>> Okay, so let's solve
some 11.3 equations right,
now we're solving equations
not just evaluating.
So, the first step
on these is usually
or maybe the easiest kind, it's
the ones where you just have
to evaluate part of it right.
So, what's log base 6 of 36?
[ Inaudible Comment ]
[ Inaudible Comment ]
Okay, so this right
is the same thing
as X equals log base 6 of 36.
[ Inaudible Comment ]
Six squared.
So.
>> It's going to
equal two [inaudible].
>> So, this left-hand
side is two right
and that's X. Do
you guys see that?
So, this wasn't really
much of a solve right
because all I really had to do
was evaluate the left-hand side.
This is as easy as
solve gets, right.
[ Inaudible Comment ]
Okay, so let's talk
about this one.
This is asking what face do I
raise to the second to get four.
So, this is saying X
squared is 4 right.
[ Inaudible Comment ]
So, X has to be 2.
So, this one actually
involved a...
[ Inaudible Comment ]
I had to undo something, right?
Do you guys see what we missed?
What did we undo?
[ Inaudible Comment ]
Yeah, here we took
a square root right.
When we take square roots
what are we responsible
for introducing?
Plus or minus.
So, I need a plus
minus square root right
which gives me not two but...
[ Inaudible Comment ]
Plus minus 2, do
you guys see that?
So, the log based 2 of 4 is 2.
The log base negative
2 of 4 is also 2 right
because I square
them both I get 4.
Do you guys see that?
[ Inaudible Comment ]
Yeah, this is maybe the
hardest thing to remember
in [inaudible] is this.
Okay when I take a square
root it's plus or minus.
When I'm just evaluating a
square root it doesn't right.
[ Inaudible Comment ]
Yeah, when I solve, I put that
thing in there I need to account
for both halves right.
That's me asking right
this thing if I thought
about this graphically
is there's a --
oops, there's a parabola
not that one.
There's a parabola in this one
right, it goes through zero zero
and I'm asking when
does it hit four.
In other words, where are
these two intersections.
There's 1 over 2 and
there's 1 over minus 2.
Do you guys see that?
You have to account for
all the intersections
on the graph that's me right.
So, when I put this root
in there I have to
account for those.
Do you guys follow that?
[ Inaudible Comment ]
Okay, so how about this guy?
[ Inaudible Comment ]
Okay, so I get, I'm looking
for this statement says X
to the what is what, the third
is eight, so X cubed is 8.
Now what do I do to both sides?
Cube root both sides.
So, I get X is two.
Why don't you need a plus or minus?
[ Inaudible Comment ]
Because odd roots are actually
inverse functions, right.
Cool, very nice guys.
How about this one,
what's this one?
[ Inaudible Comment ]
This one's a whole problem,
this one's actually pretty easy.
Why is this one pretty easy?
[ Inaudible Comment ]
Oh yeah, because they know what
log base 3 of 3 is, what is it?
[ Inaudible Comment ]
So, now it's just an
evaluate step right.
[ Inaudible Comment ]
And now I need to.
>> Add.
>> Add one to both sides right,
so I get 2 is 4 X then I.
[ Inaudible Comment ]
Divide by four, so I'll get.
[ Inaudible Comment ]
One half is X.
>> Could you write
it the other way just
because it makes
you feel better?
>> Sure.
>> Okay.
>> Yeah, swap the order if it
makes you feel better right,
X is one and [inaudible]
is also fine.
The last time we did
inverse functions right,
do you guys remember that?
So, when we did inverse
functions
if I was finding the
inverse of 4 X minus 1.
Do you guys recognize that
that's add 1 divide by 4?
That is the inverse function.
Every time I've been solving
an equation I've been manually
constructing the
inverse function
down this side, do
you guys see that?
Yeah, you've actually been
making inverse functions the
whole time.
Do you guys see that?
This one actually is
an inverse function,
this one actually is this
one right here actually is an
inverse function.
This guy isn't because I tried
to invert the squaring function
and the squaring function
doesn't have an inverse.
See I got two answers.
That was one input
for two outputs
that thing's not a function.
Yeah, it fails the -- oh,
I was doing the horizontal line
test right, I knew it failed.
It doesn't mean I can't solve
the equation, but it means
that I can't make an inverse
function, do you guys see that?
I can make an inverse
process or something,
but it's not a function.
[ Inaudible Comment ]
So, could X be negative here?
No, yeah because 6 to
the minus 2 is 1 over 36.
Do you guys see that?
Cool, you want to
try the hard one?
Okay, right.
Yeah, now this is going
to be the hard thing
because this is the
skitter [phonetic] deal.
So, we've got a log
in there we want
to get it out of there right.
Okay, so how do I normally
get something that's annoying
to me out of there?
