| Start | okay so we're gonna figure out log base 5 of square root 125
this would be a minimum competence question
so this is where we expect you to all to get this right
well put it another way, if you cannot get this right
you should be concerned about your ability to do well in calculus
that's the point of the minimum competence questions.
log base 5 square root 125, you can set that to x |
| 0:35 | and that means
that 5 raised to the x
is the square root of 125
and then you look back and say well
can I make those the same base
sure. 125 is 5 ...
125 is 5 to what power? Brandon? the third, good!
|
| 1:01 | and this is the square root also known as
the 1/2 power
and what do you do with those two powers?
5 to the 3 to the 1/2 thats 5 to the 3/2 so therefore x is 3/2 so you can expect to see something like that. i mean not that particular problem maybe i dont know, i dont really remember but you should be able to do |
| 1:31 | some problem where we ask log base 8 0f 16 same kind of idea wha is the log base 8 of 16 take a sec to figure it out and then will move on with this stuff |
| 2:20 | this you can say 8 to the x is 16 now you try and find a common base 4 is not a common base |
| 2:32 | who said that 8 is 2 cubed 16 is 2 to the 4 that means 2 to the 3x equals 2 to the 4 x is 4/3 |
| 3:04 | any other questions before i do new stuff?
cause 5 to the 3 is 125, radical 125 quick question, yes? |
| 3:32 | how do you get 4/3 so once again log base 8, 16 equals x, means 8 to the x gives you 16 thats how you use a log 8 is 2 to the 3 and 16 is 2 to the 4 so you can say that 8 is 2 to the 3 to the x 16 is 2 to the 4 you can multiply those powers and you get 2 to the 3x equals 2 to the 4 that means that 3x has to equal 4 |
| 4:00 | 3x has to equal 4, so x has to equal 4/3 3x equals 4 x is 4/3 lets do another type of problem we didnt really go through these during class, so lets make sure you understand something like this |
| 5:00 | how would we do something like that
you have the log of 4x+1 minus the log of x-2
you can use the log laws
say thats the same as single log
remember we did that
thing where we right things as one log?
expanded out we put it together and get a single log so you have log of the first term minus log of the second term thats log of this divided by that |
| 5:38 | now using the definition
we can then say that 4x+!
over x-2 equals 2 raised to the 4 what does 2 to the 4 equals to? 16 make sure you can do your small powers 3 and 4 and 5 you can recognize them |
| 6:01 | cause if we put 64, we expect you the know thats 8 squared and 4 cubed so now you got 4x+1 over x mins 2 equals 16, so cross multiply |
| 6:31 | you get 4x+1 equals 16x minus 32 do a little rearranging and you get 33 equals 12x x is 33/12 and then you always have to check something when youre done when you do these log problems |
| 7:02 | you get x is 33/12
you have to take your answer
and just check
you plug it in
your not gonna take the log of a negative number
so you take 33/12
and put it in here
divide it by 4, i get 1 that would be positive
33/12
minus 2 is that gonna be positive?
2 is just 24 over 12 so thats an acceptable answer if i plug it in and i get and i look here and i say if i take the log of a negative number |
| 7:33 | i have to throw the answer out so sometimes you do one of these problems and youll end up with no solution or youll get 2 solutions and you have to throw out one of them well this could be negative its when you plue it in you can take the log of a negative number so for example if you had log of 4-x |
| 8:00 | x can be any number less than 4 x can be a negative number, so if you had 10 that wouldnt work because 4 minus 10 is -6 and you cant take the log of -6 ill do an example where you get an invalid solution that make sense? dont confuse what x is doesnt matter if x is positive or negative it matters if youre taking log of a positive or negative |
| 8:30 | right lets do another one of these to make sure everyone gets the idea lets do this one |
| 9:08 | i have log base 3 2x+1 minus log base 3 x+5 equals 2 so i use the log laws and i get log base 3 2x plus 1 over x+5 equals 2 |
| 9:35 | alright so 2x+1
over x+5
must equal
3 squared
cause i take that base and i raise it to the 2
so 3 squared is 9
now i got 2x+1
over x+5
equals 9. so you cross multiply and you get 2x+!
