A FUNCTION E-TO ANYTHING, YOU'LL LEAVE IT ALONE AND COPY IT, BRING DOWN THE EXPONENTIAL AND DIFFERENTIATE IT. AND THAT'S JUST THE
CHAIN RULE TOO. IT'S EXACTLY THE SAME CHAIN RULE THAT'S UP THERE. HERE'S F-OF G-OF X-(ON BOARD). I LEFT G-ALL ALONE. STUDENT: SO WHAT EQUATION IS F OF X- PROFESSOR: F-OF X-IS A FUNCTION. AND I CHOSEN IT TO BE THE SAME AS IT WAS UP THERE. F-OF X-IS JUST E-TO THE X. IT'S FOR THIS PARTICULAR EXAMPLE. AND SO ITS DERIVATIVE IS UNCHANGED. WORLD'S SIMPLEST FUNCTION TO DIFFERENTIATE. ANY FUNCTION YOU CARE TO N. G-OF X-. USE THIS WHY G-IS SOME OTHER FUNCTION. LET ME ILLUSTRATE. E-TO THE POWER X-SQUARED PLUS ONE. SO G-OF X-HERE'S G-OF X. AND SO THIS IS GOING TO BE JUST COPY IT, THAT PART DOESN'T CHANGE. AND THEN I JUST HAVE TO MULTIPLY BY THE DERIVATIVE OF X-SQUARED PLUS ONE. AND MULTIPLY BY TWO X. SO THAT'S WHAT THE RULE TELLS US. LET ME DO ANOTHER ONE LIKE THAT. (ON BOARD). THERE'S THE BIG EXPONENTS. AND TO DIFFERENTIATE IT WE COPY THAT WITHOUT ANY CHANGE. THAT PART OF THE FUNCTION STAYS THE SAME AND THEN I HAVE TO MULTIPLY BY THE DERIVATIVE OF WHATEVER'S UP IN THE EXPONENT AND NOW WE USE WHAT WE KNOW ABOUT THIS, THAT'S GOING TO BE 27 X-CUBED MINUS ONE OVER X-SQUARED, PLUS ONE OVER X-SQUARED (ON BOARD). SO G-OF X-WAS THIS GUY UP HERE. SO THERE'S, THAT WAS G OF X. SO LET ME JUST WRITE DOWN THE CHAIN RULE AGAIN, JUST BECAUSE USE TWO DIFFERENT NOTATIONS 5 FOR IT IN THE PAST. STUDENT: WHY IS IT 27. PROFESSOR: DID I MULTIPLY IT BY -- GOOD FOR CATCHING THAT. THANK YOU. (ON BOARD). MAKING SURE YOU WERE AWAKE. THANK YOU. (INAUDIBLE QUESTION FROM STUDENT.) CAN EVERYONE DIFFERENTIATE THREE TIME X-CUBED NOW, GOOD, I HOPE, INCLUDING ME. LET ME
WRITE DOWN THE CHAIN RULE AGAIN USING SLIGHTLY DIFFERENT NOTATION. IT'S THE SAME THING. I WANT TO ILLUSTRATE WE WRITE IT DOWN BECAUSE WE'VE USEDS THIS NOTATION BEFORE, IF U-IS A FUNCTION OF X, TAKE D-D-X-OF E-TO THE U, AND IT'S JUST GOING TO BE E-TO THE U-TIMES D-U-D-X-. SAME RULE JUST SLIGHTLY DIFFERENT NOTATION USING U-TO REPRESENT THAT FUNCTION. THE BOOK'S USED IT BEFORE. I JUST THOUGHT I'D WRITE IT DOWN AGAIN, NO NEW IDEAS. EVERYBODY KNOWS ABOUT THE MIDTERM, SO I CAN ERASE THAT? SINCE WE'VE SHOWN THAT D-D-X-E-TO THE X-Q-(ON BOARD) THERE'S ANOTHER WAY TO SAY THAT. E-TO THE K-X, THAT FUNCTION SATISFIES SOMETHING CALLED A DIFFERENTIAL EQUATION. (ON BOARD). NEW IDEA. AN EQUATION WHERE THE UNKNOWN YOU WANT TO SOLVE FOR IS A FUNCTION. SO THIS IS AN EQUATION. IT'S SATISFIED BY -- STUDENT: SO IS THE DERIVATIVE OF E-TO THE K-X-E-TO -- STILL NEED (INAUDIBLE). PROFESSOR: WHERE DID I, THANK YOU FOR PAYING ATTENTION. THANK YOU FOR CATCHING THAT TYPO. I SHOULD SLOW DOWN. ANY YES. OTHER TYPOS. I LEFT THE K-OUT OF THE EXPONENTS. WHENEVER YOU DIFFERENTIATE YOU ALWAYS COPY IT OVER. AND THEN YOU PULL OUT THE 6 DERIVATIVE. THANK YOU. SO HERE, I'VE WRITTEN DOWN AN EQUATION. IT'S SATISFIED BY SOME FUNCTION, Y-OF X-. WE DON'T KNOW THE FUNCTION Y. SO HERE'S AN EQUATION TO SOLVE NOT FOR UNKNOWN AVAILABLE. WE DID THAT ALL THE TIME BUT FOR AN UNKNOWN FUNCTION. Y-OF X. SO I'M TELLING YOU THAT THIS AND WRITING DOWN THIS DIFFERENTIAL EQUATION BECAUSE IT COMES UP ALL THE TIME WHEN YOU WRITE DOWN EQUATIONS TO UNDERSTAND HOW THE WORLD WORKS AND THE ANSWER IS RIGHT THERE. THE ANSWER IS Y-OF X-IS E-TO THE
