WEBVTT
Kind: captions
Language: en
00:00:05.060 --> 00:00:08.660
okay so we're gonna figure out log base 5 of square root 125
00:00:08.660 --> 00:00:10.840
this would be a minimum competence question
00:00:10.840 --> 00:00:14.200
so this is where we expect you to all to get this right
00:00:15.200 --> 00:00:18.480
well put it another way, if you cannot get this right
00:00:18.480 --> 00:00:21.140
you should be concerned about your ability to do well in calculus
00:00:23.740 --> 00:00:25.020
that's the point of the minimum competence questions.
00:00:28.900 --> 00:00:31.940
log base 5 square root 125, you can set that to x
00:00:35.280 --> 00:00:36.160
and that means
00:00:37.240 --> 00:00:38.600
that 5 raised to the x
00:00:40.180 --> 00:00:41.780
is the square root of 125
00:00:46.000 --> 00:00:47.340
and then you look back and say well
00:00:47.860 --> 00:00:49.780
can I make those the same base
00:00:51.300 --> 00:00:52.100
sure. 125 is 5 ...
00:00:54.120 --> 00:00:55.240
125 is 5 to what power? Brandon? the third, good!
00:01:01.460 --> 00:01:04.100
and this is the square root also known as
00:01:06.600 --> 00:01:07.480
the 1/2 power
00:01:07.480 --> 00:01:09.640
and what do you do with those two powers?
00:01:10.420 --> 00:01:11.620
5 to the 3 to the 1/2
00:01:12.400 --> 00:01:13.520
thats 5 to the 3/2
00:01:16.320 --> 00:01:17.680
so therefore x is 3/2
00:01:20.520 --> 00:01:25.800
so you can expect to see something like that. i mean not that particular problem
00:01:25.920 --> 00:01:28.640
maybe i dont know, i dont really remember
00:01:29.800 --> 00:01:31.560
but you should be able to do
00:01:31.620 --> 00:01:33.300
some problem where we ask
00:01:35.980 --> 00:01:36.540
log base
00:01:37.620 --> 00:01:38.120
8 0f
00:01:41.440 --> 00:01:41.940
16
00:01:43.620 --> 00:01:44.740
same kind of idea
00:01:49.280 --> 00:01:50.880
wha is the log base 8 of 16
00:01:51.140 --> 00:01:52.820
take a sec to figure it out
00:01:52.840 --> 00:01:55.240
and then will move on with this stuff
00:02:20.140 --> 00:02:22.060
this you can say 8 to the x is 16
00:02:22.280 --> 00:02:24.440
now you try and find a common base
00:02:25.760 --> 00:02:27.120
4 is not a common base
00:02:32.160 --> 00:02:33.040
who said that
00:02:39.980 --> 00:02:40.700
8 is 2 cubed
00:02:44.280 --> 00:02:45.160
16 is 2 to the 4
00:02:48.020 --> 00:02:49.380
that means 2 to the 3x
00:02:51.740 --> 00:02:52.780
equals 2 to the 4
00:02:53.620 --> 00:02:54.120
x is 4/3
00:03:04.760 --> 00:03:07.560
any other questions before i do new stuff?
00:03:18.760 --> 00:03:21.000
cause 5 to the 3 is 125, radical 125
00:03:28.560 --> 00:03:30.000
quick question, yes?
00:03:32.320 --> 00:03:33.440
how do you get 4/3
00:03:34.420 --> 00:03:36.180
so once again log base 8, 16
00:03:36.660 --> 00:03:38.180
equals x, means 8 to the x
00:03:38.500 --> 00:03:39.300
gives you 16
00:03:39.420 --> 00:03:40.860
thats how you use a log
00:03:41.940 --> 00:03:42.740
8 is 2 to the 3
00:03:42.900 --> 00:03:44.020
and 16 is 2 to the 4
00:03:45.380 --> 00:03:47.860
so you can say that 8 is 2 to the 3 to the x
00:03:49.220 --> 00:03:50.100
16 is 2 to the 4
00:03:50.100 --> 00:03:51.700
you can multiply those powers
00:03:52.000 --> 00:03:54.400
and you get 2 to the 3x equals 2 to the 4
00:03:55.180 --> 00:03:56.380
that means that 3x
00:03:57.220 --> 00:03:58.100
has to equal 4
00:04:00.620 --> 00:04:03.020
3x has to equal 4, so x has to equal 4/3
00:04:08.060 --> 00:04:08.780
3x equals 4
00:04:11.960 --> 00:04:12.460
x is 4/3
00:04:20.840 --> 00:04:22.920
lets do another type of problem
00:04:25.340 --> 00:04:29.640
we didnt really go through these during class, so lets make sure you understand something like this
00:05:00.260 --> 00:05:02.580
how would we do something like that
00:05:04.840 --> 00:05:07.640
you have the log of 4x+1 minus the log of x-2
00:05:08.920 --> 00:05:10.440
you can use the log laws
00:05:12.380 --> 00:05:14.460
say thats the same as single log
00:05:18.120 --> 00:05:19.480
remember we did that
00:05:19.480 --> 00:05:21.480
thing where we right things as one log?
00:05:21.480 --> 00:05:24.660
expanded out we put it together and get a single log
00:05:24.660 --> 00:05:27.880
so you have log of the first term minus log of the second term
00:05:28.060 --> 00:05:30.220
thats log of this divided by that
00:05:38.160 --> 00:05:39.840
now using the definition
00:05:40.920 --> 00:05:42.520
we can then say that 4x+!
