DIFFERENTIATE SOMETHING AT THE VERY BEGINNING THE COURSE I'LL ADD YOU QUESTIONS USING THEM. BUT PARTICULAR QUESTIONS AS YOU'LL SEE


 Emery Cooper
 2 years ago
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1 1 MATH 16A LECTURE. OCTOBER 28, 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 BTO THE XWILL BE SOME FUNCTION BIGGER THAN 2
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 DDXTO THE EXWAS ETO 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 BTO THE XWHEN BIS GREATER THAN ONE. IF YOU KEEP MULTIPLIED BY B IT KEEPS GET BEING BIGGER. WE ALSO DREW THIS PICTURE. AND THIS IS BTO THE XWHEN BWAS 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 ETO THE XTHAT ARE VARIATIONS ON ETO THE X. USE ONE MAGIC NUMBER 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 ETO THE X. SO LET'S START BY USING THE CHAIN RULE TO DIFFERENTIATE
3 ETO THE KTIMES XWHERE KIS 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 FOF GOF XWHERE FOF X, THE FIRST THING I DO IS TAKE XAND MULTIPLY IT BY K. THAT'S THE FUNCTION G AND THEN I TAKE THAT AND I EXPONENTIATE IT. SO IF I DO FOF GOF XI GET ETO THE KX. IF I WANT TO USE THE CHAIN RULE, DDXETO THE KX I KNOW THE CHAIN RULE TELLS ME THIS. (ON BOARD). I BETTER FIGURE OUT WHAT GPRIME AND FPRIME ARE. FPRIME IS MEANT TO BE PARTICULARLY EASY. THAT FUNCTION DOESN'T CHANGE WHEN DIFFERENTIATE IT, THAT'S WHY WE CHOSE ETHE WAY WE DID. GPRIME IS JUST A CONSTANT TIMES X. SO IF I PLUG THAT ALL IN I GET THAT THIS IS JUST ETO THE KXTIMES K. (ON BOARD). THERE'S K. AND THERE'S ETO THE KX. 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. DDXTO ETO THE MINUS XTHAT APPLIES HERE BECAUSE I HAVE TO THINK THAT MINUS XIS, OF COURSE, NEGATIVE ONE TIMES X. AND APPLY THE RULE AND YOU GET NEGATIVE ONE TIMES ETO THE MINUS X. SO THERE'S THE EXAMPLE. AND IT'S NOT MUCH FARTHER TO WRITE DOWN THE GENERAL CHAIN RULE FOR TAKING DDXTHE DERIVATIVE OF ETO THE ANY OLD FUNCTION GOF X. WE'RE GOING TO BE DOING THAT. 4 THAT TELLS US IT'S THE SAME RULE AS UP THERE. FPRIME OF GOF X. THERE'S THE FPRIME OF GOF XTIMES GPRIME. SO DIFFERENTIATE A FUNCTION ETO ANYTHING, YOU'LL LEAVE IT ALONE AND COPY IT, BRING DOWN THE EXPONENTIAL AND DIFFERENTIATE IT. AND THAT'S JUST THE
4 CHAIN RULE TOO. IT'S EXACTLY THE SAME CHAIN RULE THAT'S UP THERE. HERE'S FOF GOF X(ON BOARD). I LEFT GALL ALONE. STUDENT: SO WHAT EQUATION IS F OF X PROFESSOR: FOF XIS A FUNCTION. AND I CHOSEN IT TO BE THE SAME AS IT WAS UP THERE. FOF XIS JUST ETO 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. GOF X. USE THIS WHY GIS SOME OTHER FUNCTION. LET ME ILLUSTRATE. ETO THE POWER XSQUARED PLUS ONE. SO GOF XHERE'S GOF 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 XSQUARED 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 XCUBED MINUS ONE OVER XSQUARED, PLUS ONE OVER XSQUARED (ON BOARD). SO GOF XWAS 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 XCUBED NOW, GOOD, I HOPE, INCLUDING ME. LET ME