K-TIMES X-TIMES ANY CONSTANTS. WHEN I SAY WHAT IS ANY POSSIBLE FUNCTION THAT SATISFIES THAT EQUATION, THE ANSWER IS E-TO THE K-X-BUT YOU GET TO MULTIPLY BY ANY CONSTANT YOU LIKE AND IT STILL WORKS. LET ME TRY THAT. LET'S VERIFY THAT. LET ME VERIFY. TAKE D-D-X-TIMES SOME CONSTANT E-TO THE K-X. I CAN FACTOR CONSTANT IF I TAKE THE DERIVATIVE. SO I HAVE TO DIFFERENTIATE THAT PART. THAT'S HOW CONSTANTS WORK. NOW I GET CONSTANT TIMES K-E-TO THE K-X. IF I JUST REVERSE K-AND C, I CAN MULTIPLY NUMBERS IN ANY ORDER I LIKE, IT'S THE SAME THING, I SEE I GET K-TIMES Y-OF X. AND THAT IS WHAT I WANTED TO SHOW. I WANTED TO SHOW D-D-X-OF Y-OF X-IS THIS CONSTANT K-TIMES Y-OF X. SO IT WORKS. SO THAT'S THE FUNCTION. SO FOR EXAMPLE, I CAN TAKE ANY OLD 37 E-TO THE K-X-SATISFIES THE EQUATION IN THE BOX. SO HOW DO WE PICK C? YOU NEED SOME, YOUR PROBLEM YOU'RE TRYING TO SOLVE HAS TO HAVE SOME MORE INFORMATION. SO FOR EXAMPLE, FIND A SOLUTION TO THAT SAME EQUATION, I'LL WRITE IT DOWN AGAIN, TO D-D-X-Y-OF X-EQUALS K-TIMES Y-OF X-SATISFYING, I CAN GIVE YOU 7 MORE INFORMATION, Y-OF ZERO EQUALS FOUR. OKAY. THERE'S A LITTLE BIT OF EXTRA INFORMATION. IN ADDITION TO THAT AND I WANT TO SOLVE THIS BUT I'LL TELLING YOU THE FUNCTION, THE ANSWER HAS TO GO THROUGH THAT PARTICULAR POINT. SO HOW DO WE DO THAT? WE KNOW THAT Y-OF X-HAS TO EQUAL SOME CONSTANTS. TIMES E-TO THE K-X-FOR SOME CONSTANTS C-WHICH WE DON'T KNOW YET. HOW DO I SOLVE FOR IT? PLUG IN X-EQUALS ZERO. I GET THAT Y-OF ZERO WHICH HAD BETTER EQUAL FOUR, IS GOING TO BE THIS CONSTANT TIMES E-TO THE K-TIMES ZERO, AND SO, AND THAT'S CONSTANT E-TO THE ZERO, AND E-TO THE ZERO IS? ONE. SO IT'S C. THAT HAS TO EQUAL, THEREFORE, FOUR
BECAUSE THAT'S THE EXTRA INFORMATION I GAVE IN THE PROBLEM. SO THE ANSWER IS Y-OF X-IS FOUR TIME E-TO THE K-X. IF YOU GIVE ME ONE POINT ON THE CURVE I CAN FIGURE OUT C- STUDENT: WHAT ABOUT K- PROFESSOR: SO K-WAS GIVEN TO ME AS PART OF PROBLEM. IF THAT HAD BEEN SEVEN, I JUST DO IT FOR A GENERAL K. SO FOR THE PROBLEM TO WORK YOU HAVE TO HAVE POINT ON THE CURVE, K, WHICH IS CONSTANT. STUDENT: WHAT IF X-WASN'T THERE. PROFESSOR: LET'S DO THAT. LET'S SEE, LET'S SEE HOW THAT WORKS. I MIGHT NEED SOMETHING FROM SECTION 4.4. LET'S SOLVE THE SAME THING. D-D-X-Y-OF X-EQUALS, I THINK CONSTANT MAYBE SPECIFIC SEVEN TIMES Y-OF X. AND Y-OF, SAY ONE EQUALS FOUR JUST TO BE SPECIFIC. SHOULD I TRY THAT ONE? THAT I KNOW THAT Y-OF X-IS GOING TO BE SOME CONSTANTS TIMES E-TO THE POWER SEVEN X. I KNOW THAT FOR SURE. WHAT I HAVE TO PICK IS A CONSTANT C. SO I PLUG IN AND GET 8 Y-OF ONE, THAT BETTER BE FOUR, AND THAT'S C-TIMES E-TO THE SEVEN TIMES ONE. I THINK YOU CAN SOLVE THAT FOR C. SO SOLVE THAT FOR C. AND WHAT DO I GET? I GET THAT C-IS GOING TO BE FOUR DIVIDED BY E-TO THE SEVEN TIMES ONE OR FOUR DIVIDED BY E-TO THE SEVENTH POWER. SOME NUMBER WE CAN COMPUTE. SO THERE'S HOW I WOULD SOLVE THAT ONE. WE CAN SOLVE THOSE TOTALLY GENERALLY. THEY COME UP A LOT IN PRACTICE. STUDENT: I DON'T SEE WHERE YOU GET IN BOTH OF THOSE THE POWER OF FOUR. WHERE DID THE FOUR COME? PROFESSOR: THIS IS JUST EXTRA INFORMATION THAT WHEREVER THE PROBLEM COMES FROM YOU NEED TO KNOW THIS. THIS SAYS WHEN YOU DRAW THE CURVE BEING SOMEBODY HAS TO GIVE YOU A POINT ON IT, AND
I'M TELLING YOU IT HAS TO GO THROUGH THE POINT ONE WHICH IS FOUR. THE STATEMENT OF THE PROBLEM HAS TO TELL YOU THAT OTHERWISE YOU CAN'T DO IT. YOU HAVE TO KNOW ONE POINT ON THE CURVE AND THEN YOU CAN FIGURE OUT THIS CONSTANT. SO LET ME, OKAY, SO I'VE SAID ALL THAT. WHAT I WANT TO DO NOW, IT'S JUST A COUPLE OF LINES, IS TELL YOU WHY THAT'S THE ONLY ANSWER. SO THAT THE SOLUTION IS UNIQUE. THERE'S ONLY ONE ANSWER. AND IT'S GOING TO BE A CONSTANT TIMES E-TO THE K-X. LET ME JUST, SO WHY AM I DOING THIS? NOT EVERY EQUATION THAT YOU CAN WRITE DOWN HAS ONE SOLUTION. FOR EXAMPLE, HOW ABOUT X-SQUARED EQUALS TWO. THERE'S AN EQUATION WHERE YOU DON'T KNOW THE X-. THERE'S TWO SOLUTIONS, PLUS THE SQUARE ROOT OF TWO OR MINUS THE SQUARE ROOT OF TWO. BUT THAT PARTICULAR EQUATION I'M GOING TO PROVE THAT YOU NOW THERE'S 9 EXACTLY ONE SOLUTION. YOU DON'T HAVE TO WORRY ABOUT ANY OTHER ONE. BUT THIS ONE HAS, ONLY HAS A SOLUTION Y-OF X-IS SOME CONSTANT WHICH WE STILL HAVE TO PICK, TIMES E-TO THE K-X. THAT'S THE ONLY SOLUTION. THAT'S WHAT I WANT TO SHOW YOU. THERE CAN NOT BE ANY OTHER FUNNY STUFF WHICH CAN HAPPEN. SO HERE, LET ME TRY TO PROVE THAT NOW. IT'S VERY EASY. SO LET'S Y-OF X-BE SOME SOLUTION TO THIS DIFFERENTIAL EQUATION AND I WANT TO PROVE THERE'S ONLY THAT ONE WAY TO GET THE ANSWER. HERE'S THE IDEA. I'M GOING TO LET F-OF X-BE A NEW FUNCTION WHICH I'M GOING TO GET FROM THIS SOLUTION, MULTIPLIED BY E-TO THE MINUS K-X. I'M GOING TO PROVE TO YOU NOW THAT THIS FUNCTION HERE HAS TO BE A CONSTANT. IT'S GOING TO BE THE CONSTANT C. THAT'S GOING TO BE THE GAME PLAN. SO GOAL HERE IS SHOW THAT F-OF X-IS REALLY A CONSTANT. HOW DO YOU KNOW IF A FUNCTION IS CONSTANT? WHAT'S ITS DERIVATIVE