00:05:44.720 --> 00:05:45.280
over x-2
00:05:46.400 --> 00:05:47.920
equals 2 raised to the 4
00:05:51.640 --> 00:05:53.560
what does 2 to the 4 equals to?
00:05:55.360 --> 00:05:55.860
16
00:05:55.860 --> 00:05:57.800
make sure you can do your small powers
00:05:58.720 --> 00:06:00.960
3 and 4 and 5 you can recognize them
00:06:01.700 --> 00:06:04.740
cause if we put 64, we expect you the know thats
00:06:05.440 --> 00:06:06.800
8 squared and 4 cubed
00:06:10.500 --> 00:06:11.380
so now you got
00:06:13.120 --> 00:06:13.620
4x+1
00:06:14.920 --> 00:06:15.720
over x mins 2
00:06:16.600 --> 00:06:18.520
equals 16, so cross multiply
00:06:31.840 --> 00:06:32.640
you get 4x+1
00:06:34.640 --> 00:06:35.360
equals 16x
00:06:36.700 --> 00:06:37.260
minus 32
00:06:44.200 --> 00:06:45.800
do a little rearranging
00:06:47.800 --> 00:06:49.400
and you get 33 equals 12x
00:06:53.500 --> 00:06:54.140
x is 33/12
00:06:55.220 --> 00:06:59.140
and then you always have to check something when youre done
00:06:59.140 --> 00:07:00.840
when you do these log problems
00:07:02.320 --> 00:07:03.440
you get x is 33/12
00:07:03.440 --> 00:07:04.940
you have to take your answer
00:07:05.360 --> 00:07:06.320
and just check
00:07:07.080 --> 00:07:07.960
you plug it in
00:07:07.960 --> 00:07:10.900
your not gonna take the log of a negative number
00:07:11.860 --> 00:07:12.980
so you take 33/12
00:07:13.620 --> 00:07:14.740
and put it in here
00:07:15.220 --> 00:07:18.100
divide it by 4, i get 1 that would be positive
00:07:19.220 --> 00:07:19.720
33/12
00:07:19.720 --> 00:07:21.960
minus 2 is that gonna be positive?
00:07:23.760 --> 00:07:24.960
2 is just 24 over 12
00:07:24.960 --> 00:07:26.820
so thats an acceptable answer
00:07:27.840 --> 00:07:28.720
if i plug it in
00:07:29.200 --> 00:07:29.760
and i get
00:07:29.800 --> 00:07:33.720
and i look here and i say if i take the log of a negative number
00:07:33.720 --> 00:07:35.200
i have to throw the answer out
00:07:35.200 --> 00:07:37.420
so sometimes you do one of these problems
00:07:37.420 --> 00:07:39.300
and youll end up with no solution
00:07:39.300 --> 00:07:43.060
or youll get 2 solutions and you have to throw out one of them
00:07:49.980 --> 00:07:51.820
well this could be negative
00:07:51.820 --> 00:07:54.900
its when you plue it in you can take the log of a negative number
00:07:55.380 --> 00:07:57.380
so for example if you had log of
00:07:58.420 --> 00:07:58.920
4-x
00:08:00.600 --> 00:08:02.520
x can be any number less than 4
00:08:03.460 --> 00:08:06.180
x can be a negative number, so if you had 10
00:08:06.980 --> 00:08:09.700
that wouldnt work because 4 minus 10 is -6
00:08:09.700 --> 00:08:11.620
and you cant take the log of -6
00:08:13.500 --> 00:08:16.860
ill do an example where you get an invalid solution
00:08:19.760 --> 00:08:22.320
that make sense? dont confuse what x is
00:08:22.320 --> 00:08:24.700
doesnt matter if x is positive or negative
00:08:24.700 --> 00:08:27.080
it matters if youre taking log of a positive or negative
00:08:30.060 --> 00:08:32.340
right lets do another one of these to make sure everyone gets the idea
00:08:59.820 --> 00:09:00.860
lets do this one
00:09:08.780 --> 00:09:10.540
i have log base 3 2x+1 minus
00:09:12.360 --> 00:09:13.240
log base 3 x+5
00:09:14.500 --> 00:09:15.060
equals 2
00:09:15.120 --> 00:09:16.400
so i use the log laws
00:09:17.360 --> 00:09:18.560
and i get log base 3
00:09:19.900 --> 00:09:20.460
2x plus 1
00:09:22.200 --> 00:09:22.760
over x+5
00:09:25.460 --> 00:09:26.020
equals 2
00:09:35.520 --> 00:09:36.560
alright so 2x+1
00:09:38.300 --> 00:09:38.860
over x+5
00:09:40.080 --> 00:09:40.800
must equal
00:09:41.880 --> 00:09:42.520
3 squared
00:09:42.520 --> 00:09:44.920
cause i take that base and i raise it to the 2
00:09:52.000 --> 00:09:53.040
so 3 squared is 9
00:09:54.040 --> 00:09:54.920
now i got 2x+1
00:09:56.620 --> 00:09:57.180
over x+5
00:09:58.440 --> 00:10:01.560
equals 9. so you cross multiply and you get 2x+!