|
| 10:03 | is 9 times x plus 5 2x+1 is 9x + 45 so negative 4 is 7 x, 44 sorry x |
| 10:31 | is negative 44/7 if i take 2 times negative 44/7 and i add 1, i get a negative number if i take -44/7 and add 5 i get a negative number so theres no solution its not because this is negative because it makes this negative okay its like absolute value of x then that wouldnt be a problem, in fact in calculus |
| 11:01 | youll see you have to use the absolute value doesnt matter any of them are negative, one of them are negative one of them are positive if any of them are negative you have to throw the answer out because youre doing something with value if you plug it in and you are taking the log of negative so if you plug it in to this part that comes out negative then you throw out the answer |
| 11:32 | you cant take the log of a negative number so for example log base 3 x+5 has to be positive x has to be bigger then negative 5 if you have any answer less then negative 5, so negative one would work -1 to plus 5 is 4 thats fine negative 4 would work cause you would take log of 1 but if you get negative anything less than 5 you take the log of negative and thats no good little tricky, people have trouble with this kind of stuff |
| 12:02 | its not what the answer is, its what the log if it is okay lets do another one |
| 12:47 | so i can rewrite this as log base 5 of 16x-2/x-3 equals 2 that means 5 to the 2 equals this.. so 16 |
| 13:01 | x-2 over x-3 5 squared has to be 25 so then 16x-2 you cross multiply is 25 times x minus 3 |
| 13:30 | 16x-2 is 25x -75 r 73 is 9x x is 73/9 and then you can say if you take 73/9 subtract 3 and still get a positive number then if i take 16 times 73/9 you get a huge number minus 2 its still gonna be a positive number so either way im safe, so thats the answer |
| 14:05 | howd we do on this one? good?
ill give you another one can i cover this up? |
| 14:40 | how bout that?
alright so now we have addition so this time instead of dividing were gonna multiply were gonna have log base 4 x-10 times x+10' equals 3 |
| 15:08 | so thats log base 4 x-10, times x+10 is x squared - 100 thats the difference of 2 squares that equals 3 so that means 4 raised to the 3 is x squared minus 100 |
| 15:37 | and whats square root of 3 64 x squared is 164 and x would b plus or minus square root of 164 lets see can it be plus or minus the square root of 64 sure both of those, take away 10 and your fine |
| 16:02 | how bout minus the square root of 164
no thats gonna give you a problem right here
so
its just
the square root of 164
you see why i throw the one out?
okay now you guys try one |
| 17:05 | so first we multiply the 2 together first we multiply the 2 together we get log base 5 64-x squared equals 3 so that means 64- minus x squared has to equal 5 cubed |
| 17:32 | which is 125 but now we have a problem if we solve this we will get x squared equals -61 we cant get square root of negative number so youre done no solution so how many go there? good how many messed up and got 189 okay be careful lets try another |
| 19:01 | that works so we write this as a single log 9x squared -x+1 on top x squared minus 4 on the bottom equals 2 minus not plus you guys were multiplying that out |
| 19:32 | so that means 9x squared minus x + 1 over x squared minus 2 is 3 squared qhich is 9 and now you cross multiply you get 9x squared minus x plus 1 equals 9 times x squared minus 2 |
| 20:02 | 9x squared minus x plus 1 is nine x squared minus 18, 9x square cancel and you get 19 equals x now plug in 19 well 19 squared is a big positive number so we are okay 9 times 19 is a big positive number so we are okay so thats a valid answer remember when you are doing these |
| 20:31 | in algebra and we have square root problems and you solve it you get 2 answers and you throw one ou you forgot you lost points on your test your teacher said remember blah blah blah same kind of thing okay the key is when you plug in x you cant take the log of a negative number its not about what x is its what your taking the log of lets practice some other stuff |
| 21:00 | how bout one of those word problems
half life kind of things
we good with that?