5 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 UIS A FUNCTION OF X, TAKE DDXOF ETO THE U, AND IT'S JUST GOING TO BE ETO THE UTIMES DUDX. SAME RULE JUST SLIGHTLY DIFFERENT NOTATION USING UTO 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 DDXETO THE XQ(ON BOARD) THERE'S ANOTHER WAY TO SAY THAT. ETO THE KX, 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 ETO THE KXETO  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 KOUT 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, YOF 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. YOF 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 YOF XIS ETO THE
6 KTIMES XTIMES ANY CONSTANTS. WHEN I SAY WHAT IS ANY POSSIBLE FUNCTION THAT SATISFIES THAT EQUATION, THE ANSWER IS ETO THE KXBUT 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 DDXTIMES SOME CONSTANT ETO THE KX. 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 KETO THE KX. IF I JUST REVERSE KAND C, I CAN MULTIPLY NUMBERS IN ANY ORDER I LIKE, IT'S THE SAME THING, I SEE I GET KTIMES YOF X. AND THAT IS WHAT I WANTED TO SHOW. I WANTED TO SHOW DDXOF YOF XIS THIS CONSTANT KTIMES YOF X. SO IT WORKS. SO THAT'S THE FUNCTION. SO FOR EXAMPLE, I CAN TAKE ANY OLD 37 ETO THE KXSATISFIES 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 DDXYOF XEQUALS KTIMES YOF XSATISFYING, I CAN GIVE YOU 7 MORE INFORMATION, YOF 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 YOF XHAS TO EQUAL SOME CONSTANTS. TIMES ETO THE KXFOR SOME CONSTANTS CWHICH WE DON'T KNOW YET. HOW DO I SOLVE FOR IT? PLUG IN XEQUALS ZERO. I GET THAT YOF ZERO WHICH HAD BETTER EQUAL FOUR, IS GOING TO BE THIS CONSTANT TIMES ETO THE KTIMES ZERO, AND SO, AND THAT'S CONSTANT ETO THE ZERO, AND ETO THE ZERO IS? ONE. SO IT'S C. THAT HAS TO EQUAL, THEREFORE, FOUR
7 BECAUSE THAT'S THE EXTRA INFORMATION I GAVE IN THE PROBLEM. SO THE ANSWER IS YOF XIS FOUR TIME ETO THE KX. IF YOU GIVE ME ONE POINT ON THE CURVE I CAN FIGURE OUT C STUDENT: WHAT ABOUT K PROFESSOR: SO KWAS 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 XWASN'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. DDXYOF XEQUALS, I THINK CONSTANT MAYBE SPECIFIC SEVEN TIMES YOF X. AND YOF, SAY ONE EQUALS FOUR JUST TO BE SPECIFIC. SHOULD I TRY THAT ONE? THAT I KNOW THAT YOF XIS GOING TO BE SOME CONSTANTS TIMES ETO 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 YOF ONE, THAT BETTER BE FOUR, AND THAT'S CTIMES ETO 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 CIS GOING TO BE FOUR DIVIDED BY ETO THE SEVEN TIMES ONE OR FOUR DIVIDED BY ETO 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
8 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 ETO THE KX. LET ME JUST, SO WHY AM I DOING THIS? NOT EVERY EQUATION THAT YOU CAN WRITE DOWN HAS ONE SOLUTION. FOR EXAMPLE, HOW ABOUT XSQUARED 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 YOF XIS SOME CONSTANT WHICH WE STILL HAVE TO PICK, TIMES ETO THE KX. 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 YOF XBE 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 FOF XBE A NEW FUNCTION WHICH I'M GOING TO GET FROM THIS SOLUTION, MULTIPLIED BY ETO THE MINUS KX. 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 FOF XIS REALLY A CONSTANT. HOW DO YOU KNOW IF A FUNCTION IS CONSTANT? WHAT'S ITS DERIVATIVE
