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
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!
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
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
this you can say 8 to the x is 16
now you try and find a common base
4 is not a common base
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
any other questions before i do new stuff?
cause 5 to the 3 is 125, radical 125
quick question, yes?
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
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
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
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
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
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
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
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
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
right lets do another one of these to make sure everyone gets the idea
lets do this one
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
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+!
is 9 times x plus 5
2x+1
is 9x
+ 45
so negative 4 is 7
x, 44 sorry
x
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
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
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
its not what the answer is, its what the log if it is
okay lets do another one
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
x-2
over x-3
5 squared has to be 25
so then 16x-2 you cross multiply
is 25
times x minus 3
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
howd we do on this one? good?
ill give you another one
can i cover this up?
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
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
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
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
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
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
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
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
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
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
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
so suppose
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
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
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
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
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?
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
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
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
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
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
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
8 is 2 cubed so this
2 cubed, x+3
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?
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
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
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
8 x
these are equivalent
put the x+3 in front
now you divide by log 8
and subtract 3
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
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
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
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
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
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
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
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
how bout that one?
alright take the log of both sides
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
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
-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
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
some of you are gonna be business students
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
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
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+
.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
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?
so i started with 1000 dollars
you use this formula
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
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
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