HAVE TO BE? ZERO. SO LET ME JUST GO AHEAD AND COMPUTE THE DERIVATIVE OF THAT FUNCTION AND CONFIRM THAT IT'S ZERO. AND IT HAS TO BE A CONSTANT. SO LET'S CONFIRM THAT THE DERIVATIVE OF THIS NEW FUNCTION IS ZERO. THAT WILL DO IT. THAT WILL SHOW IT'S A CONSTANT. THAT MEANS I HAVE TO DO D-D-X-OF F-OF X-WHICH IS JUST THE PRODUCT RULE. GOING TO USE THE PRODUCT RULE HERE. (ON BOARD). SO HERE'S, WRITE DOWN THE PRODUCT RULE. (ON BOARD). AND NOW I JUST HAVE TO USE THE INFORMATION THAT I HAVE, D-D-X-OF Y-OF X, WELL THAT, I'M TOLD WHAT THAT IS. I HAVE NO CHOICE. THIS THING HERE IS K-TIME Y-OF X. THAT WAS GIVEN TO ME. AND WHAT'S D-D-X-OF E-TO THE MINUS K-TIMES X? BY THE CHAIN RULE? SO 10 WE JUST TALKED ABOUT THAT. SO D-D-X-OF THIS E-TO THIS PARTICULAR CONSTANT TIMES X-IS -- I JUST PULLED CONSTANT OUT IN FRONT. SO IT'S GOING TO BE THIS CONSTANT WHICH IS MINUS K-TIMES E-TO THE MINUS K-X. THAT'S JUST THE RULE I HAD BEFORE. SO I HAVE K-TIMES Y-OF X-TIMES E-TO THE MINUS K-X-MINUS K-TIMES Y-OF X-TIMES E-TO THE MINUS K-X-. THESE TWO TERMS ARE THE SAME. I SUBTRACT THEM AND GETS ZERO. SO THE FUNCTIONS'S DERIVATIVE IS ZERO. AND SO IT'S GOT TO BE A CONSTANT. SO F-OF X-IS SOME CONSTANT, LET ME GIVE IT A NAME. CALL IT C. BECAUSE THE ONLY FUNCTION WHO'S DERIVATIVE IS ZERO IS A CONSTANT. THAT'S THE WAY CONSTANTS ARE. AND SO FINALLY LET ME WRITE IT DOWN. I'LL FINISH THIS PROOF RIGHT UNDERNEATH THE STATEMENT. SO WHAT DO I HAVE? I HAVE THAT THIS CONSTANT IS WHAT I GET WHEN I MULTIPLY Y-OF X-TIMES E-TO THE MINUS K-X. THAT'S WHAT'S SATISFIED BY Y-OF X. SO LET ME SOLVE FOR Y-OF X. (ON BOARD). LET ME USE THE LAW OF EXPONENTS HERE TO SIMPLIFY THIS. THAT'S THE CONSTANT TIMES E-TO THE MINUS K-X-TO
THE MINUS ONE. THAT'S WHAT HAPPENS WHEN I TAKE IT TO THE DENOMINATOR. THE NEXT LAW OF EXPONENTS TELLS ME MULTIPLY THOSE TOGETHER. MINUS ONE TIMES THE OTHER EXPONENT. AND THAT'S E-TO THE PLUS K-X. WHICH IS WHAT I WANT. YEAH, THERE IT IS. SO THAT'S JUST REASSURING THAT WHENEVER YOU SEE THIS DIFFERENTIAL EQUATION UP THERE THERE'S ONLY ONE ANSWER TO WRITE DOWN WHICH IS NOT TRUE OF EVERY EQUATION. ALSO GOOD PRACTICE. ANY QUESTIONS ABOUT THAT? BEFORE I GO ON? SO LET'S TALK SOME MORE ABOUT WHAT THESE GRAPHS LOOK LIKE. 11 FIRST I'LL REMIND YOU THE PROPERTIES OF E TO THE X-THEN I WANT TO DO GENERAL E-TO THE X. WHAT CAN THE GRAPH OF THE FUNCTION E-TO THE K-TIMES X-LOOK LIKE? WELL, IF K-IS GREATER THAN OR EQUAL TO ZERO, THAT'S GOING TO BE INCREASING FUNCTION OF X. AND IT WILL LOOK AS IT HAS OVER HERE, WE'VE DRAWN THIS PICTURE MANY TIMES, E-TO THE K-X-WHERE K-IS POSITIVE. SO E-TO THE K-X-GOES TO INFINITY AS X-GOES TO PLUS INFINITY. BUT IF I GO IN THE OTHER DIRECTION, AS X-GOES TO MINUS INFINITY, AS X-GOES THIS WAY, THE FUNCTION GETS SMALLER AND SMALLER AND APPROACHES ZERO, SO IT HAS AN ASYMPTOTE. THIS DIRECTION HERE, AN ASYMPTOTE, THE X-AXIS. AND, OF COURSE, IT'S ALWAYS POSITIVE. NEVER CROSSES THE X-AXIS. AND THE SLOPE HERE, THE SLOPE IS WHAT? WE BUILD IT HERE SO IT'S EASY TO FIGURE OUT. WHAT IS THE SLOPE IF I DO D-D-X-TO E-TO THE K X-IS THAT K-TIME E TO THE K-X. IF I PLUG THIS IN AS I WANT THE SLOPE AT X EQUALS ZERO, K-TIMES E-TO THE K-ZERO IS K. SO THE SLOPE THERE IS JUST K. JUST READ IT OFF THE FUNCTION. WHEN K-IS LESS THAN ZERO YOU GETS A MIRROR IMAGE LITERALLY OF EVERYTHING ABOUT THE X-AXIS. LET ME WRITE DOWN AGAIN WHAT IT LOOKS LIKE.
