Welcome, ladies and gentlemen.
So what I'd like to do is show you
how to evaluate a logarithm without a calculator.
Now, we can do this and hopefully, you
watched my least evaluated logarithm kind
of using with simple ones.
So therefore, I talk about what exactly
a logarithm means and represents.
In this case, I always like to use the one-to-one property
when I'm evaluating logarithms, especially
logarithms that maybe might not come
to me as quickly in my head.
So this one-to-one property basically
is, if x raised to the a is equal to x
raised to b, then a equals b.
Now, that might make sense, might not make sense,
so I gave you an example.
An example is, if I have 3 raised
to some number x and that's equal to 3 squared,
well, it would kind of makes sense that if 3 to the x
is equal to 3 squared, then that x has to equal 2.
So remember when we're looking at a logarithm,
at lease for all these, we're basically
asking us 27 raised to what power is going to equal 3?
So I think it's kind of beneficial
when doing these problems to kind of rewrite it
in that exponential format.
So I can say 27 raised to what number is going to equal to 3?
Now, I don't know what number 27 raised up to 3.
It's not as simple as like the last video
I did where three raised to what number gives you 9?
It's 2.
But in this, it's not very transparent.
So what I can do though is can use the one-to-one property.
Basically, if I can get this to be an expression or an equation
where I have the base that's exactly the same,
then I know my powers are going to be equal to each other.
Well, 3 and 27 are related to each other
because 3 to the first power is 3, 3 squared is 9,
and 3 cubed is going to equal 27.
So therefore, I can rewrite 3 cubed raised to the x equals 3.
Now, I can use my power rule and I'm not
going to do this for all of them,
but I've kind of run out of space here.
Maybe I'll do it for the last ones.
Anyway, so therefore now, my bases are exactly the same,
so therefore, I have-- now this, you
could say, well, what is this raise to?
Well, you could just say that's raised to the first power.
So now you could say 3x is equal to 1
whereas, x is equal to the 1/3 power.
Kind of running out of space here,
so that's why I kind of went over from there.
So if we do the next example, again, I
would do the same thing.
I could say 16 raised to what value is going to equal 2?
Now, hopefully, we understand that 2, we have 2
to the first part equals 2.
2 to the second power equals 2.
2 to the third power is 8.
2 to the fourth power is 16.
So I could rewrite this as 2 to the fourth
raised to the x equals 2.
Now, basically what I didn't show over here
is, you should use the power rule at least
to show that that's going to be 2 to the 4x equals 2,
and then you can raise that to the first power.
Now you can divide them out and we
could say x equals to the 1/4.
Kind of running out of space on those, aren't I?
For the next one, again, we can just do it one last time,
so I could do 16 raised to what value
is going to leave me with 8?
So therefore, x equals 1/4.
I should write these in there.
It's not really mathematically the way I want to do it.
So log base 27 of 3 is 1/3.
Log base 16 of 2 is 1/4.
So log base 16 of 8, I can rewrite 16 and 8.
Now, you can see that these-- I can't write 16
to a power that's going to give me 8.
So I got to look back into my values
and think, well, what base thing can I write?
Could I do 4 because I know 4 squared is 16,
but I can't make 4 and raise it to a power to give me 8.
I can always go back to my 2's though.
Because 2 to the fourth power is 16, and 2 to the third power
is 8.
So in this case, I'm going to do both of them.
So I'm going to do 2 to the fourth power, x equals 2 cubed.
Now, using my one-to-one property,
I'm left with 4x is equal to 3.
Divide by 4, divide by 4, and I get x equals 3/4.
So I have x equals 3/4 in that example.
Now, for the next one, I'm going to use green.
Actually, I'm going to go over here and use my blue just
so you guys can see the difference
and I'm not getting everything crowded here.
All right.
So now I'm going to do 9 over to the 27th.
So you have 9 to the x equals 27.
Well, again, 9 squared is 81, so obviously, it's not go work,
however, I can rewrite 3 squared is 9 and 3 cubed is 27.
So I could do 3 squared raised to the x equals 3 cubed.
Now, I can use my one-to-one property
and say 2x is equal to 3.
So therefore, x will equal 3/2.
So we need to divide by 2 on both sides, so x equals 3/2.
Over here, again, I have 4 raised to what power equals 32?
Well, 4 squared is 16, 4 cubed is 64,
so that's not going to work, but I can rewrite this out.
So I'll rewrite this as 4 raised to the x equals 32.
However, looking at my 2's here, 2 to the fifth power
is equal to 32.
So I can rewrite this as 2 squared raised to the x
equals 2 to the fifth power.
Therefore, now I can see that x-- so I have 2x
equals 5 divided by 2 divided by 2.
x equals 5/2.
So log base 4 of 32 equals 5/2.
For 16, again, looking at this, you
got to look at your numbers, which numbers go up
to there and in this case, you can
see that 4 to the first power is 4, 4 squared is 16,
and 4 cubed is 64.
So therefore, by writing this I, could say 16
raised to what value equals 64?
Well, 16 to the 16 squared, I don't know what it is,
but it's not 64.
But I can write 16 as 4 squared raised to the x equals 4 cubed.
And then, hopefully, you'll see we
did another couple of these which would be exactly
the same just like over here.
