Let's try another
problem, which is
extremely similar, but perhaps
a little more straightforward.
So we want to find
all values of x
such that the square
root of x plus 1
divided by the cube root of x
plus 1 is equal to the number
4.
So before we do anything,
it's a good idea just
to make sure we understand
maybe some potential problem
points in this expression,
especially the left-hand side.
So on the left-hand side
we put a square root.
And if I have a square
root, I know what?
Well, I know that
I can't be taking
square roots negative numbers.
Not in the world
of real numbers.
So we have to have x plus 1.
It's got to be positive.
So it's got to be bigger than 0.
You can take the square
root of 0, of course.
It's just equal to 0.
But if you had
that situation, you
would end up with a 0 in
the denominator, which
doesn't make sense.
So in this story, x plus
1 is going to be positive.
So that might be
useful a little bit.
So the next thing to do is
to take the left-hand side
and try to simplify it
using my laws of exponents.
So this guy is equal to what?
It's equal to, by definition,
x plus 1 to the power of 1/2,
divided by, of course, x
plus 1 to the power 1/3.
So now, x plus 1 is positive.
So I can apply my
law of exponents,
where I take the difference
between the two exponents.
So this becomes x plus 1
to the power 1/2 minus 1/3.
So just to reiterate,
what we're doing here
is the general rule,
which is b divided
by a divided by b
divided by b to the c is
equal to b to the a minus c.
That's what we're using here.
So this is a general property.
But you do need to
be a bit careful.
So for example,
if b is negative,
then this may not make
sense. b can be, for example,
taking a negative number
to the power 1/2, which
doesn't make any sense.
So we're only really confident
in this when b is positive.
If b is negative, it
does work sometimes.
For example, if a or c are
positive whole numbers.
But if they're not,
like in this example,
it wouldn't make sense.
So that's just something
to be aware of.
But anyway, we have this now.
And 1/2 minus 1/3 is 1/6.
So I get x plus 1
to the power 1/6,
So putting everything
together, what do we have?
Well, we're trying to find
all the values of x such
that this equality holds.
So clearly, this is
if and only if now.
Well, we can replace the
left-hand side with x plus 1
to the power 1/6.
And this has to be equal to 4.
So now how can one do this?
Well, it's basically saying
that the sixth root of x plus 1
has to be equal to 4.
So this now is going to
be if and only if what?
Well, let's take the
sixth power of both sides.
So I'm just going to do
this incredibly carefully.
Now, this is if
and only if what?
Well, now I can use
my law of exponents
where I multiply them together.
And 1/6 times 6 is equal to 1.
So this becomes x
plus 1 on this side.
And this becomes what?
Well, 4 to the power
of 6 is 4,096, I think.
So the property we're
using here is what?
Well, we're using the fact that,
well, b to the a to the power
c is equal to b
to the a times c.
That's what we'll be using.
And again, we're assuming
b is positive here.
Otherwise, we could run
into all sorts of problems.
But it is by assumption, because
x plus 1 has to be positive.
Otherwise, this
wouldn't make any sense.
And I suppose we're also
using the fact that b to the 1
is equal to b, if you want
to be really strict about it.
So finally, we
can conclude what?
Well, that's positive.
It's all good.
This is of course if and
only if x is equal to 4,095.
4,095.
So it's a good example.
You've got to really be on
top of your laws of exponents
to do these types of things.
But once you've got a bit of
practice, they're all the same.
