We've just looked
at how to do algebra
with quite complicated
expressions involving
exponents.
In all those examples, the
exponents were constant.
And it turned out that
using the exponent laws
was a good way to try to
manipulate these expressions.
So now let's think
about scenarios
where the exponent
isn't constant.
And what we're going
to find is using
logarithms is a more sensible
approach in this case.
So let's do some examples.
I want to do everything
here by example, really.
So example, find all x find all
x such that 3 to the power x
squared minus 1-- so
that any exponent--
is equal to 16.
Right, so let's take a
think about this problem.
First of all, first of
all, what do I know?
I know that, I know I know that
16 isn't a simple power of 3.
3 to the power 1 is 3.
3 to the power 2 is 9.
3 to the power of 3 is 27.
It jumps over 16.
So there's no real
hope of easily
spotting a solution here.
One thing I noticed as well
is that the exponent here
is no longer constant.
The exponent is the variable.
And what this
means, really, is I
think I'm going to have
to use logarithm laws.
So I really should
use logarithms.
So let's take a moment to think
about what I mean by that.
So first of all,
if I have this guy,
well, this now, if I apply
log to the base 3 of this,
it's sensible to do this, right?
Log to the base 3 as a
sensible thing to do,
because of the fact it's a
power 3 on the left hand side.
So 3 x squared minus 1 is equal
to log to the base 3 of 16.
OK, so now what do I want to do?
Well, I want to
simplify, don't I?
And the left hand side is
actually very straightforward,
right?
So this is now if and only if.
So the left hand side is
literally x squared minus 1.
That's literally the answer.
Why is this true?
Maybe I should do
the equals first.
Log 3 16.
So this is the definition.
This is one of my
cancellation laws, really.
Log to the base b of b
to the c is equal to c.
So my fundamental
properties, right?
So now what I have, at this
point, it's quite a bit easier.
Now, I just have to do some
much more basic algebra.
So now this is if
and only if x squared
is equal to log base 16 plus 1.
And I suppose now, x, I
could take a square root.
But I could take a square root.
But do I need to be
careful about this?
I do need to be a bit careful.
Because taking the square
root will only give me
the positive possibility.
There is a chance that--
so if I take the square root, I
only end up with the positive.
But of course the
negative would also work.
So maybe, before I
do that, I'll want
to label the pieces of
algebra at each stage.
This here is using the
fact that a minus b
is equal to c if and
only if a is equal to b
plus c, that piece of algebra.
And then now, if I
going to do, if I'm
going to undo the
square, I need to have
either the plus or minus
of the square root.
So it's no good just
having one of the two.
Something going to be
a little bit careful of
is, am I taking the
square root of a negative
or anything like that?
Well, no, I don't think so.
Log to the base 3 of 16 is
certainly, it's certainly,
it's got to be bigger than 0,
because 16 is bigger than 1.
3 squared, for example, is 9.
So log to the base
3 of 16 is going
to have to be bigger than 2.
So you not taking the
square root of a negative.
Something as well
that's worth saying
is, we can't really
simplify this any further.
This guy here, I mean, it's not
like 16 is a nice pair of 3.
So this is a good example
of something we know exists.
But we can't be completely sure
what it is, not by hand anyway.
We can't really simplify.
Can't be simplified.
So there we go.
That's the way to solve this.
Applying log to the
base 3 on both sides
was sensible, because
of the power here being,
the base being 3.
Actually, though that's, not
strictly speaking, necessary.
So let me kind of show you
a slight variation of this,
so observation.
This this pretty confusing
when you first see it.
But it's important.
We could actually have done this
with, well any base, really.
Do this with any base.
It's a weird thing to say.
But let me just show
you what I mean.
So if I take 3 to the power of x
squared minus 1 is equal to 16,
well, I could take any base.
Why I don't I do it with base 2?
Might as well.
Let's just see what happens.
Log to the base 2 of 3 to
the power x squared minus 1
is equal to log to
the base 2 of 16.
Now there is a reason log to
the base to 16 isn't a plan,
or base 2 isn't a bad plan.
It's because 16 is
the fourth power of 2.
So that means it's
going to be fairly
straightforward on
the right hand side.
And the left hand side is
a bit problematic, though.
So this is if and only if what?
Well, let's remember the
following important log law.
The following log law
says that log to the base
b of a to the power c is equal
to c times log to the base
b of a.