00:10:03.320 --> 00:10:04.440
is 9 times x plus 5
00:10:14.540 --> 00:10:15.040
2x+1
00:10:17.860 --> 00:10:18.360
is 9x
00:10:19.620 --> 00:10:20.120
+ 45
00:10:22.480 --> 00:10:23.600
so negative 4 is 7
00:10:25.040 --> 00:10:25.760
x, 44 sorry
00:10:27.020 --> 00:10:27.520
x
00:10:31.000 --> 00:10:32.120
is negative 44/7
00:10:34.780 --> 00:10:36.780
if i take 2 times negative 44/7
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and i add 1, i get a negative number
00:10:40.360 --> 00:10:43.400
if i take -44/7 and add 5 i get a negative number
00:10:43.600 --> 00:10:45.040
so theres no solution
00:10:52.420 --> 00:10:54.580
its not because this is negative
00:10:54.640 --> 00:10:56.720
because it makes this negative
00:10:57.220 --> 00:10:59.380
okay its like absolute value of x
00:10:59.380 --> 00:11:01.540
then that wouldnt be a problem, in fact in calculus
00:11:01.540 --> 00:11:03.940
youll see you have to use the absolute value
00:11:03.940 --> 00:11:09.260
doesnt matter any of them are negative, one of them are negative one of them are positive
00:11:09.260 --> 00:11:11.840
if any of them are negative you have to throw the answer out
00:11:11.840 --> 00:11:13.940
because youre doing something with value
00:11:19.860 --> 00:11:20.900
if you plug it in
00:11:20.900 --> 00:11:22.820
and you are taking the log of negative
00:11:23.740 --> 00:11:25.740
so if you plug it in to this part
00:11:26.180 --> 00:11:27.780
that comes out negative
00:11:27.780 --> 00:11:29.360
then you throw out the answer
00:11:32.040 --> 00:11:34.760
you cant take the log of a negative number
00:11:34.920 --> 00:11:35.880
so for example
00:11:36.980 --> 00:11:37.860
log base 3 x+5
00:11:37.860 --> 00:11:38.960
has to be positive
00:11:39.100 --> 00:11:41.260
x has to be bigger then negative 5
00:11:41.340 --> 00:11:45.340
if you have any answer less then negative 5, so negative one would work
00:11:46.460 --> 00:11:48.140
-1 to plus 5 is 4 thats fine
00:11:49.640 --> 00:11:51.280
negative 4 would work cause you would take log of 1
00:11:52.020 --> 00:11:54.900
but if you get negative anything less than 5
00:11:54.900 --> 00:11:57.080
you take the log of negative and thats no good
00:11:59.820 --> 00:12:02.120
little tricky, people have trouble with this kind of stuff
00:12:02.820 --> 00:12:06.100
its not what the answer is, its what the log if it is
00:12:07.420 --> 00:12:09.020
okay lets do another one
00:12:47.040 --> 00:12:49.360
so i can rewrite this as log base 5 of
00:12:52.840 --> 00:12:53.560
16x-2/x-3
00:12:55.060 --> 00:12:55.620
equals 2
00:12:58.500 --> 00:13:01.060
that means 5 to the 2 equals this.. so 16
00:13:01.920 --> 00:13:02.420
x-2
00:13:04.420 --> 00:13:04.980
over x-3
00:13:07.660 --> 00:13:09.020
5 squared has to be 25
00:13:15.260 --> 00:13:17.420
so then 16x-2 you cross multiply
00:13:22.060 --> 00:13:22.560
is 25
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times x minus 3
00:13:30.740 --> 00:13:31.240
16x-2
00:13:34.280 --> 00:13:34.780
is 25x
00:13:36.100 --> 00:13:36.600
-75
00:13:39.580 --> 00:13:40.140
r 73 is 9x
00:13:43.120 --> 00:13:43.680
x is 73/9
00:13:46.520 --> 00:13:48.840
and then you can say if you take 73/9
00:13:48.840 --> 00:13:50.780
subtract 3 and still get a positive number
00:13:51.720 --> 00:13:54.840
then if i take 16 times 73/9 you get a huge number
00:13:54.840 --> 00:13:57.380
minus 2 its still gonna be a positive number
00:13:57.460 --> 00:14:00.180
so either way im safe, so thats the answer
00:14:05.880 --> 00:14:07.720
howd we do on this one? good?
00:14:10.080 --> 00:14:11.680
ill give you another one
00:14:13.120 --> 00:14:14.400
can i cover this up?
00:14:40.400 --> 00:14:41.360
how bout that?