i think thats everything thats gonna be on the exam |
| 21:50 | so suppose |
| 22:40 | so suppose roth pond has a various square inches of goo anyone gone to roth pond yet, lot of you are freshman its fun at the end of the year in may theres a thing called roth rogatta yyou race boats across the pond, you build boats from cardboard and duck tape |
| 23:00 | its very entertaining most people dont get very far some people get about that far you get in and it goes right down to the stone and theresz a lot of stuff in that pond so the surface of the pond has 200 inches of algae on it and 4 days later it has 600 square algae on it] how not very much cause its square inches |
| 23:32 | will there be on the pond after 10 days so you should be able to do something like this suppose a pond, whenever you see these problems the first thing you want to do is write this as an exponential equation |
| 24:06 | you want to say y=a times b to the x exponential equation and you have some information we know that time 0 200 square of algae and at time 4 theres 600 square inch of algae how much are we gonna have at time 10 if this were linear youd find the slope and put it in the equation of a line |
| 24:31 | and then you would solve
but this isnt linear its exponential
okay
at time 0
theres 200 square inch of algae
that means a is gonna be 200
youre gonna have y
equals 200
b to the x
you all rememberwhy that comes out to 200?
|
| 25:02 | this is a part 2 type of question or problem not problems you all understand that a is 200 okay so now we can say that 600 is 200 b to the 4th so that becomes 3 equals b to the 4 where b is |
| 25:30 | 3to the 1/4, so again we know initially were 200 square inch of algae and 4 days later were at 600 square algae so thats why we have these points y=a times b to the x so this when you hae the initial condition, when you have a time 0 this number i always a its always a because when we plug in 0, we get 1 and you solve for a |
| 26:00 | so now i go to this equation, i take a and replace it so now i have y=200 b to the x so ive gotten one of the two variables in the equation its like when you find the equation of a line, y=mx+b you have to find a and then b here you have to find a and then find b so now i use the second piece of information that when x is 4, y is 600 now youre just gonna have to solve for b so 600 is 200 b tot he 4th |
| 26:30 | divide by 200 and you get 3 equals b to the 4 take the 4th root and b is 3 to 1/4 so that tells me that my equation is y= 200 3 to the 1/4 to the x or y=200 |
| 27:00 | times 3 to the x/4 okay and thats because x times 1/4 gives youx/4 so now we just need to find out how much we have on day 10 should i bring this back down so you just plug in 10 and to anticipate your questions yes you can leave it like that because what is 3 to the 10/4 |
| 27:31 | i dont know right
its 9 radical 3
so thats 8ish
go the idea?
so we were able to do the pond algae problem okay next thing |
| 28:23 | 8 is 2 cubed so this 2 cubed, x+3 |
| 28:31 | 4 is 2 squared
2 squared to the 2x-1
so now we canmultiply the powers
you get 2 to the 3x+9
2 to the 4x-2
whcih means 3x+9 equasl 4x-2
and you solve that and you get x=11
anybody get 11?
|
| 29:01 | thats such a feeling of warmth could be heart burn but it feels just warm and squishy insides now lets make this a little harder |
| 29:35 | so that was 8 to the 2x-1
what if 8 to the x+3, just equals 11
how bout this one?
so now the problem is you cant take 11 and make it into similar base with 8 cause 8 is 2 to the something but 11 no base in common so we can take the log of both sides so you could take the log base 8 of this, theres 2 ways you can solve this |
| 30:02 | one you can take log base 8 8 to the x+3 equals log base 8 of 11 that would give you x plus 3 log base 8 of 11 x is log base 8 of 11 minus 3, thats one way to do it or you can do it the way i like to do it which is take the log of both sides |
| 30:31 | 8 x these are equivalent put the x+3 in front now you divide by log 8 and subtract 3 |
| 31:09 | howd you do on this one?