9 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 DDXOF FOF XWHICH 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, DDXOF YOF X, WELL THAT, I'M TOLD WHAT THAT IS. I HAVE NO CHOICE. THIS THING HERE IS KTIME YOF X. THAT WAS GIVEN TO ME. AND WHAT'S DDXOF ETO THE MINUS KTIMES X? BY THE CHAIN RULE? SO 10 WE JUST TALKED ABOUT THAT. SO DDXOF THIS ETO THIS PARTICULAR CONSTANT TIMES XIS  I JUST PULLED CONSTANT OUT IN FRONT. SO IT'S GOING TO BE THIS CONSTANT WHICH IS MINUS KTIMES ETO THE MINUS KX. THAT'S JUST THE RULE I HAD BEFORE. SO I HAVE KTIMES YOF XTIMES ETO THE MINUS KXMINUS KTIMES YOF XTIMES ETO THE MINUS KX. 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 FOF XIS 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 YOF XTIMES ETO THE MINUS KX. THAT'S WHAT'S SATISFIED BY YOF X. SO LET ME SOLVE FOR YOF X. (ON BOARD). LET ME USE THE LAW OF EXPONENTS HERE TO SIMPLIFY THIS. THAT'S THE CONSTANT TIMES ETO THE MINUS KXTO
10 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 ETO THE PLUS KX. 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 XTHEN I WANT TO DO GENERAL ETO THE X. WHAT CAN THE GRAPH OF THE FUNCTION ETO THE KTIMES XLOOK LIKE? WELL, IF KIS 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, ETO THE KXWHERE KIS POSITIVE. SO ETO THE KXGOES TO INFINITY AS XGOES TO PLUS INFINITY. BUT IF I GO IN THE OTHER DIRECTION, AS XGOES TO MINUS INFINITY, AS XGOES THIS WAY, THE FUNCTION GETS SMALLER AND SMALLER AND APPROACHES ZERO, SO IT HAS AN ASYMPTOTE. THIS DIRECTION HERE, AN ASYMPTOTE, THE XAXIS. AND, OF COURSE, IT'S ALWAYS POSITIVE. NEVER CROSSES THE XAXIS. 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 DDXTO ETO THE K XIS THAT KTIME E TO THE KX. IF I PLUG THIS IN AS I WANT THE SLOPE AT X EQUALS ZERO, KTIMES ETO THE KZERO IS K. SO THE SLOPE THERE IS JUST K. JUST READ IT OFF THE FUNCTION. WHEN KIS LESS THAN ZERO YOU GETS A MIRROR IMAGE LITERALLY OF EVERYTHING ABOUT THE XAXIS. LET ME WRITE DOWN AGAIN WHAT IT LOOKS LIKE.
11 KXGOING TO BE DECREASING. SO THIS PICTURE IS THE MIRROR IMAGE OF THAT PICTURE. ETO THE KXIS GOING TO GO TO ZERO AS XGOES TO PLUS INFINITY. (ON BOARD). IT'S STILL TRUE THAT ETO THE KXIS (ON BOARD). JUST THE IMAGE IS REVERSED. ONE WAY YOU CAN THINK ABOUT THIS FUNCTION ETO THE KXWHEN KIS NEGATIVE, WHAT IF I WRITE IT THIS WAY? (ON BOARD). SORRY, LET ME WRITE IT THIS WAY, ETO THE MINUS KTIMES MINUS X. THAT DOESN'T CHANGE THE 12 FUNCTION. SO IF KIS NEGATIVE, THEN MINUS KIS POSITIVE. SO THERE I HAVE A POSITIVE THING. BUT NOW I JUST HAVE A FUNCTION WHERE I'VE CHANGED XTO MINUS X. SO IT'S GOING TO LOOK LIKE THAT EXCEPT I CHANGED XTO MINUS X. AND WHAT HAPPENS WHEN I PLOT A FUNCTION BY CHANGE THE ARGUMENT FROM XTO MINUS X. HOW DO YOU CHANGE THE GRAPH? IF YOU HAVE THE GRAPH OF ANY FUNCTION YOU LIKE, FOF X, I WILL DRAW ANY OLD FUNCTION. THEN THE GRAPH OF FOF MINUS XIS GOTTEN BY, WELL IT'S SIMPLY JUST THE MIRROR IMAGE IN THE YAXIS. SIMPLY DO IT BACKWARDS. SO THAT'S WHAT HAPPENS WHEN YOU CHANGE XTO 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 ETO THE KXIS ALL YOU NEED TO KNOW TO GRAPH ANY OTHER EXPONENTIAL FUNCTION. SO THIS OVER THERE IS ALL WE NEED TO GRAPH YEQUALS BTO THE XFOR ANY BGREATER THAN ZERO. SO LET ME SHOW YOU HOW THAT WORKS. SO LET ME DRAW ETO THE XFOR 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 YAXIS. 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 ETO THE X, IT GETS AS CLOSE TO