00:14:43.140 --> 00:14:45.220
alright so now we have addition
00:14:45.220 --> 00:14:47.760
so this time instead of dividing were gonna multiply
00:14:47.880 --> 00:14:49.560
were gonna have log base 4
00:14:50.800 --> 00:14:51.300
x-10
00:14:52.360 --> 00:14:53.160
times x+10'
00:14:55.580 --> 00:14:56.140
equals 3
00:15:08.680 --> 00:15:09.880
so thats log base 4
00:15:11.620 --> 00:15:13.860
x-10, times x+10 is x squared - 100
00:15:15.960 --> 00:15:18.200
thats the difference of 2 squares
00:15:19.000 --> 00:15:19.880
that equals 3
00:15:25.680 --> 00:15:26.640
so that means 4
00:15:27.420 --> 00:15:28.380
raised to the 3
00:15:28.400 --> 00:15:29.840
is x squared minus 100
00:15:37.560 --> 00:15:39.240
and whats square root of 3
00:15:40.100 --> 00:15:40.600
64
00:15:43.940 --> 00:15:44.580
x squared
00:15:47.120 --> 00:15:47.620
is 164
00:15:48.640 --> 00:15:49.440
and x would b
00:15:50.300 --> 00:15:51.180
plus or minus
00:15:52.360 --> 00:15:53.560
square root of 164
00:15:53.920 --> 00:15:57.280
lets see can it be plus or minus the square root of 64
00:15:57.700 --> 00:15:58.200
sure
00:15:58.840 --> 00:16:01.480
both of those, take away 10 and your fine
00:16:02.020 --> 00:16:04.420
how bout minus the square root of 164
00:16:04.420 --> 00:16:06.120
no thats gonna give you a problem right here
00:16:06.540 --> 00:16:07.040
so
00:16:08.300 --> 00:16:08.860
its just
00:16:10.740 --> 00:16:12.180
the square root of 164
00:16:21.000 --> 00:16:23.000
you see why i throw the one out?
00:16:26.880 --> 00:16:28.480
okay now you guys try one
00:17:05.980 --> 00:17:08.300
so first we multiply the 2 together
00:17:08.300 --> 00:17:10.340
first we multiply the 2 together
00:17:12.560 --> 00:17:13.600
we get log base 5
00:17:15.620 --> 00:17:16.500
64-x squared
00:17:17.600 --> 00:17:18.160
equals 3
00:17:22.540 --> 00:17:23.420
so that means
00:17:26.240 --> 00:17:27.520
64- minus x squared
00:17:28.300 --> 00:17:29.580
has to equal 5 cubed
00:17:32.080 --> 00:17:32.880
which is 125
00:17:33.120 --> 00:17:34.720
but now we have a problem
00:17:37.400 --> 00:17:40.440
if we solve this we will get x squared equals -61
00:17:40.440 --> 00:17:43.140
we cant get square root of negative number
00:17:43.840 --> 00:17:44.720
so youre done
00:17:45.720 --> 00:17:46.520
no solution
00:17:48.140 --> 00:17:49.820
so how many go there? good
00:17:51.380 --> 00:17:53.300
how many messed up and got 189
00:17:54.340 --> 00:17:54.840
okay
00:17:56.760 --> 00:17:58.600
be careful lets try another
00:19:01.340 --> 00:19:02.060
that works
00:19:04.280 --> 00:19:06.280
so we write this as a single log
00:19:07.840 --> 00:19:08.560
9x squared
00:19:09.740 --> 00:19:10.460
-x+1 on top
00:19:11.400 --> 00:19:13.400
x squared minus 4 on the bottom
00:19:15.500 --> 00:19:16.060
equals 2
00:19:16.680 --> 00:19:17.640
minus not plus
00:19:17.640 --> 00:19:19.760
you guys were multiplying that out
00:19:32.280 --> 00:19:33.880
so that means 9x squared
00:19:34.980 --> 00:19:35.620
minus x + 1
00:19:36.460 --> 00:19:37.900
over x squared minus 2
00:19:39.160 --> 00:19:40.600
is 3 squared qhich is 9
00:19:46.700 --> 00:19:48.460
and now you cross multiply
00:19:49.500 --> 00:19:50.700
you get 9x squared
00:19:51.660 --> 00:19:52.540
minus x plus 1
00:19:53.680 --> 00:19:54.640
equals 9 times
00:19:55.680 --> 00:19:56.800
x squared minus 2
00:20:02.140 --> 00:20:02.860
9x squared
00:20:04.340 --> 00:20:05.220
minus x plus 1
00:20:05.760 --> 00:20:06.880
is nine x squared
00:20:07.920 --> 00:20:09.680
minus 18, 9x square cancel
00:20:13.740 --> 00:20:15.180
and you get 19 equals x
00:20:16.720 --> 00:20:17.600
now plug in 19
00:20:17.740 --> 00:20:21.260
well 19 squared is a big positive number so we are okay
00:20:21.840 --> 00:20:24.960
9 times 19 is a big positive number so we are okay
00:20:24.960 --> 00:20:26.260
so thats a valid answer
00:20:28.820 --> 00:20:31.060
remember when you are doing these
00:20:31.060 --> 00:20:33.200
in algebra and we have square root problems
00:20:33.200 --> 00:20:35.800
and you solve it you get 2 answers and you throw one ou
00:20:36.100 --> 00:20:38.660
you forgot you lost points on your test
00:20:38.660 --> 00:20:40.780
your teacher said remember blah blah blah
00:20:41.260 --> 00:20:42.780
same kind of thing okay
00:20:43.860 --> 00:20:45.620
the key is when you plug in x
00:20:45.620 --> 00:20:48.200
you cant take the log of a negative number
00:20:48.200 --> 00:20:51.520
its not about what x is its what your taking the log of
00:20:56.840 --> 00:20:58.920
lets practice some other stuff
00:21:00.280 --> 00:21:02.600
how bout one of those word problems
00:21:02.640 --> 00:21:04.240
half life kind of things
00:21:04.940 --> 00:21:06.140
we good with that?