K-X-GOING TO BE DECREASING. SO THIS PICTURE IS THE MIRROR IMAGE OF THAT PICTURE. E-TO THE K-X-IS GOING TO GO TO ZERO AS X-GOES TO PLUS INFINITY. (ON BOARD). IT'S STILL TRUE THAT E-TO THE K-X-IS (ON BOARD). JUST THE IMAGE IS REVERSED. ONE WAY YOU CAN THINK ABOUT THIS FUNCTION E-TO THE K-X-WHEN K-IS NEGATIVE, WHAT IF I WRITE IT THIS WAY? (ON BOARD). SORRY, LET ME WRITE IT THIS WAY, E-TO THE MINUS K-TIMES MINUS X. THAT DOESN'T CHANGE THE 12 FUNCTION. SO IF K-IS NEGATIVE, THEN MINUS K-IS POSITIVE. SO THERE I HAVE A POSITIVE THING. BUT NOW I JUST HAVE A FUNCTION WHERE I'VE CHANGED X-TO MINUS X. SO IT'S GOING TO LOOK LIKE THAT EXCEPT I CHANGED X-TO MINUS X. AND WHAT HAPPENS WHEN I PLOT A FUNCTION BY CHANGE THE ARGUMENT FROM X-TO MINUS X-. HOW DO YOU CHANGE THE GRAPH? IF YOU HAVE THE GRAPH OF ANY FUNCTION YOU LIKE, F-OF X, I WILL DRAW ANY OLD FUNCTION. THEN THE GRAPH OF F-OF MINUS X-IS GOTTEN BY, WELL IT'S SIMPLY JUST THE MIRROR IMAGE IN THE Y-AXIS. SIMPLY DO IT BACKWARDS. SO THAT'S WHAT HAPPENS WHEN YOU CHANGE X-TO MINUS X, IT JUST REVERSES. SO THAT FUNCTION IS JUST THE REVERSE OF THAT. IT'S AN EASIER WAY TO REMEMBER, IF I LIKE. OKAY. THE REASON I'M REMINDING YOU OF E-TO THE K-X-IS ALL YOU NEED TO KNOW TO GRAPH ANY OTHER EXPONENTIAL FUNCTION. SO THIS OVER THERE IS ALL WE NEED TO GRAPH Y-EQUALS B-TO THE X-FOR ANY B-GREATER THAN ZERO. SO LET ME SHOW YOU HOW THAT WORKS. SO LET ME DRAW E-TO THE X-FOR THE MOMENT. AND LET ME TAKE THE NUMBER B. I DON'T KNOW WHERE IT IS. IT'S SOMEWHERE AND BRING THERE, PLOT IT ON THE Y-AXIS. THERE'S B. I'M GOING TO DRAW A HORIZONTAL LINE THROUGH B. THAT'S EASY ENOUGH TO DO. BECAUSE OF THE WAY THIS FUNCTION E-TO THE X, IT GETS AS CLOSE TO
ZERO AS YOU LIKE HERE, IT GOES OFF TO INFINITY OVER THERE. SO THIS HORIZONTAL LINE HAS TO INSECT SOMETHING. IT DOESN'T MATTER WHAT B-IS. IF I TAKE B-DOWN HERE, IT MIGHT INTERSECT OVER THERE. BUT SOMEWHERE THAT HORIZONTAL LINE HAS TO INTERSECT. WHY DO I CARE ABOUT THAT? BECAUSE THERE'S AT SOME POINT, ONCE YOU FIGURE 13 OUT THIS POINT YOU GO STRAIGHT DOWN AND THAT GIVES A YOU A K. AND SO E-TO THE K-EQUALS WHAT? SO THIS IS THE GRAPH THE CURVE OF THE FUNCTION E-TO THE X. THERE'S K. THERE'S B. SO B, E-TO THE K-EQUALS B? OKAY. SO FOR ANY B-THERE'S ALWAYS A K. IT MIGHT BE DOWN HERE. OR IT MIGHT BE OVER THERE. WHEREVER, THERE'S ALWAYS GOING TO BE A K. SO WHAT ABOUT THE FUNCTION B-TO THE X? B-I JUST EXPLAINED TO YOU WAS E-TO THE K-TIMES X. AND NOW WHAT DOES THE LAW OF EXPONENTS TELL ME? TELLS ME MULTIPLY THE TWO EXPONENT TOGETHER. SO IN OTHER WORDS THE GRAPH OF THE FUNCTION B-TO THE X-IS THE SAME AS THE GRAPH OF E-TO THE K-X-FOR SOME K-. THIS IS THE ONLY FUNCTION YOU NEED TO UNDERSTAND. E-TO THE K-X, IT TELLS YOU EVERYTHING. ANY QUESTIONS. SO THIS IS ONE OF THE THINGS THAT MATH IS GOOD FOR, TAKE THIS WHOLE FAMILY OF FUNCTIONS AND SAYS I ONLY HAVE TO UNDERSTAND IT FOR ONE CASE, FOR THE VALUE OF E. SO LET'S DO A COUPLE OF EXAMPLES TO ILLUSTRATE THAT. I'LL USE MONEY JUST TO GET YOUR ATTENTION AGAIN. SO LET ME DO AN EXAMPLE. YOU INVEST A THOUSAND DOLLARS AT 5 PERCENT INTEREST, IF YOU'RE LUCKY I GUESS THESE DAYS, POSITIVE INTEREST THAT IS, PER YEAR. AND HOW MUCH DO YOU HAVE AFTER T-YEARS? LET'S WRITE DOWN BASED ON KNOWLEDGE OF WHAT IT MEANS TO EARN INTEREST. HERE'S I'LL USE M FOR MONEY, MONEY IS A FUNCTION OF TIME. START WITH A THOUSAND DOLLARS AND EVERY YEAR MULTIPLY
IT BY ONE PLUS 5 PERCENT. AND AFTER T-YEARS THAT'S WHAT IT EQUALS. SO THERE'S THE FAMILIAR FORMULA. I IMAGINE. AND SO I'M GOING TO WRITE THIS AS B-TO THE T-WHERE B- IS 1.05. SO 14 THERE'S THE FUNCTION. AND THIS IS IF I ASK, IF I WRITE B-AS E-TO THE K, WHAT'S K, YOU CAN DO THAT PARTICULAR ARITHMETIC OR LOOK UP IN A TABLE AND K-IS.0488. IT HAPPENS TO BE THAT. THAT TELLS YOU HOW MUCH MONEY YOU HAVE AS A FUNCTION OF TIME. IT'S AN EXPONENTIAL FUNCTION, 1,000 E-TO THE POWER.0488 T. THERE'S YOUR BANK ACCOUNT. LET ME CHANGE THE PROBLEM SLIGHTLY. AND DO IT AGAIN BUT WHERE THEY COMPOUND INTEREST DAY. THIS IS ONE A YEAR YOU GET PAID 5 PERCENT. LET'S CHANGE IS VERY SLIGHTLY TO DO DAILY COMPOUNDING AND SEE HOW THAT CHANGES THE ANSWER. SO WHAT IS M OF T-IF INTEREST IS COMPOUNDED DAILY? SO I'LL WRITE M OF T-AGAIN. START WITH A THOUSAND DOLLARS. AND NOW, WHAT DOES COMPOUNDED DAILY MEAN? IT MEANS YOU GET ONE DAY'S WORTH OF YOUR INTEREST EVERYDAY. SO THAT MEANS