12 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 BIS. IF I TAKE BDOWN 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 ETO THE KEQUALS WHAT? SO THIS IS THE GRAPH THE CURVE OF THE FUNCTION ETO THE X. THERE'S K. THERE'S B. SO B, ETO THE KEQUALS B? OKAY. SO FOR ANY BTHERE'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 BTO THE X? BI JUST EXPLAINED TO YOU WAS ETO THE KTIMES 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 BTO THE XIS THE SAME AS THE GRAPH OF ETO THE KXFOR SOME K. THIS IS THE ONLY FUNCTION YOU NEED TO UNDERSTAND. ETO THE KX, 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 TYEARS? 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
13 IT BY ONE PLUS 5 PERCENT. AND AFTER TYEARS THAT'S WHAT IT EQUALS. SO THERE'S THE FAMILIAR FORMULA. I IMAGINE. AND SO I'M GOING TO WRITE THIS AS BTO THE TWHERE B IS SO 14 THERE'S THE FUNCTION. AND THIS IS IF I ASK, IF I WRITE BAS ETO THE K, WHAT'S K, YOU CAN DO THAT PARTICULAR ARITHMETIC OR LOOK UP IN A TABLE AND KIS 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 ETO 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 TIF INTEREST IS COMPOUNDED DAILY? SO I'LL WRITE M OF TAGAIN. 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 BTO THE POWER 365 TWHERE BIS ONE PLUS.05. OVER 365. SO THAT'S I'M GOING TO WRITE ETO THE K. SO BEQUALS ETO THE K. BUT NOW 365 T. AND I'LL TELL YOU WHAT KIS FINALLY. KIS 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 ETO THE KTIMES 365 TIMES T. MULTIPLY ALL THE THE EXPONENT TOGETHER. RULE OF EXPONENTS. AND SO WHAT I WANT TO DO IS FIGURE OUT WHAT
14 IS KTIMES 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 TVERSUS ETO 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 TEQUALS 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 ETO 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 ETO 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 ETO THE KXAND YOU CAN DO ALL THE OTHER EXPONENTIAL FUNCTIONS. GOOD QUESTION. STUDENT: HOW DO YOU FIGURE OUT KAGAIN. 16
15 PROFESSOR: SO THAT'S SECTION 44. BUT IF YOU DO IT PICTORIALLY. GRAPH ETO THE X, DRAW A HORIZONTAL LINE THROUGH BAND 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 XCOMMA YON THE CURVE. HERE'S THE XCOORDINATE AND YCOORDINATE. AND NATURAL QUESTION IS GIVEN THE YCOORDINATE, WHAT IS X? HOW DO YOU FIND THAT? WHAT IS THAT FUNCTION? AND THE DEFINITION IS IF YEQUAL ETO THE XTHEN XIS CALLED THE NATURAL LOGARITHM OF YOR SOMETIMES THE LOGARITHM BASE EOF 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 XEQUALS L NY. 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 YIS ETO THE XIS ETO THE NATURAL LOG OF Y. SO THAT'S TRUE FOR ANY OLD Y. AS LONG AS YGREATER THAN ZERO. THE LOGARITHM FUNCTION IS ONLY, MAKE SENSE FOR YGREATER 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
16 LOG OF WHAT? TRANSLATING THAT, SOMETHING IS THE LOG OF SOMETHING. SO ZERO IS THE LOG OF ONE. ETO THE ZERO IS ONE. OKAY. LET ME DO ANOTHER EXAMPLE. HOW ABOUT TWO ESQUARED IS ON THE GRAPH. OF ETO THE XOVER THERE. IN OTHER WORDS, ESQUARED EQUAL ESQUARED. THAT'S NOT SO EXCITING. SO SOMETHING IS THE LOG OF SOMETHING. TWO IS THE LOG OF ESQUARED. AND FINALLY LET ME SUMMARIZE THIS PROPERTY. WHICH IS THAT SINCE XCOMMA ETO THE XIS ON THE GRAPH FOR ANY X, IF YOU TAKE THE LOG OF ETO THE X, YOU GET XBACK. 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 ETO THE XAND TAKE THE LOG TO GET XBACK. 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 YEQUALS XCUBED. 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