00:21:16.420 --> 00:21:19.780
i think thats everything thats gonna be on the exam
00:21:50.460 --> 00:21:51.180
so suppose
00:22:40.500 --> 00:22:44.100
so suppose roth pond has a various square inches of goo
00:22:47.040 --> 00:22:50.960
anyone gone to roth pond yet, lot of you are freshman its fun
00:22:51.360 --> 00:22:55.440
at the end of the year in may theres a thing called roth rogatta
00:22:55.640 --> 00:22:59.880
yyou race boats across the pond, you build boats from cardboard
00:22:59.880 --> 00:23:00.640
and duck tape
00:23:00.640 --> 00:23:03.940
its very entertaining most people dont get very far
00:23:04.680 --> 00:23:06.680
some people get about that far
00:23:07.860 --> 00:23:10.740
you get in and it goes right down to the stone
00:23:13.500 --> 00:23:15.980
and theresz a lot of stuff in that pond
00:23:17.920 --> 00:23:21.440
so the surface of the pond has 200 inches of algae on it
00:23:22.200 --> 00:23:25.160
and 4 days later it has 600 square algae on it]
00:23:27.520 --> 00:23:28.020
how
00:23:28.020 --> 00:23:30.440
not very much cause its square inches
00:23:32.360 --> 00:23:33.080
will there
00:23:36.420 --> 00:23:36.920
be on
00:23:41.460 --> 00:23:42.900
the pond after 10 days
00:23:44.660 --> 00:23:47.700
so you should be able to do something like this
00:23:53.860 --> 00:23:57.060
suppose a pond, whenever you see these problems
00:23:57.060 --> 00:24:01.480
the first thing you want to do is write this as an exponential equation
00:24:06.840 --> 00:24:09.000
you want to say y=a times b to the x
00:24:09.120 --> 00:24:10.640
exponential equation
00:24:12.180 --> 00:24:14.180
and you have some information
00:24:14.700 --> 00:24:15.900
we know that time 0
00:24:17.420 --> 00:24:18.700
200 square of algae
00:24:19.820 --> 00:24:20.620
and at time 4
00:24:21.040 --> 00:24:23.120
theres 600 square inch of algae
00:24:23.540 --> 00:24:25.220
how much are we gonna have
00:24:26.160 --> 00:24:26.800
at time 10
00:24:26.840 --> 00:24:29.400
if this were linear youd find the slope
00:24:29.400 --> 00:24:31.420
and put it in the equation of a line
00:24:31.420 --> 00:24:32.940
and then you would solve
00:24:33.320 --> 00:24:35.800
but this isnt linear its exponential
00:24:38.720 --> 00:24:39.220
okay
00:24:42.400 --> 00:24:42.960
at time 0
00:24:42.960 --> 00:24:44.960
theres 200 square inch of algae
00:24:45.580 --> 00:24:47.340
that means a is gonna be 200
00:24:48.120 --> 00:24:49.320
youre gonna have y
00:24:49.740 --> 00:24:50.460
equals 200
00:24:52.720 --> 00:24:53.280
b to the x
00:24:57.400 --> 00:25:00.200
you all rememberwhy that comes out to 200?
00:25:02.840 --> 00:25:04.920
this is a part 2 type of question
00:25:05.480 --> 00:25:07.080
or problem not problems
00:25:09.940 --> 00:25:12.020
you all understand that a is 200
00:25:13.820 --> 00:25:14.540
okay so now
00:25:15.660 --> 00:25:16.860
we can say that 600
00:25:18.260 --> 00:25:18.760
is 200
00:25:20.800 --> 00:25:21.520
b to the 4th
00:25:22.420 --> 00:25:23.460
so that becomes
00:25:24.900 --> 00:25:26.020
3 equals b to the 4
00:25:27.120 --> 00:25:27.760
where b is
00:25:30.060 --> 00:25:30.940
3to the 1/4, so again
00:25:32.520 --> 00:25:35.640
we know initially were 200 square inch of algae
00:25:36.260 --> 00:25:38.900
and 4 days later were at 600 square algae
00:25:38.900 --> 00:25:40.300
so thats why we have these points
00:25:41.960 --> 00:25:43.160
y=a times b to the x
00:25:44.160 --> 00:25:44.660
so this
00:25:44.860 --> 00:25:48.140
when you hae the initial condition, when you have
00:25:49.280 --> 00:25:49.780
a time 0
00:25:50.560 --> 00:25:52.000
this number i always a
00:25:54.420 --> 00:25:57.380
its always a because when we plug in 0, we get 1
00:25:57.560 --> 00:25:58.760
and you solve for a
00:26:00.160 --> 00:26:03.440
so now i go to this equation, i take a and replace it
00:26:04.660 --> 00:26:07.300
so now i have y=200 b to the x so ive gotten
00:26:07.300 --> 00:26:09.360
one of the two variables in the equation
00:26:09.360 --> 00:26:12.020
its like when you find the equation of a line, y=mx+b
00:26:12.580 --> 00:26:14.340
you have to find a and then b
00:26:14.480 --> 00:26:16.880
here you have to find a and then find b
00:26:17.640 --> 00:26:20.520
so now i use the second piece of information
00:26:23.320 --> 00:26:24.840
that when x is 4, y is 600
00:26:24.840 --> 00:26:26.680
now youre just gonna have to solve for b
00:26:27.900 --> 00:26:29.420
so 600 is 200 b tot he 4th
00:26:30.440 --> 00:26:33.160
divide by 200 and you get 3 equals b to the 4
00:26:34.360 --> 00:26:35.480
take the 4th root
00:26:36.340 --> 00:26:37.300
and b is 3 to 1/4
00:26:38.600 --> 00:26:41.080
so that tells me that my equation is y=
00:26:41.600 --> 00:26:42.100
200
00:26:43.500 --> 00:26:44.000
3
00:26:47.180 --> 00:26:47.820
to the 1/4
00:26:50.100 --> 00:26:50.740
to the x or
00:26:58.420 --> 00:26:58.920
y=200
00:27:00.240 --> 00:27:00.740
times 3
00:27:02.020 --> 00:27:02.660
to the x/4
00:27:03.240 --> 00:27:04.760
okay and thats because
00:27:06.140 --> 00:27:06.860
x times 1/4
00:27:08.620 --> 00:27:09.500
gives youx/4
00:27:09.500 --> 00:27:11.900
so now we just need to find out how much we have on day 10
00:27:12.800 --> 00:27:14.720
should i bring this back down
00:27:17.580 --> 00:27:18.940
so you just plug in 10
00:27:22.580 --> 00:27:24.820
and to anticipate your questions
00:27:24.820 --> 00:27:26.400
yes you can leave it like that
00:27:27.220 --> 00:27:29.060
because what is 3 to the 10/4
00:27:31.720 --> 00:27:32.840
i dont know right
00:27:37.400 --> 00:27:38.360
its 9 radical 3
00:27:42.760 --> 00:27:43.640
so thats 8ish
00:27:48.500 --> 00:27:49.300
go the idea?