i can say this is an incompetence i would say this is an incompetence question maybe not, that definitely an incompetence question this maybe i just dont remember cause you kno |
| 31:30 | im old, i dont remember what i had for breakfast suppose to say youre not old to late okay now lets make this just a little bit harder i have no choice, i wish i didnt have to what if i give you that |
| 32:00 | lets do this one
so lets take the log of both sides
you get log to the 3 to the x+3
log
11 to the x
so far so good? got a couple credit points with that?
take the x+3 and put it in front |
| 32:30 | so far so good and distribute x log 8 plus 3 log 8 is x log 11 so now you look and say you have the x of the left and x on the right, what am i gonna do?' put all the terms of x on the same sides and put all the terms without x on the other side you get 3 log 8 |
| 33:02 | is x log 11 minus x log 8 groiup all the terms that contain an x on one side and all the terms that donnot contain an x on the other side then you can factor out x |
| 33:30 | 3log
8
over log 11-logg
8 and that equals x
anyone get that
anyone else manage to get that?
one person? wow okay so lets go through this again take the log of both sides you bring the power in front |
| 34:01 | the power is just x, x+3, so you just distribute and you get x tiimes log8 + 3 times log8 each of these is multiplied by log 8 now i want to isolate that the way you isolate this is always the same technique first you group the terms then you factor then you divide so you group everything that contains an x |
| 34:30 | goes on the right side, or the left it doesnt matter and anything that does not contain an x goes on the other sides then you factor out the x then you divide remember this technique, its very useful youll see it in calculus again lets do another one so you guys can get good at it |
| 35:17 | how bout that one?
alright take the log of both sides |
| 35:31 | by the way, can i take the natural log of both sides?
sure why not, i can do the log anyway i want i can do base 3, in fact ill do base 1 now ill put the 2x-1 in front and get x in front okay distribute 2x log 3 |
| 36:02 | minus log3 equals x log 45 now i will group the terms that contain x on one side and group the terms that do not contain x on the other side and it doesnt matter which one you do they come out the same so i get minus log 3 is x log 25 |
| 36:30 | -2xlog3 or i could put the xlog25 on this sides and the log 3 on the other side, that would be fine in the end theyll be identical factor out the x and divide |
| 37:09 | and you can run that through the calculator and get some number
alright compound interest proble
how do we do compound interest problems?
did i hear someone say horrible? thats really sad cause you got to go to the bank and put money in |
| 37:33 | some of you are gonna be business students |
| 39:04 | you deposit a 1000 bucks for 6 years at 12% how much would you have if the interest compound annually monthly continuously you deposit 1000 bucks for 6 years at 12% so if you dont remember the formula is f |
| 39:32 | equals p(1+r t the amount you have in the future is 1000 dollars times 1+.12 to the 6, thats the answer, thats all you need to do you dont know what 1.22 to the 6 is 6 years |
| 40:02 | and if i said monthly you take the interest rate and divide it by 12 because your gonna get 1/12 every month and youre gonna multiply the number of years by 12 because you can get 72 months instead of 6 years so thatd be 1000 1+ |
| 40:30 | .01 to the 72 and again theres nothing you can do with that, you can run it through a calculator certainly you wont have calculators so you dont how bout continuously? continuously is a different formula continuously compound formula is e so here it would be 1000 dollars |
| 41:00 | e
.12
times 6
so how can we make this just a little harder?
not a lot just a little suppose i gave you the exact same problem and i put the range back to 12% compound continuously, how long will it take to dounle? |
| 41:53 | so i started with 1000 dollars you use this formula |
| 42:02 | f=pe to the rt so i want to have 2000 dollars right now i have 1000 dollars plug everything in one of those first day of class problems maybe the second day divide by 1000 you get 2 |
| 42:34 | is e to the .12t and take the natural log of both sides .12t but and youre done, because again you dont know what the log of 2 divided by.12 is |
| 43:12 | they call it rule 72 the rule 72, you divide it by the interest rate so whats the natural log of 2/.12 natural log of 2 thats it |