17 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 YEQUALS L NOF X. OKAY. SO THE PROPERTY IS GOING TO BE THAT IF XCOMMA YIS GOING TO BE ON THE GRAPH OF YEQUALS E TO THE XBECAUSE 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. XIS THE LOGARITHM OF Y. SO WHAT POINTS IS GOING TO BE ON THAT GRAPH? SO IF I TELL YOU X COMMA YIS ON THE GRAPH OF ETO THE X, THEN WHAT, HOW DO I GET A POINT ON THE OTHER GRAPH? WHAT IS THE LOG OF WHAT? XIS THE LOG OF Y. SO XIS THE LOG OF X. THAT POINT'S GOING TO BE ON THE GRAPH OF XARE OF THE OTHER FUNCTION. IN OTHER WORDS LET ME DRAW IT HERE. SO HERE IS ETO 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 ETO THE ONE? THAT'S ON THE EXPONENTIAL GRAPH SO I'M CLAIMING THAT THE POINT ETO THE ONE 19 COMMA ONE IS GOING TO BE ON THE LOG GRAPH BECAUSE LOG OF ETO THE ONE IS ONE. AND IF I TAKE SEVEN COMMA ETO THE SEVEN AND THERE'S GOING TO BE SOME POINT ETO 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
18 COORDINATES ARE XCOMMA WHY, FOR ANY POINT HERE WHO'S COORDINATES ARE XCOMMA YI'LL GET A COORDINATE DOWN HERE WHAT'S COORDINATES ARE YCOMMA X BECAUSE IF YIS THE EXPONENTIAL OF XTHEN  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 XEQUALS YAND I GET THIS GRAPH. THAT'S WHAT IT MEANS TO TAKE EVERY POINT XYAND FLIP IT TO YX. FOLD THE PIECE OF PAPER OVER, AND YOU GET THE PLOT. SO THAT'S JUST TAKE THE MIRROR IMAGE OF THE CURVE YEQUALS ETO THE XTO GET THE LOG PLOT. IS THAT SOMETHING, I THINK YOU'VE DONE THIS BEFORE. IF I PLOT YEQUALS XSQUARED, WE KNOW HOW TO PLOT THAT. WHAT IS THE GRAPH OF XEQUALS YSQUARED? 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 YEQUALS ETO THE X. (ON BOARD). SO LET ME WRITE DOWN, SUMMARIZE HERE THE PROPERTIES OF THE CURVE YEQUALS E TO THE X. AND THEN JUST BY REALIZING I'M TAKING A MIRROR THERE, READ OFF THE PROPERTIES OF Y, WELL, XEQUALS LOG OF Y. THIS IS THE SAME AS SAYING THIS. JUST SAME EQUATION WRITTEN DOWN SLIGHTLY DIFFERENTLY. ETO THE XIS ALWAYS GREATER THAN ZERO. ALWAYS ABOVE THE AXIS. THAT MEANS THAT THIS
19 FUNCTION IS ONLY DEFINED FOR YGREATER THAN ZERO. THAT CURVE IS ONLY DEFINED OVER THERE. YOU CAN ONLY TAKE THE LOG OF A POSITIVE NUMBER. SO BY, IF I TAKE XLESS THAN ZERO THAT MEANS THAT ETO THE XIS LESS THAN ONE. IF I LOOK AT, IF I TAKE XLESS THAN ZERO ETO THE XIS LESS THAN ONE. SO THAT'S RIGHT OFF. THE EQUIVALENT PROPERTY OVER THERE IS IF THE LOGARITHM OF YIS LESS THAN ZERO, IF AND ONLY IF, THESE TWO ARE EQUIVALENT TO ONE ANOTHER, IF I'M, IF THE ARGUMENT THERE, IF XIS LESS THAN ONE, SORRY, IF YIS LESS THAN ONE. SO I HAVE TO BE OVER HERE. AND IT'S GOING TO BE LESS THAN ZERO. XEQUALS ZERO, I GET ETO THE XIS ONE, ETO 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 XGREATER 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, ETO THE XIS 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 ETO THE NATURAL LOG OF FOUR, YOU GET FOUR. BECAUSE THESE CANCELING OUT.