00:27:49.720 --> 00:27:52.520
so we were able to do the pond algae problem
00:27:57.660 --> 00:27:58.700
okay next thing
00:28:23.120 --> 00:28:24.320
8 is 2 cubed so this
00:28:29.020 --> 00:28:29.820
2 cubed, x+3
00:28:31.140 --> 00:28:32.020
4 is 2 squared
00:28:35.800 --> 00:28:37.160
2 squared to the 2x-1
00:28:39.100 --> 00:28:41.260
so now we canmultiply the powers
00:28:42.640 --> 00:28:43.920
you get 2 to the 3x+9
00:28:46.620 --> 00:28:47.420
2 to the 4x-2
00:28:49.560 --> 00:28:51.480
whcih means 3x+9 equasl 4x-2
00:28:56.900 --> 00:28:58.680
and you solve that and you get x=11
00:28:59.620 --> 00:29:00.120
anybody get 11?
00:29:01.400 --> 00:29:03.400
thats such a feeling of warmth
00:29:04.580 --> 00:29:06.100
could be heart burn but
00:29:06.100 --> 00:29:08.380
it feels just warm and squishy insides
00:29:12.980 --> 00:29:15.220
now lets make this a little harder
00:29:35.180 --> 00:29:36.700
so that was 8 to the 2x-1
00:29:37.920 --> 00:29:40.160
what if 8 to the x+3, just equals 11
00:29:40.920 --> 00:29:42.120
how bout this one?
00:29:43.640 --> 00:29:46.040
so now the problem is you cant take 11
00:29:47.000 --> 00:29:49.320
and make it into similar base with 8
00:29:49.680 --> 00:29:51.920
cause 8 is 2 to the something but 11
00:29:52.700 --> 00:29:53.820
no base in common
00:29:53.820 --> 00:29:55.800
so we can take the log of both sides
00:29:56.620 --> 00:29:58.620
so you could take the log base 8
00:29:58.680 --> 00:30:01.320
of this, theres 2 ways you can solve this
00:30:02.040 --> 00:30:03.720
one you can take log base 8
00:30:05.640 --> 00:30:06.360
8 to the x+3
00:30:07.660 --> 00:30:08.780
equals log base 8
00:30:09.460 --> 00:30:09.960
of 11
00:30:10.600 --> 00:30:12.360
that would give you x plus 3
00:30:12.760 --> 00:30:13.260
log
00:30:14.640 --> 00:30:15.360
base 8 of 11
00:30:18.600 --> 00:30:19.800
x is log base 8 of 11
00:30:20.500 --> 00:30:22.420
minus 3, thats one way to do it
00:30:22.420 --> 00:30:24.700
or you can do it the way i like to do it
00:30:25.880 --> 00:30:28.120
which is take the log of both sides
00:30:31.180 --> 00:30:31.680
8 x
00:30:37.640 --> 00:30:39.080
these are equivalent
00:30:40.160 --> 00:30:41.440
put the x+3 in front
00:30:49.900 --> 00:30:51.340
now you divide by log 8
00:30:59.480 --> 00:31:00.440
and subtract 3
00:31:09.880 --> 00:31:11.400
howd you do on this one?