So that exponent being inside,
and we brought out the front.
So that means the following.
That means the left
hand side is equal to x
squared minus 1 times
log to the base 2 of 3.
So that's exactly this.
And this now is equal to what?
It's equal 4, because of the
fact 2 to the 4 is equal to 16.
So now, I'm kind of in the same
type of situation as before.
But a little bit different.
I need to divide through
by log to the base 2 of 3.
So x squared minus 1 is going
to be equal to 4 divided
by log to the base 2 of 3.
So now, this is if and
only if-- so that's
a standard piece of algebra,
just dividing through.
And we're going to label
that, because we've seen it so
many times.
Have x now is going
to be equal to,
will exactly the same
logic as before, plus minus
the square root It's a bit of
a mess, this way of doing it.
But it's not wrong--
4 over log to the
base 2 3 plus 1.
So this and this are actually
completely identical.
It's not obvious they are.
But that and that
are exactly the same.
Like I said, it's
not totally obvious.
But it is true.
OK, so very nice.
So moral of the story,
because of this property here,
because of that property, the
base doesn't matter too much.
It was nice to choose
Base 3 in the top,
because it was so, so simple.
That the jump from there.
Today was really nice.
But we don't actually
need to use base 3.
You could use any base
at all that you choose,
as long as it's greater
than 0 and not equal to 1,
because you don't have
logarithms for those,
basically.
OK, so let's do another
example, shall we?
So find x such that--
OK.
this is a bit more tricky.
To the power x plus 2 to
the power x plus 1 is equal,
let's say 4 to x.
Now in this situation,
it's very tempting to say,
OK, let's do what we did before.
Let's just apply
log to both sides,
and simplify, and hey, presto.
So that's definitely not
the correct thing to do.
So let me show you why.
If I choose, let's say base 2--
it shouldn't matter too much.
But let's choose base
2, because it's obvious.
So if I choose base 2,
apply it to both sides,
so that would have
to be equal to what?
That would have to be equal
to log to the base 2 of 4x.
Sorry, what am I doing?
4 to the power of x--
so that's true.
But the problem now is,
you cannot simplify this.
So this cannot be simplified,
cannot be easily simplified.
In particular, this
guy is not equal,
as you might pray, to this, as
an example we've seen before.
Because if that were
true, this would be
equal to something really nice.
Would be x plus x minus
1, which is equal to 2x.
And what would it be?
It would be 2x minus 1.
So something has gone
horribly wrong, right?
Well, it's not obvious
it's gone horribly wrong.
But this is not correct at all.
So if you were to replace
the right hand, side
and the left hand
side, sorry, with this,
it's just totally wrong.
It is not true.
What is not true
is the following.
Log to the base b of a plus c
does not equal log to the base
b of a plus log to
the base b of c.
And that's kind of the
crucial thing here.
They are not equal
to each other.
This does not work.
I cannot easily simplify a sum.
So what's the fix here?
Well, the bottom line is,
the laws of logarithms
require some, they require
products, quotients,
and powers.
So to apply the logarithm
laws, we can't have sums.
We need products,
quotients, and powers.
So what I've got to do
is express the left hand
side as something not a
sum, basically, right?
So let's try it.
Do 2 to the x plus
2 to the x plus 1.
Well, I do know some
laws of exponents--
2 to the x plus 2 the
x times 2 to the 1, OK?
So this is just my law that
says, what does it say?
So it's just, I'm
just doing it to this
is a statement that b to
the a times b to the c
is equal to b to the a plus c.
But next, in doing this, there's
a 2x factor from each piece.
This becomes 2x times 1
plus 2 to the 1, which is 2.
And this is equal to what?
This is equal to 2x times 3.
So I've taken the left hand side
of my original equation, which
was a nasty sum, and now I've
expressed it as a product.
Because it's a
product, I'm going
to be able to apply
my laws of logarithms.
So let's try it so as
before, this is me just
looking at the left hand side.
So 2x plus 2 to the x plus
1 is equal to 4 to the x.
Pretty nasty example here.
This is if and only
if 2 to the x times 3
is equal to 4 to the x, OK?
So now, I've got
a product, no sum.
I can apply logarithms,
and try to simplify.
So now this is if and only
if log-- let's do base 2.
It seems logical, doesn't it?
To the x times 3 is equal
to log base 2 of 4 to the x.