[ Inaudible Comment ]
Yeah, kind of do the
opposite of it right.
So, there is an opposite
for logarithms it's
raise it to the base.
Do you guys see that?
So, I'm going to
apply it to this side,
to this guy to both sides
and I've been doing this
and I just didn't tell you.
I'm going to apply two to the
eighth whatever to this thing.
So, I'm going to feed both
sides of this into the two
to the whatever function.
So, I'm going to
get 2 to the log 2
of 3 X minus 4 is
2 to the third.
So, this is literally
just me putting both sides
as exponents on 2.
Cool with that?
Now why would I do this?
Okay, so earlier we
were canceling twos
and log twos the
other way, right?
Like when I had this I
wo9uld cancel the log six
and the six right.
They cancel the other way too
because they're inverse
functions.
So, this two and this
log base two cancel
and all this junk falls down.
So, I get 3 X minus
4 is 2 cubed.
Why not three?
Where, here?
[ Inaudible Comment ]
No, I started with
a three over there
and then I raised
both sides to the 2.
So, I put a 2 down here and
then put both these sides
on the exponent.
Yeah and now I can
just solve for X
because I know what 2 cubed is.
>> It's eight.
>> Two cubed is eight.
So, I'm looking at
3 X minus 4 is 8,
so what do I do to both sides?
[ Inaudible Comment ]
Add four, I got 3
X is 12 then I.
[ Inaudible Comment ]
Divide by 3 and get X is 4.
Do you guys want to see
a really cool connection?
>> Hey what do we call that
when we bring the two down?
>> This thing?
>> Yeah.
>> What should we?
>> I would call that
exponentiation,
but some people would
call that a skitter.
Right that's the two skitter
it gets rid of log base twos.
>> Okay.
>> Do you guys see that, it
drags them out of there?>
[ Inaudible Comment ]
So, it won't work if you have,
so if you had a log base
three raising everything
to the two won't help you out,
but raising everything
above a three would.
>> Okay, so we're just
matching [inaudible].
>> Yeah, we're matching
the base.
>> Okay.
>> Cool? So, you guys want
to see the cool thing?
If we had a function of X right
that's this thing that's log
base 2 of 3 X minus
4 okay, right.
If I think about function
[inaudible] what's the first
thing I do to an X when
I stick it in there?
[ Inaudible Comment ]
Yeah, multiply by 3 right,
but that gets me to this,
so I started with an
X that gets me to 3 X.
[ Inaudible Comment ]
Then I should subtract 4
that gets me to 3 X minus 4.
Then this function is hard
to write down, what is it?
What's the next thing I do?
[ Inaudible Comment ]
I already subtracted the four,
so now I know what
this number is now.
[ Inaudible Comment ]
No, then I evaluate
the log on it right.
So, I ask what's the log base
two of whatever number I had.
So, that's applying the
log base two function.
So, I get over here
and now I'm done
with my function evaluation.
I've got log base
2 of 3 X minus 4.
If I want to go backwards
what's unlogging?
[ Inaudible Comment ]
Raising it to the two.
What's unsubtracting four?
Add four. What's
unmultiply by three?
[ Inaudible Comment ]
Divide by three.
Do you guys see that's
what we did down this side?
So, if I want to write down
the inverse function right,
F inverse of X.
Cool. I start with an X
if I put a two under it.
Then I add four, then I divide
the whole thing by three.
Do you guys see that?
I just found the inverse
function for that.
Are you guys cool with that?
So, my point here with this is
that these inverse functions
that we were doing
the other day right,
they're not really magic they're
the same thing I've been doing
to solve this whole time.
I've been manually
making an inverse function
because what I'm trying to
do is I'm trying to say okay,
here's a function of X right.
So, in this case it's a
logarithm that's shifted
around a bit so it looks
something like this.
And then here's a level
right and I'm looking
for this intersection.
Do you guys see that?
So, this is a picture roughly
of log base 2 of 3 X minus 4
and this is 3 right,
Y equals 3 maybe.
[ Inaudible Comment ]
No, this arc is the log base
two because it's what happens
when I take an exponential
right,
exponentials look like this.
When I do the inverse
trick to them I flip them
over Y equals X they
look like that.
[ Inaudible Comment ]
Yeah, so I started with
this is 2 to the X maybe
and then I flip it over.
If I was being really honest
I would say that's the graph
of this one right.
And then I flip it over I
end up with an arc like this
and I'm wondering where's
this intersection that's what
solves [inaudible].
[ Inaudible Comment ]
You guys see that?
So, the X value is this guy
wherever the heck x-axis is.
Cool?
>> So it relates to?
>> Everything.
Cool?
>> If you can do
another one like that
that would be [inaudible].
>> Yeah, let's find another one.