00:31:15.400 --> 00:31:17.560
i can say this is an incompetence
00:31:19.140 --> 00:31:22.100
i would say this is an incompetence question
00:31:23.400 --> 00:31:26.360
maybe not, that definitely an incompetence question
00:31:26.680 --> 00:31:27.400
this maybe
00:31:27.620 --> 00:31:28.980
i just dont remember
00:31:29.640 --> 00:31:30.520
cause you kno
00:31:30.780 --> 00:31:33.900
im old, i dont remember what i had for breakfast
00:31:35.780 --> 00:31:37.620
suppose to say youre not old
00:31:38.480 --> 00:31:38.980
to late
00:31:40.060 --> 00:31:43.180
okay now lets make this just a little bit harder
00:31:45.660 --> 00:31:48.140
i have no choice, i wish i didnt have to
00:31:56.920 --> 00:31:58.360
what if i give you that
00:32:00.060 --> 00:32:01.100
lets do this one
00:32:02.320 --> 00:32:04.480
so lets take the log of both sides
00:32:07.200 --> 00:32:09.040
you get log to the 3 to the x+3
00:32:10.040 --> 00:32:10.540
log
00:32:12.260 --> 00:32:12.900
11 to the x
00:32:14.760 --> 00:32:18.200
so far so good? got a couple credit points with that?
00:32:19.080 --> 00:32:21.080
take the x+3 and put it in front
00:32:30.160 --> 00:32:31.040
so far so good
00:32:33.100 --> 00:32:34.140
and distribute
00:32:36.940 --> 00:32:37.440
x log 8
00:32:39.160 --> 00:32:39.800
plus 3 log
00:32:40.160 --> 00:32:40.660
8
00:32:41.500 --> 00:32:42.000
is x
00:32:44.100 --> 00:32:44.600
log 11
00:32:46.420 --> 00:32:46.920
so now
00:32:46.920 --> 00:32:51.260
you look and say you have the x of the left and x on the right, what am i gonna do?'
00:32:51.380 --> 00:32:53.860
put all the terms of x on the same sides
00:32:53.940 --> 00:32:55.940
and put all the terms without x
00:32:55.940 --> 00:32:56.940
on the other side
00:32:58.960 --> 00:32:59.840
you get 3 log 8
00:33:02.660 --> 00:33:03.300
is x log 11
00:33:04.580 --> 00:33:05.080
minus
00:33:06.540 --> 00:33:07.040
x log 8
00:33:09.520 --> 00:33:12.720
groiup all the terms that contain an x on one side
00:33:12.720 --> 00:33:16.080
and all the terms that donnot contain an x on the other side
00:33:16.760 --> 00:33:18.360
then you can factor out x
00:33:30.980 --> 00:33:31.480
3log
00:33:31.760 --> 00:33:32.260
8
00:33:35.500 --> 00:33:36.620
over log 11-logg
00:33:38.120 --> 00:33:39.320
8 and that equals x
00:33:41.920 --> 00:33:42.960
anyone get that
00:33:45.020 --> 00:33:47.100
anyone else manage to get that?
00:33:47.720 --> 00:33:48.760
one person? wow
00:33:54.000 --> 00:33:56.240
okay so lets go through this again
00:33:56.240 --> 00:33:57.840
take the log of both sides
00:33:58.400 --> 00:34:00.240
you bring the power in front
00:34:01.060 --> 00:34:04.180
the power is just x, x+3, so you just distribute
00:34:04.600 --> 00:34:07.080
and you get x tiimes log8 + 3 times log8
00:34:07.080 --> 00:34:09.260
each of these is multiplied by log 8
00:34:11.360 --> 00:34:13.040
now i want to isolate that
00:34:13.040 --> 00:34:15.720
the way you isolate this is always the same technique
00:34:15.840 --> 00:34:17.520
first you group the terms
00:34:20.440 --> 00:34:21.480
then you factor
00:34:23.400 --> 00:34:24.440
then you divide
00:34:27.840 --> 00:34:30.640
so you group everything that contains an x
00:34:30.640 --> 00:34:33.000
goes on the right side, or the left it doesnt matter
00:34:33.000 --> 00:34:35.720
and anything that does not contain an x goes on the other sides
00:34:37.520 --> 00:34:39.120
then you factor out the x
00:34:40.480 --> 00:34:41.520
then you divide
00:34:42.500 --> 00:34:47.300
remember this technique, its very useful youll see it in calculus again
00:34:49.780 --> 00:34:52.900
lets do another one so you guys can get good at it
00:35:17.040 --> 00:35:18.240
how bout that one?
00:35:21.360 --> 00:35:23.600
alright take the log of both sides
00:35:31.820 --> 00:35:35.180
by the way, can i take the natural log of both sides?
00:35:36.400 --> 00:35:39.120
sure why not, i can do the log anyway i want
00:35:40.260 --> 00:35:42.500
i can do base 3, in fact ill do base 1
00:35:43.360 --> 00:35:45.200
now ill put the 2x-1 in front
00:35:49.060 --> 00:35:50.180
and get x in front
00:35:56.160 --> 00:35:57.280
okay distribute
00:35:59.880 --> 00:36:00.380
2x log 3
00:36:02.080 --> 00:36:02.800
minus log3
00:36:05.480 --> 00:36:06.280
equals x log
00:36:08.060 --> 00:36:08.560
45
00:36:09.040 --> 00:36:12.400
now i will group the terms that contain x on one side
00:36:12.400 --> 00:36:15.480
and group the terms that do not contain x on the other side
00:36:15.760 --> 00:36:18.160
and it doesnt matter which one you do
00:36:18.580 --> 00:36:20.020
they come out the same
00:36:21.720 --> 00:36:22.920
so i get minus log 3
00:36:25.580 --> 00:36:26.220
is x log 25
00:36:30.320 --> 00:36:30.880
-2xlog3
00:36:32.600 --> 00:36:33.100
or
00:36:33.500 --> 00:36:35.820
i could put the xlog25 on this sides
00:36:35.820 --> 00:36:38.260
and the log 3 on the other side, that would be fine
00:36:39.800 --> 00:36:41.800
in the end theyll be identical
00:36:42.720 --> 00:36:43.760
factor out the x
00:36:56.300 --> 00:36:57.020
and divide
00:37:09.580 --> 00:37:13.740
and you can run that through the calculator and get some number
00:37:14.480 --> 00:37:16.800
alright compound interest proble
00:37:16.800 --> 00:37:19.500
how do we do compound interest problems?