INSTEAD OF GETTING FIVE PERCENT YOU GET, LET'S SAY THERE'S 365 DAYS IN THE YEAR YOU GET 5 PERCENT DIVIDED BY 365 BUT YOU GET IT 365 TIMES PER YEAR. IF IT WERE MONTHLY IT WOULD BE 12 AND 12. BUT THAT'S THE GENERAL IDEA. SO THAT'S GOING TO BE 1,000 TIMES B-TO THE POWER 365 T-WHERE B-IS ONE PLUS.05. OVER 365. SO THAT'S I'M GOING TO WRITE E-TO THE K. SO B-EQUALS E-TO THE K. BUT NOW 365 T. AND I'LL TELL YOU WHAT K-IS FINALLY. K-IS.000137. YOU CAN FIGURE OUT WHAT THAT IS AGAIN USING A CALCULATOR OR TABLE. NOW I'M GOING TO USE THE LAW OF EXPONENTS. WRITE THAT AS E-TO THE K-TIMES 365 TIMES T. MULTIPLY ALL THE THE EXPONENT TOGETHER. RULE OF EXPONENTS. AND SO WHAT I WANT TO DO IS FIGURE OUT WHAT
IS K-TIMES 365. THAT'S GOING TO BE THE EXPONENT I HAVE UP IN 15 THERE. AND IT'S GOING TO BE.05002, FOUR DIGITS. (ON BOARD). THAT'S HOW FAST YOUR MONEY GROWS. NOW WE CAN COMPARE A THOUSAND TIME E- TO THE.0500 T-VERSUS E-TO THE.0488 T. IT GROWS A LITTLE FASTER BUT NOT A HECK OF A LOT. HERE FILL IT IN, HERE, HOW MUCH MONEY DO I HAVE AFTER ONE YEAR? SO WE DON'T HAVE TO ACTUALLY WORK VERY HARD. YOU JUST PLUG IN T-EQUALS ONE TO THE ORIGINAL FORMULA AND HOW MUCH MONEY DO YOU HAVE? 5 PERCENT MORE? AND THE ANSWER IS, WHAT'S 5 PERCENT OF A THOUSAND BUCKS? FIFTY BUCKS. THAT'S WHAT YOU GET IN THAT REGIME. LET'S COMPARE IT. AND SEE HOW MUCH FARTHER AHEAD WE COME OUT. NOW M OF ONE HERE, PLUG IN ONE AND E-TO THE POWER.05 AND THE ANSWER IS YOU'RE NOT GOING TO BE MUCH RICHER. THAT'S WHAT DAILY COMPOUNDING GIVES YOU. STUDENT: WHAT'S THE POINT OF PUTTING IT INTO FORM OF E-TO THE ONE WHEN YOU CAN KEEP IT. PROFESSOR: THERE ARE TWO REASONS. ONE IS I CAN COMPARE THIS TO THAT. THIS FUNCTION TO THAT FUNCTION MORE EASILY THAN THIS FUNCTION TO THAT FUNCTION. SO I CAN INDEED, WHEN THEY'RE ACTUALLY COMPUTED ON THE CALCULATOR OR THE SPREADSHEET THIS IS HOW IT'S REPRESENT INTERNALLY. WOULD YOU KNOW IF FROM STARTING OUT THIS HOW MUCH BIGGER FROM THAT? NOT SO OBVIOUS. OTHERWISE IT'S JUST TO PROVE TO YOU YET AGAIN THAT ALL YOU HAVE TO DO IS UNDERSTAND ONE EXPONENTIAL FUNCTION WHICH IS E-TO THE K-X-AND YOU CAN DO ALL THE OTHER EXPONENTIAL FUNCTIONS. GOOD QUESTION. STUDENT: HOW DO YOU FIGURE OUT K-AGAIN. 16
PROFESSOR: SO THAT'S SECTION 44. BUT IF YOU DO IT PICTORIALLY. GRAPH E-TO THE X, DRAW A HORIZONTAL LINE THROUGH B-AND DROP THE VERTICAL LINE IT K. THERE'S A NAME, ANOTHER NAME FOR K, IT'S CALLED THE LOGARITHM. THAT'S WHAT IT IS. THIS IS THE LOGARITHM. OKAY. INDEED THAT IS THE NEXT SECTION THAT I'M ABOUT TO DO. SECTION 4.4. THERE'S MORE THAN ONE LOGARITHM. I'M GOING TO START BY TALKING ABOUT K. IN PARTICULAR IT'S CALLED THE NATURAL LOGARITHM. NOW DRAW THE PICTURE AGAIN. RELABEL IT. (INAUDIBLE). CALL IT X-COMMA Y-ON THE CURVE. HERE'S THE X-COORDINATE AND Y-COORDINATE. AND NATURAL QUESTION IS GIVEN THE Y-COORDINATE, WHAT IS X? HOW DO YOU FIND THAT? WHAT IS THAT FUNCTION? AND THE DEFINITION IS IF Y-EQUAL E-TO THE X-THEN X-IS CALLED THE NATURAL LOGARITHM OF Y-OR SOMETIMES THE LOGARITHM BASE E-OF Y. YOU MAY HAVE ENCOUNTERED LOGARITHMS BASE TEN BEFORE. YOU CAN DO ANY BASE BUT I'M GOING TO START WITH BASE E. SO THE MAIN PROPERTY THAT WE JUST, AND IT'S WRITTEN, SO THE NOTATION IS X-EQUALS L N-Y. LOG NATURAL OR SOMETHING LIKE THAT. L N. THAT'S THE NOTATION. SO THE MAIN PROPERTY OF LOGARITHM I CAN WRITE DOWN THIS WAY. JUST REPEATING WHEN I SAID THERE. WHICH IS Y-IS E-TO THE X-IS E-TO THE NATURAL LOG OF Y. SO THAT'S TRUE FOR ANY OLD Y. AS LONG AS Y-GREATER THAN ZERO. THE LOGARITHM FUNCTION IS ONLY, MAKE SENSE FOR Y-GREATER THAN ZERO. YOU CAN ONLY GET POSITIVE NUMBERS FROM THE EXPONENTIAL FUNCTION. SO LET ME DO SOME EXAMPLE. ZERO COMMA ONE IS ON THE GRAPH. HERE IT IS. ZERO 17 COMMA ONE. WE KNOW THAT ONE EQUAL E TO THE ZERO, SO WHAT IS THE