So therefore, I have 2x equals 3 divided by 2 divided by 2.
x equals 3/2.
So the important thing I want to show the reason why
I'm using this green here is because I'm setting up
this exponential equation to help me solve, to help
me think through the problem, but this
is a logarithmic expression.
Remember we're evaluating.
We're trying to find the value of that.
So I'm writing equals 1/3, equals 3/2
because that's what the value is going to be
equal to or equivalent to.
But it's not actually an equation.
It's a logarithm, we're just evaluating.
I'm using these exponential equations though
to help me think through the problems.
All right.
The next one here is 16.
Now we're dealing with fractions,
and I know everybody hates fractions.
So the main important thing with the fractions is a simple rule.
x to the m is equal to 1 over x to the negative m.
Well, let's not write it that way, let's write it this way.
x to the negative m equals 1 over x to the m.
So when we were first learning about our negative powers,
one of the things that we talked about
was this negative power rule.
And x raised to a negative exponent
is equal to 1 over x to the m.
So what's important about that is,
you guys can see that all of them
are raised to the first power.
However, they're in the denominator.
I want to get them off the denominator,
I have to put them in the numerator,
but you just can't throw a number
from the denominator to the numerator.
To do that, if you have a variable or any term
raised to a power, to put them in the numerator,
now when your exponent or your power is going to be negative.
So therefore, I'm going to rewrite this as log base 16
of 4 to the negative first power.
Now, in reality what I'm asking is,
16 raised to what number equals 4
raised to the negative first power?
Well, you can see here 16 you could write is squared.
So I could rewrite this as 4 squared raised to the x
equals 4 to the negative first power.
Now, you can see that my 4's are going to divide out,
and I'll be left with 2x equals negative 1
divided by 2 divided by 2. x equals negative 1/2.
In the next example, we can do kind
of like the exact same thing.
I can rewrite this as log 125 of 25 to the negative first power.
Ran out of space there.
So therefore, when doing that, I am basically asking myself
125 raised to what power equals 25 to the negative first power?
Well, again, we want to see, well,
what can I raise a number to to give you 125
as well as to give me 25?
125 is actually 5 cubed x, and this is 5 squared.
So now I can divide these out, and I'm
left with 3x equals negative 2.
So therefore, x equals a negative 3/2.
So I'm just going to have to circle or square
my answer over there since I kind of ran on space.
Over here, another example.
So instead of equaling it, I'm just
going to say 81 is equal to 27 to the negative first power.
I automatically see the 81, 27.
I know it's probably going to be as factors of 3, which
is correct, which is going to be 3 to the fourth power.
That's to the x.
So that's 3 to the fourth to the x equals
3 cubed to the negative first.
Therefore, now, I can use my one-to-one property
and therefore, I have 4 to the x equals
negative 3 divided by 4 divided by 4.
x equals negative 3/4.
All right.
Over here now I have the fraction at the bottom,
but it doesn't really matter.
It's like the exact same thing that you're going to be doing.
We're not really changing anything.
What I'm going to do is, I'm going
to rewrite this as 1/3 raised to what power is equal to 81?
Well, again, I need to rewrite this
as with the 3 in the numerator.
So it's 3 to the negative first power
raised to the x equals 81.
And again, 81 is 3 to the fourth power.
So in reality, it's 3 to the negative 1 times x
using the power rule.
You multiply the negative 1 in the x.
So therefore, you have 3 to the negative x is equal to 3
to the fourth power.
Well now I have negative x equals 4.
So therefore, divide by negative 1, divided by negative 1.
x equals negative 4.
Over here, we're going to do the same thing,
but I'm just going to rewrite this as 11.
So it's 11 to the negative first power, so 11 raised
to the first power of x is going to equal 121.
Well, 121 is the same thing as 11 squared.
So now I'll just use my one-to-one property,
and I could say negative x is equal to 2,
so therefore, x is going to equal negative 2.
And last but not least, I have log base a of 1 over 64 of 128.
So let's do 64 to the negative first power raised
to the x power equals 128 to the negative first power.
What I liked about this problem is,
again, you have them both as fractions,
so you got to raise them both to the negative power.
Now you've got to think, well, what number
can I rewrite there, 2, that's going to be raising 64 and 128?
Well, I know that 2 to the sixth power--
probably write that in there.
2 to the sixth power is going to be 64,
so 2 to the seventh power is going to be 128.
So therefore, this is 2 to the sixth,
so raised to the sixth, raised to the negative first,
raised to the x equals 2 to the seventh raised
to the negative first, raised to the x.
So therefore, by multiplying all of these out,
my 2's now divide out, and that's a 7,
so therefore, I am left with-- man, I'm running out of space.
Where did my blue go?
So dividing that out, I have a negative 6x
equals negative 7x or negative 7.
Where did that x come from?
Negative 7.
So therefore, divided by negative 6, divided
by negative 6, x equals 7/6.
So in this case, x equals 7 over 6.
So there you go, ladies and gentlemen.
Hope you stayed all the way to the end.
I know this is a lot of work, but that
is how you evaluate logarithms with a calculator,
and the more and more you do and the more and more practice
you get, the faster and faster you will get.
Thanks.