20 HOW ABOUT ETO 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 ETO THE POWER XMINUS 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 XMINUS THREE, EQUALS NATURAL LOG OF 28, AND FINALLY I GET XEQUALS 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
21 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 XPLUS SEVEN EQUALS ZERO. TWO NATURAL LOG OF XEQUALS MINUS SEVEN (ON BOARD). SO XIS ETO 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 BTHERE WAS SOME NUMBER KSO THAT BEQUALED ETO THE K. I DREW THAT PICTURE A WHOLE BUNCH OF TIMES. SO WHAT DOES THAT MEAN? IF I TAKE BTO THE XTHAT'S GOING TO BE ETO THE KX. AND THAT'S BY THE LAW OF EXPONENTS THAT'S ETO THE KX. IF I COMPUTE ETO THE SOMETHING YOU CAN COMPUTE BTO THE SOMETHING BY USING IF I CAN FIGURE OUT THAT MAGIC NUMBER K. NOW WE WANT TO KNOW WHAT KIS. SO KIS WHAT? USING THE FUNCTION WE JUST DEFINED FOR THE
22 LAST FIVE MINUTES. IF BEQUALS ETO THE KKEQUALS NATURAL LOG OF B. THAT'S HOW YOU GET K. SO THIS IS EQUAL TO ENATURAL LOG OF BTIMES 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 NXLOG 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 EAND 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 XMEANS. WHATEVER NUMBERS WORKS THERE. NOW WHAT IS TEN? HOW DO I WRITE TEN ETO SOMETHING? WHAT GOES UNHERE? IF I JUST TEN TO ETO THE POWER, ANYBODY? NATURAL LOG OF TEN, THAT'S WHAT THE NATURAL LOG DOES. THAT'S X. AND THE OTHER WAY I CAN WRITE XIS ETO THE POWER NATURAL LOG OF X. THAT'S, SO THIS IS ETO THE POWER NATURAL LOG OF TEN, SOME NUMBER, TIMES LOG BASE TEN OF XWHICH IS THE THING I DON'T KNOW. AND I ALSO KNOW IT'S ETO THE NATURAL LOG OF
23 XBECAUSE THAT'S ANOTHER WAY TO WRITE X. SO IF THAT, IF ETO THIS NUMBER EQUALS ETO 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 ETO THE LOG? THESE TWO THINGS CANCEL. SO THIS IS JUST LOG X. MADE EGO 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 XAND 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 BOF X. SO THIS ONLY WORK IF BIS GREATER THAN ZERO. SO I HAVE THIS FUNCTION. I'M GOING TO WRITE ETO THE B. SO BIS GOING TO BE WRITTEN AS ETO THE LOG B. AND THIS IS GOING TO BE, WHICH IS XSO. THAT'S ETO THE LOG OF X. SO THAT EXPONENT AND
24 THIS EXPONENTS, THEY HAVE TO AGREE WITH ONE ANOTHER. SAME AS BEFORE. ETO THE THAT EQUALS ETO THE THAT. THESE TWO EXPONENT HAVE TO BE EQUAL. SO I LET LOG BTIMES LOG BOF X(ON BOARD). AGAIN REALLY SIMPLE RELATIONSHIP. I GET THAT THE LOG BASIC BOF XFOR ANY OLD BIS ONE OVER NATURAL LOG OF BTIMES 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.
25
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