LOG OF WHAT? TRANSLATING THAT, SOMETHING IS THE LOG OF SOMETHING. SO ZERO IS THE LOG OF ONE. E-TO THE ZERO IS ONE. OKAY. LET ME DO ANOTHER EXAMPLE. HOW ABOUT TWO E-SQUARED IS ON THE GRAPH. OF E-TO THE X-OVER THERE. IN OTHER WORDS, E-SQUARED EQUAL E-SQUARED. THAT'S NOT SO EXCITING. SO SOMETHING IS THE LOG OF SOMETHING. TWO IS THE LOG OF E-SQUARED. AND FINALLY LET ME SUMMARIZE THIS PROPERTY. WHICH IS THAT SINCE X-COMMA E-TO THE X-IS ON THE GRAPH FOR ANY X, IF YOU TAKE THE LOG OF E-TO THE X, YOU GET X-BACK. SO THESE ARE, THAT PROPERTY AND THIS PROPERTY ARE THE TWO MAIN PROPERTIES OF LOGS AND EXPONENTS. LET ME WRITE THEM DOWN AGAIN. HERE WE HAVE THE TWO MAIN PROPERTIES ARE GOING TO BE IF I TAKE E-TO THE X-AND TAKE THE LOG TO GET X-BACK. THE OTHER ONE SAYS TAKE THE LOG AND EXPONENTIATE IT, YOU GET THE NUMBER BACK AS WELL. ANOTHER WAY TO SAY THIS, IS THAT LOGARITHM AND EXPONENTIAL FUNCTIONS, THREES TWO FUNCTIONS ARE INVERSES, THEN UNDO ONE ANOTHER AM IF YOU DO THE LOG AND EXPONENTS AND GET BACK TO WHERE YOU STARTED. IF YOU DO THE EXPONENTS AND THE LOG YOU GET BACK TO WHERE YOU STARTED. THEY'RE INVERSES. THEY UNDO ONE ANOTHER. SO CAN YOU GIVE, WE'VE SEEN FUNCTIONS LIKE THIS BEFORE THAT UNDO ONE ANOTHER. CAN WE THINK OF SOME EXAMPLES THAT WE'VE SEEN BEFORE? TWO FUNCTIONS THAT UNDO ONE ANOTHER THAT GET YOU BACK TO WHERE YOU STARTED FROM? WHAT IF YOU DO THE FUNCTION Y-EQUALS X-CUBED. WHAT IS THE INVERSE OF CUBING, CUBED ROOT. IF I CUBE 18 SOMETHING AND THEN TAKE THE CUBE ROOT I GET BACK TO WHERE I STARTED. OR I CAN TAKE THE CUBE ROOT FIRST, AND THEN CUBE IT I ALSO GET BACK TO WHERE I STARTED. THEY UNDO ONE ANOTHER, EITHER
OR. SO THIS IS OBVIOUS STUFF. IT'S, BUT FOR THE LOG AND THE EXPONENTS A LITTLE MORE INTERESTING. MORE COMPLICATED FUNCTIONS. THEY DO INVERSES OF ONE ANOTHER. LET US USE THAT FACT TO GRAPH THE LOGARITHM. LET'S USE THIS INVERSE PROPERTY TO GRAPH THE FUNCTION Y-EQUALS L N-OF X. OKAY. SO THE PROPERTY IS GOING TO BE THAT IF X-COMMA Y-IS GOING TO BE ON THE GRAPH OF Y-EQUALS E TO THE X-BECAUSE WE KNOW WHAT THAT GRAPH LOOKS LIKE. WE'VE DRAWN THAT A DOZEN TIMES. THEN THERE'S SOME OTHER POINT THAT'S GOING TO BE REALLY EASY FOR US TO WRITE DOWN, WHICH IS GOING TO BE ON THE GRAPH OF SOMETHING EQUALS LOGARITHM. BECAUSE THESE TWO ARE EQUIVALENT TO ONE ANOTHER. X-IS THE LOGARITHM OF Y. SO WHAT POINTS IS GOING TO BE ON THAT GRAPH? SO IF I TELL YOU X COMMA Y-IS ON THE GRAPH OF E-TO THE X, THEN WHAT, HOW DO I GET A POINT ON THE OTHER GRAPH? WHAT IS THE LOG OF WHAT? X-IS THE LOG OF Y. SO X-IS THE LOG OF X. THAT POINT'S GOING TO BE ON THE GRAPH OF X-ARE OF THE OTHER FUNCTION. IN OTHER WORDS LET ME DRAW IT HERE. SO HERE IS E-TO THE X. AND SO THERE'S A TWO POINTS. IT'S GOING TO BE ON THE EXPONENTIAL ONE BUT THE OTHER ONE, ONE COMMA ZERO, THAT'S GOING TO BE ON THE LOG GRAPH BECAUSE THE LOG OF ONE IS ZERO. HOW ABOUT THE.1 COMMA E-TO THE ONE? THAT'S ON THE EXPONENTIAL GRAPH SO I'M CLAIMING THAT THE POINT E-TO THE ONE 19 COMMA ONE IS GOING TO BE ON THE LOG GRAPH BECAUSE LOG OF E-TO THE ONE IS ONE. AND IF I TAKE SEVEN COMMA E-TO THE SEVEN AND THERE'S GOING TO BE SOME POINT E-TO THE SEVEN COMMA SEVEN THAT'S GOING TO BE ON THE LOGARITHM. IF I FILL THAT IN, IT'S GOING TO LOOK LIKE THIS. SO FOR EVERYONE POINTS UP HERE, I'M GOING TO GET, WHO'S
COORDINATES ARE X-COMMA WHY, FOR ANY POINT HERE WHO'S COORDINATES ARE X-COMMA Y-I'LL GET A COORDINATE DOWN HERE WHAT'S COORDINATES ARE Y-COMMA X- BECAUSE IF Y-IS THE EXPONENTIAL OF X-THEN -- THEY UNDO ONE ANOTHER. SO THAT MEANS THAT I'M JUST TAKING EVERY POINT AND FLIPPING IT. REVERSING ITS COORDINATES. DID YOU AT SOME POINT LEARN IF SOMEBODY GIVES YOU A GRAPH HOW TO FLIP ITS COORDINATES BY DRAWING A PICTURE? JUST BY DOING SOME SORT OF FLIPPING OR FOLDING OF A PIECE OF PAPER. IF YOU DREW IT ON A PIECE OF PAPER, HOW DO YOU DO IT? I TAKE A MIRROR IMAGE OF THE GRAPH. REMEMBER THIS? THIS GRAPHY JUST FLIP IT IN THE DIAGONAL, THE LINE X-EQUALS Y-AND I GET THIS GRAPH. THAT'S WHAT IT MEANS TO TAKE EVERY POINT X-Y-AND FLIP IT TO Y-X. FOLD THE PIECE OF PAPER OVER, AND YOU GET THE PLOT. SO THAT'S JUST TAKE THE MIRROR IMAGE OF THE CURVE Y-EQUALS E-TO THE X-TO GET THE LOG PLOT. IS THAT SOMETHING, I THINK YOU'VE DONE THIS BEFORE. IF I PLOT Y-EQUALS X-SQUARED, WE KNOW HOW TO PLOT THAT. WHAT IS THE GRAPH OF X-EQUALS Y-SQUARED? ANOTHER PARABOLA, LOOKS LIKE THIS. AND THIS IMAGE IS JUST THE MIRROR IMAGE IN THAT DIAGONAL LINE. JUST FLIP THEM. IT'S REALLY EASY TO THINK ABOUT HOW TO GET FROM THIS GRAPH TO THAT GRAPH. SO THIS THING HERE IS THE LOGARITHM. 20 THERE'S THE GRAPH OF Y-EQUALS E-TO THE X. (ON BOARD). SO LET ME WRITE DOWN, SUMMARIZE HERE THE PROPERTIES OF THE CURVE Y-EQUALS E TO THE X. AND THEN JUST BY REALIZING I'M TAKING A MIRROR THERE, READ OFF THE PROPERTIES OF Y, WELL, X-EQUALS LOG OF Y-. THIS IS THE SAME AS SAYING THIS. JUST SAME EQUATION WRITTEN DOWN SLIGHTLY DIFFERENTLY. E-TO THE X-IS ALWAYS GREATER THAN ZERO. ALWAYS ABOVE THE AXIS. THAT MEANS THAT THIS
