[ Music ]
>> All right so why don't you guys
go ahead and try one of the types
with the fractions is the argument
and we'll go over it together.
So by now if you've seen a pattern, because
what's really sticking out here is the 49
with this log expression
here is saying is 7 raised
to what power is going to give me 1 over 49.
And we know if we just wanted 49,
we know 7 squared would be 49.
So by now if you've noticed when it's in
this kind of a format with the fractions
and what not, it's just going to be
negative of what your answer was.
So the answer to this would be
negative 2 because if we had 7 squared
that would give us 49, but we
want the reciprocal of that
so we just apply a negative exponent and
that would cause the exponent to be positive
on the bottom and that would give
us what we're looking for 1 over 49.
So when I raised 7 to the negative 2
power -- I got what I was looking for.
So in this case x is equal to negative 2.
If you want to look at it the exponential
form way and the answer is negative 2 here.
Makes sense anyway, right?
Because logs spit out exponents.
And we were looking for an exponent here.
So either way it's negative 2.
OK, so why don't you go ahead
and try another one
with the fractions and I'll go over it with you.
So now here we've got a base as a
fraction and an argument as a fraction.
You haven't seen that before
but we're just going to stick
to what we know and see how this goes.
So what this log expression is saying -- if I
want to switch from log form to exponential form
and I know this thing is going
to spit out an exponent right?
So I don't know what that exponent is right now,
because that's what logs
do, they spit out exponents.
But what this is saying in exponential
form is 2/3 raised to what power is going
to give me 3/2, which is just
the reciprocal of 2/3, right?
OK, so if I have 2/3 here and basically what
I want is just to raise it to an exponent
that pretty much just makes it flip, right?
So that's just going to come
from our negative exponent laws,
bar raises to the negative 1
power, it would cause it to flip
because this negative would get
distributed to the top and bottom.
Get 2 to the negative 1 power, 3
to the negative 1 power, right?
And then a negative exponent is a positive
on the bottom and a negative exponent
on the bottom is a positive on top.
I think I'm explaining this too much but you
end up getting 3/2, just because we remember
from our exponent laws if you have a to the
b and that's raised to the negative 1 power,
that's equal to b over a. So our exponent that
we were looking for and that the log is going
to evaluate for us, we saw
that it is negative 1.
So x would equal negative 1.
OK, so here's another one.
What's interesting about this
one is we have 0 as the argument.
So I see log here so I know eventually this
thing is going to spit out an exponent.
But what this is saying is 3 to
what power is going to give me 0?
So we just go ahead and analyze
that for a second.
There's really no power you
could raise 3 to, to get 0.
If you raise 3 to 1 we get 3,
if we raise 3 to 0 we get 1.
I mean there's really no way about it.
So just thinking logically, when you raise stuff
to an exponent, you got to get something out.
You'd never get 0 out.
So there's actually not an answer for this,
and later we will see graphically why you can't
take the log of 0 or anything less than that.
So in the beginning I think I believe I said you
can't have anything negative in the argument,
well you also can't have 0 in there.
So I guess this is a special case
and there's no answer for this.
OK, just another example here -- what sticks
out about this one is you have a negative
inside the argument and in the beginning
of the video we already mentioned that you
couldn't have negative inside the argument
for log, that's just something we can't have.
We'll see graphically why later but you know,
just investigate this a little bit if we want.
This is saying give me an exponent.
That's what logs do.
And if we're to write this in exponential form,
its saying 3 to what power gives me negative 9?
And you know, when you raise something
to a power you don't really
never get negative element.
So I guess that's a good enough
reason for us just to accept for now,
just algebraically why you can't
have a negative inside the argument.
So this is another one that
we don't have an answer for.
So we can't evaluate this.
OK, so here's another one.
Log of some stuff, so we know it's
going to spit out an exponent.
What this is really saying is 3 to
what power is going to give me 15?
So remember we said how we're
going to do some log stuff by hand
or we might have to use a calculator.
This turns out to be the calculator case.
We have to use the calculator because
you come off to the side and you're like,
OK 3 to the power of 2 gives me 9.
But that's not enough.
But then you're like let's try 3
-- 3 to the third power, that's 27.
And that's too much.
So then you know your answer is
going to lie between 2 and 3.
It's going to be 2 point something.
So I mean I can't think of
that off the top of my head
so we're going to have to use a calculator.
OK, so we're going to have to
resort to using a calculator here.
But most calculators only
come with a common base of 10.
So that's not going to be useful for
us here because we have a base of 3,
so we're going to resort to the change of base
formula, which states that if you have log
with a base of a and a argument of b, then that
is equivalent to log of b divided by log of a.
So we could use that to our advantage here,
so this is going to become using that change
of base formula is going to become the log of
our argument divided by the log of our base.
And make sure not to make
an algebraic mistake here.
This is not the same as log of 15 divided by
3 or try to reduce this to be 5 or something.
You have to input this on
the calculator individually.
Log 15, enter and then divide that by log 3.
So let's see what I get when I do that.
So I'm going to press log 15 divided
by log 3 and I got approximately 2.46.
And I just rounded off so
you're going to have --
be asked to round off to however
many decimal places they ask you to.
OK, so another one we're going to have to use a
calculator for, because what this is saying is 2
to what power is going to give me 17.
And this log is going to go
ahead and spit out an exponent.
It's just that exponent doesn't
happen to be a nice looking number.
So we're going to use our calculator
and remember from the change
of base formula that we can do this with it.
So here, we're going to rewrite this as
-- we're going to apply this to it --
the change of base formula --
so we're going to get the log
of the argument divided by the log of the base.
And then we just input this into
our calculator individually.
We're not going to think that we could combine
the arguments or anything like that, so log 17,
divided by log 2, I got approximately
4.1 and I rounded that off.
So the exponent that this will
approximately spit out is 4.1.
OK, so why don't you guys go ahead and give
this a shot and then I'll go over it with you.
Got this, because I know I'm going to have
to use my calculator because you know,
6 to the second power will give
me 36, but this isn't 36 it's 35.
So I know it's going to be some long looking
decimal type answer, so I'm going to apply --
whenever I use my calculator I'm going to
apply the change of base formula to it.
So this would become log
of 35, divided by log of 6.
And I just enter those individually
into my calculator
and see what kind of answer I get and round.
We're just going with the
flow with this right now.
So log 35 divided by log
6, I'll just go with 1.98.
So this is approximately 1.98 and that
makes sense because it's smaller than 2,
because we saw that 6 to the second power
would give us 36 and that's a little too high,
so we know our answer has to be a
little bit less than 2 -- So 1.98.