OK, so now let's start
during my log laws.
First of all, I can break apart
the left hand side of the sum.
Log to 2x plus log
base 2 of 3 Is equal--
I'm going to do
one at a time here.
I'm not going to simplify
too much at any stage.
So this is the statement
that log base b of a times c
is equal to log base b of
a plus log base b of c.
Assuming and b are
positive-- or sorry,
assuming that a
and c are positive,
which they are in this case.
So now, we can do a bunch more.
I'm going to really just use
my laws of logarithms here.
So this is if and only if.
I've got a power.
I can bring it out the
front, log to the base
2 of 2 plus log
to the base 2 of 3
is equal to x times
log to the base to 4.
So this is now just
my law of logarithms--
we've already
mentioned it before.
Log to the base b
of a to the power c
is equal to c times
log to the base
b of a, my other log property.
Finally, though, log
to the base 2 of 2
is just equal to 1,
my absolute base case.
And a log to the base
to a 4 is equal to 2.
So this guy is to the 2,
and this guy, of course,
is 2 to the 1.
So this tells you x plus log
2 of 3 has to be equal to 2,
times x.
Ooh, another easy mistake there.
And now I just completely
elementary algebra.
Moving the x to the
side, this would give me
x is equal to log
to the base 2 of 3.
It's nice to write
it that way around.
Very nice.
So this is good example of
using log laws carefully.
We couldn't just immediately
attack it from the get go,
because it was a sum.
But if I cleverly interpreted
it as a product, then applied
logarithms, I was
able to get somewhere.
So the conclusion of this
is the following, really.
When dealing with the
variable exponents,
you want to manipulate-- if
you're trying to kind of undo
exponents and things
like this, you
want to manipulate into
products, quotients,
and powers, before applying
log laws, before applying logs.
Manipulate into products--
like we did, quotients--
that's one of the
other log laws,
and powers before applying
logarithms and logarithm laws,
and their laws.
That's kind of a
broad statement.
It's not completely,
universally true.
But it's something
to be aware of.
And I just want to really
stress, the laws of logarithms
are incredibly important.
Because like I've said
many times before,
you're very likely to be
asked to do certain things
with specific expressions.
And as they are
presented to you,
it could be
incredibly difficult.
But if you apply a law of
logarithms or something
like that, they'll become
much more straightforward.
So these laws of logarithms
are totally vital.
Let's do an example, actually,
which is very similar to that.
So here's an example.
So express, let's say 2 log
base 2 of x minus 1 minus 3
of log base 2 of x squared plus
1, in the form, so let's do it,
log base 2 of, log base 2 of
a rational expression in x.
Let's do that.
This is a rational.
This is a good example, right?
Perhaps, as it stands,
you can't work out what's
going on with this function.
You need to apply,
you need to manipulate
it to put in this format.
So first of all, let's kind of
do our laws of logs backwards.
Here, I can see there's
a 2 at the front.
I can bring that inside the
brackets, right, as a square.
It's one of my
laws of logarithms.
And I can do the same thing
with the 3, but with a cube.
So too so this
expression, let's get it.
So this expression
is equal to log
to the base 2 of x minus 1 or
squared, minus log to the base
2 of x squared plus 1
or cubed there, right?
So this now is me applying
the log law, which
says log to the
base b of a to the c
is equal to c times
log to the base b of a.
That's the log law
I'm using here.
But now I've got a difference.
And the difference of two
logs is equal to what?
Well, this is equal
to the log base
2 of the ratio of each of them.
And now this log law, of
course, is the following.
Log to the base b
of, well, it would
be sort of c is equal to log
to the base b of a minus log
to the base b of c.
That's how the logs
work for quotients.
This is a good example.
So we're done, right?
We've put it in exactly
the right format.
This now would be
your px over qx.
That's a very common
use of log laws
that you're going
to see in calculus,
simplifying things or
manipulating things
using laws of logarithms.
They're terribly important.
In fact, really, for
things like calculus,
finding specific values
for logarithms and things,
like that it's not really that.
Important.
What matters is deploying log
laws to manipulate expressions
accurately.
So log laws are
difficult to use.
But there's only
three of them, right?
Remember, there's logs of
products, logs of quotients,
and logs of powers.
That's all you ever use.
So if you internalize those,
you're kind of fireproof.
Because that's the only three
things you need to use ever.