FUNCTION IS ONLY DEFINED FOR Y-GREATER THAN ZERO. THAT CURVE IS ONLY DEFINED OVER THERE. YOU CAN ONLY TAKE THE LOG OF A POSITIVE NUMBER. SO BY, IF I TAKE X-LESS THAN ZERO THAT MEANS THAT E-TO THE X-IS LESS THAN ONE. IF I LOOK AT, IF I TAKE X-LESS THAN ZERO E-TO THE X-IS LESS THAN ONE. SO THAT'S RIGHT OFF. THE EQUIVALENT PROPERTY OVER THERE IS IF THE LOGARITHM OF Y-IS LESS THAN ZERO, IF AND ONLY IF, THESE TWO ARE EQUIVALENT TO ONE ANOTHER, IF I'M, IF THE ARGUMENT THERE, IF X-IS LESS THAN ONE, SORRY, IF Y-IS LESS THAN ONE. SO I HAVE TO BE OVER HERE. AND IT'S GOING TO BE LESS THAN ZERO. X-EQUALS ZERO, I GET E-TO THE X-IS ONE, E-TO THE ZERO IS ONE. OVER HERE I GET THAT LOGARITHM OF THE NUMBER ZERO IS THE SAME THING AS SAYING THE NUMBER IS EQUAL TO ONE. THAT MEANS I'M, THERE'S THE POINT THERE. THE LOG OF, THE ARGUMENT IS ONE. AND X-GREATER THAN ZERO MEANS THAT THE EXPONENTIAL IS BIGGER THAN ONE. AND THAT'S LIKE SAYING THE LOGARITHM IS POSITIVE IF THE ARGUMENT BIGGER THAN ONE. THAT SAYS I'M OVER HERE. ALL THOSE PROPERTIES GO TOGETHER BY TAKING THESE TWO MIRROR IMAGE OF ONE ANOTHER. THIS IS INCREASING 21 FUNCTION. IT GOES UP. AND THIS IS AN INCREASING FUNCTION, IT GOES UP. THAT'S SORT OF OBVIOUS FROM THE MIRROR PROPERTIES. BUT HERE'S SOMETHING THAT WORK OUT BACKWARDS, E-TO THE X-IS CONCAVE WHICH WAY? UP. AND HOW ABOUT THE LOGARITHM FUNCTION? CONCAVE DOWN. SO THAT PROPERTY WORKS KINDS OF THE OPPOSITE. IF THERE ARE NO QUESTIONS ABOUT THOSE BASIC PROPERTIES LET'S USE A FEW. AND DO SOME COMPUTATIONS AND CALCULATIONS OF LOGARITHMS. THIS IS JUST MAKE SURE WE KNOW THE INVERSE PROPERTIES. SO TAKE E-TO THE NATURAL LOG OF FOUR, YOU GET FOUR. BECAUSE THESE CANCELING OUT.
HOW ABOUT E-TO THE NATURAL LOG OF FOUR PLUS TWO TIME NATURAL LOG OF THREE? NOW, THERE ARE PROPERTIES OF LOGARITHMS. WE'LL GET TO THEM BUT I WANT TO USE THE LAW OF EXPONENTS. THE FIRST LAW OF EXPONENTS TELLS ME I CAN DO THIS. THAT'S JUST THE LAW OF EXPONENTS. BECAUSE IF I HAVE A SUM UP HERE I CAN BREAK INTO A PRODUCT. AND NOW, I CAN WRITE THAT AS, THIS IS FOUR, WE JUST FIGURED THAT OUT. THIS ONE, I'M GOING TO WRITE THIS WAY, DOESN'T MATTER WHAT ORDER I MULTIPLY THOSE TWO SCALERS IN. DOESN'T CHANGE THINGS. AND NOW I CAN USE THE LAW OF EXPONENTS AGAIN. WRITE IT THAT WAY. AND FINALLY I GET FOUR, NOW I CAN SEE THAT'S A THREE IN THERE AND NOW ALL THE LOGS AND EXPONENTS HAVE GONE AWAY AND I HAVE FOUR TIMES NINE OR 36. THE OTHER WAY YOU COULD HAVE DONE THIS ONE, WE RECALL WHICH PROPERTIES OF LOGARITHM I COULD HAVE RECOGNIZED THAT THIS EXPONENTS WAS ACTUALLY THE LOG OF 36 BUT WE'LL DO THAT EVENTUALLY. HOW ABOUT THIS ONE, SOLVE FIVE E-TO THE POWER X-MINUS THREE EQUALS FOUR. SO LET'S TRY TO SOLVE 22 THAT ONE. DIVIDE BOTH SIDE BY FIVE. MAKE THE EXPONENTIAL GO AWAY. SO TAKE THE LOG OF BOTH SIDES. LOG OF EXPONENTS IS WHAT? WHAT IS THIS? THIS TURNS INTO X-MINUS THREE, EQUALS NATURAL LOG OF 28, AND FINALLY I GET X-EQUALS THREE PLUS THE NATURAL LOG OF.8. AND WE CAN DO THAT, 2.78, SOMETHING LIKE THAT. DOT DOT DOT. STUDENT: HOW DO YOU KNOW (INAUDIBLE). PROFESSOR: SO I HAVEN'T TAUGHT US HOW TO COMPUTE THE DERIVATIVE YET. I WILL. AND RIGHT NOW I'M JUST GOING BY THE PICTURE WHICH SAYS IF I TAKE SOMETHING THAT'S CONCAVE UP, AND FLIP IT UPSIDE DOWN, IT'S CONCAVE DOWN. BUT I WILL COMPUTE THE SECOND
DERIVATIVE IN DUE COURSE. IT WILL BE EASY TO SEE WHETHER IT'S POSITIVE OR NEGATIVE. STUDENT: SO (INAUDIBLE). PROFESSOR: YOU'RE GOING TO BE DOING ALGEBRAIC MANIPULATIONS OF THIS. ONCE YOU GOT TO HERE, YOU'D BE DONE. I JUST FILLED IN THE REST FOR FUN. YOU WOULD BE DONE BY THE TIME YOU'VE DONE THE MANIPULATIONS TO GET THAT. ANY OTHER QUESTIONS? OKAY. SOLVE ANOTHER ONE HERE. TWO NATURAL LOG OF X-PLUS SEVEN EQUALS ZERO. TWO NATURAL LOG OF X-EQUALS MINUS SEVEN (ON BOARD). SO X-IS E-TO THE POWER MINUS 3.5. AT THAT POINT YOU'D BE DONE UNLESS YOU REALLY WANT TO KNOW THE ANSWER,.03. SO THIS IS AS I SAID THE LOGARITHM BASE E-. AND FROM HIGH SCHOOL OR WHENEVER YOU LAST LOOKED AT LOGARITHM YOU PROBABLY LEARNED ABOUT LOGARITHMS BASE TEN. LET ME TELL YOU THEY'RE ALL RELATED TO 23 EACH OTHER. CONSTANT. TO GET TO THIS -- ALL YOU HAVE TO DO IS MULTIPLY BY SO ALL THE LOG BASE, THEY ALL DIFFERENT BY MULTIPLYING BY (INAUDIBLE). SO THEY'RE EASY TO GET FROM ONE TO THE OTHER. OTHER LOGARITHM AND EXPONENTIAL FUNCTIONS. SO LET ME JUST REVIEW WHAT I SAID BEFORE. EARLIER, WE SAID THAT GIVEN ANY POSITIVE B-THERE WAS SOME NUMBER K-SO THAT B-EQUALED E-TO THE K. I DREW THAT PICTURE A WHOLE BUNCH OF TIMES. SO WHAT DOES THAT MEAN? IF I TAKE B-TO THE X-THAT'S GOING TO BE E-TO THE K-X. AND THAT'S BY THE LAW OF EXPONENTS THAT'S E-TO THE K-X. IF I COMPUTE E-TO THE SOMETHING YOU CAN COMPUTE B-TO THE SOMETHING BY USING IF I CAN FIGURE OUT THAT MAGIC NUMBER K-. NOW WE WANT TO KNOW WHAT K-IS. SO K-IS WHAT? USING THE FUNCTION WE JUST DEFINED FOR THE