00:37:23.380 --> 00:37:25.540
did i hear someone say horrible?
00:37:28.160 --> 00:37:32.240
thats really sad cause you got to go to the bank and put money in
00:37:33.500 --> 00:37:36.300
some of you are gonna be business students
00:39:04.340 --> 00:39:05.940
you deposit a 1000 bucks
00:39:06.500 --> 00:39:07.220
for 6 years
00:39:08.140 --> 00:39:08.640
at 12%
00:39:08.840 --> 00:39:10.360
how much would you have
00:39:10.640 --> 00:39:12.320
if the interest compound
00:39:13.760 --> 00:39:14.960
annually monthly
00:39:15.200 --> 00:39:16.160
continuously
00:39:19.780 --> 00:39:22.420
you deposit 1000 bucks for 6 years at 12%
00:39:22.420 --> 00:39:23.700
so if you dont remember
00:39:27.400 --> 00:39:28.440
the formula is f
00:39:32.620 --> 00:39:33.500
equals p(1+r
00:39:34.660 --> 00:39:35.160
t
00:39:38.500 --> 00:39:40.660
the amount you have in the future
00:39:42.480 --> 00:39:43.520
is 1000 dollars
00:39:46.740 --> 00:39:47.540
times 1+.12
00:39:47.540 --> 00:39:50.780
to the 6, thats the answer, thats all you need to do
00:39:51.140 --> 00:39:53.300
you dont know what 1.22 to the 6 is
00:39:55.500 --> 00:39:56.000
6 years
00:40:02.280 --> 00:40:03.640
and if i said monthly
00:40:05.260 --> 00:40:07.020
you take the interest rate
00:40:08.840 --> 00:40:10.040
and divide it by 12
00:40:10.620 --> 00:40:12.140
because your gonna get
00:40:12.460 --> 00:40:15.260
1/12 every month and youre gonna multiply
00:40:16.060 --> 00:40:17.660
the number of years by 12
00:40:17.760 --> 00:40:19.680
because you can get 72 months
00:40:19.760 --> 00:40:20.960
instead of 6 years
00:40:23.820 --> 00:40:24.860
so thatd be 1000
00:40:28.600 --> 00:40:29.100
1+
00:40:30.100 --> 00:40:30.600
.01
00:40:31.840 --> 00:40:32.400
to the 72
00:40:33.980 --> 00:40:34.620
and again
00:40:34.620 --> 00:40:37.500
theres nothing you can do with that, you can run it through a calculator
00:40:37.500 --> 00:40:39.800
certainly you wont have calculators so you dont
00:40:40.120 --> 00:40:44.200
how bout continuously? continuously is a different formula
00:40:45.340 --> 00:40:47.500
continuously compound formula
00:40:49.100 --> 00:40:49.600
is e
00:40:56.220 --> 00:40:58.300
so here it would be 1000 dollars
00:41:00.020 --> 00:41:00.520
e
00:41:02.060 --> 00:41:02.560
.12
00:41:05.300 --> 00:41:05.800
times 6
00:41:08.360 --> 00:41:11.240
so how can we make this just a little harder?
00:41:12.000 --> 00:41:13.440
not a lot just a little
00:41:16.780 --> 00:41:19.500
suppose i gave you the exact same problem
00:41:21.740 --> 00:41:23.660
and i put the range back to 12%
00:41:25.460 --> 00:41:29.220
compound continuously, how long will it take to dounle?
00:41:53.700 --> 00:41:55.700
so i started with 1000 dollars
00:41:57.280 --> 00:41:58.640
you use this formula
00:42:02.080 --> 00:42:02.960
f=pe to the rt
00:42:04.580 --> 00:42:06.500
so i want to have 2000 dollars
00:42:07.580 --> 00:42:09.500
right now i have 1000 dollars
00:42:16.820 --> 00:42:18.100
plug everything in
00:42:20.180 --> 00:42:22.820
one of those first day of class problems
00:42:24.800 --> 00:42:26.160
maybe the second day
00:42:29.300 --> 00:42:30.820
divide by 1000 you get 2
00:42:34.400 --> 00:42:35.360
is e to the .12t
00:42:41.040 --> 00:42:43.520
and take the natural log of both sides
00:42:48.100 --> 00:42:48.600
.12t
00:42:52.440 --> 00:42:52.940
but
00:42:57.140 --> 00:43:02.180
and youre done, because again you dont know what the log of 2 divided by.12 is
00:43:12.620 --> 00:43:13.900
they call it rule 72
00:43:13.900 --> 00:43:16.840
the rule 72, you divide it by the interest rate
00:43:20.180 --> 00:43:22.340
so whats the natural log of 2/.12
00:43:24.100 --> 00:43:25.140
natural log of 2
00:43:29.960 --> 00:43:30.520
thats it