LAST FIVE MINUTES. IF B-EQUALS E-TO THE K-K-EQUALS NATURAL LOG OF B. THAT'S HOW YOU GET K. SO THIS IS EQUAL TO E-NATURAL LOG OF B-TIMES X. SO THAT'S HOW YOU GO FROM ONE EXPONENTIAL FUNCTION TO ANOTHER. THAT'S HOW EXPONENTIALS ARE RELATED. NOW LET'S DO LOGS. SO THERE'S THE PROPERTY I'VE WRITTEN DOWN MANY TIMES. WE CALL L N-X-LOG BASE E. AND LET ME TRY WRITING DOWN ANOTHER LOGARITHM THAT'S FAMILIAR. (ON BOARD). WHAT DOES LOG BASE TEN MEAN? IT MEANS THAT THAT'S I THINK THE RELATIONSHIP THAT WE'VE LEARNED BEFORE. LOG BASE TEN. (ON BOARD). SO THIS IS (ON BOARD) AND DEPENDING ON HOW YOU LEARNED IT THIS MIGHT HAVE BEEN CALLED THE COMMON LOG. AND THIS IS NATURAL LOG. TERMINOLOGY IS 24 EASIER TO SAY BASE E-AND BASE TEN. SO YOU DON'T HAVE TO REMEMBER -- MAYBE IN HIGH SCHOOL YOU LEARNED HOW TO YOU COMPUTATIONS WITH LOG BASE TEN. THE QUESTION IS HOW ARE THEE RELATED TO ONE ANOTHER. OR FOR THAT MATTER, BASE TWO, USE THAT A LOT, FOR FOR THAT MATTER BASE B, THE GENERAL. HOW ARE ALL THESE DIFFERENT LOGARITHMS RELATED TO ONE ANOTHER. AND THE ANSWER IS IT'S EASY. LET JUST DO BASE STEP, DO THE GENERAL CASE. THERE'S THE DEFINITION OF WHAT LOG BASE TEN OF X-MEANS. WHATEVER NUMBERS WORKS THERE. NOW WHAT IS TEN? HOW DO I WRITE TEN E-TO SOMETHING? WHAT GOES UNHERE? IF I JUST TEN TO E-TO THE POWER, ANYBODY? NATURAL LOG OF TEN, THAT'S WHAT THE NATURAL LOG DOES. THAT'S X. AND THE OTHER WAY I CAN WRITE X-IS E-TO THE POWER NATURAL LOG OF X. THAT'S, SO THIS IS E-TO THE POWER NATURAL LOG OF TEN, SOME NUMBER, TIMES LOG BASE TEN OF X-WHICH IS THE THING I DON'T KNOW. AND I ALSO KNOW IT'S E-TO THE NATURAL LOG OF
X-BECAUSE THAT'S ANOTHER WAY TO WRITE X. SO IF THAT, IF E-TO THIS NUMBER EQUALS E-TO THAT NUMBER, WHAT ARE, CAN YOU TELL ME WHAT THESE TWO EXPONENTS? THEY'RE GOING TO BE THE SAME. IF, SO, IN FACT, SO IF, IF THIS IS TRUE, LET ME WRITE IT DOWN THIS WAY, WRITE DOWN EVERY STEP, I'LL BE CAREFUL, THAT'S WHAT I'VE GOTTEN SO FAR. AND SO IF THOSE TWO NUMBERS ARE EQUAL, THEN THEY'RE NATURAL LOGARITHMS ARE EQUAL. AND SO WHAT IS THE LOG OF THE E-TO THE LOG? THESE TWO THINGS CANCEL. SO THIS IS JUST LOG X. MADE E-GO AWAY. OVER HERE, LOG OF EXPONENTIAL, THESE TWO FUNCTIONS UNDO ONE ANOTHER. SO I GET (ON BOARD). WHAT DO I 25 WANT TO DO? SOLVE THIS EQUATION FOR LOG BASE TEN OF X-AND DIVIDE BY THIS CONSTANT. IT'S GOING TO BE ONE DIVIDED BY THE NATURAL LOG OF TEN, SOME NUMBER, TIMES THE NATURAL LOG OF X. SO THE LOG BASE TEN IS JUST SOME CONSTANT TIMES NATURAL LOG OF X-. THAT'S HOW THEY'RE WRITTEN. AND WHAT IS A CONSTANT? IT'S ABOUT.43 SOMETHING OR OTHER TIMES NATURAL LOG OF X. SO JUST, ONE THAT CHANGES ONE TOGETHER. IS THAT OKAY BIT OF ALGEBRA THERE TO RELATE THE TWO? YOU KNOW ONE LOGARITHM, YOU KNOW ALL OF THEM BECAUSE YOU HAVE TO JUST MULTIPLY THEM BY ONE OVER LOG OF TEN. NOW IN THAT ALGEBRA UP THERE WAS THERE ANYTHING SPECIAL ABOUT TEN THAT I USED? COULD IT HAVE BEEN ANY OTHER NUMBER? I DIDN'T REALLY USE ANY PROPERTIES WITH THE NUMBER TEN SO LET ME DO IT WITH ANY OLD BASE B-. SO NOW DEFINE THE LOG BASE ANY POSITIVE NUMBER B-OF X. SO THIS ONLY WORK IF B-IS GREATER THAN ZERO. SO I HAVE THIS FUNCTION. I'M GOING TO WRITE E-TO THE B. SO B-IS GOING TO BE WRITTEN AS E-TO THE LOG B. AND THIS IS GOING TO BE, WHICH IS X-SO. THAT'S E-TO THE LOG OF X. SO THAT EXPONENT AND
THIS EXPONENTS, THEY HAVE TO AGREE WITH ONE ANOTHER. SAME AS BEFORE. E-TO THE THAT EQUALS E-TO THE THAT. THESE TWO EXPONENT HAVE TO BE EQUAL. SO I LET LOG B-TIMES LOG B-OF X-(ON BOARD). AGAIN REALLY SIMPLE RELATIONSHIP. I GET THAT THE LOG BASIC B-OF X-FOR ANY OLD B-IS ONE OVER NATURAL LOG OF B-TIMES NATURAL LOG OF X. (ON BOARD) AM SO YOU CAN COMPARE ANY BASE LOG BY MULTIPLYING BY THE SCALER OUT FRONT. ANY QUESTIONS ABOUT THAT? I THINK THAT'S A GOOD PLACE TO STOP. NEXT TIME WE'LL DIFFERENTIATE 26 LOGARITHMS.