- [Instructor] So you got
a few problems here dealing
with the graph of a logarithm.
In general, the graph of a logarithm
looks like this.
That would be y equals,
this is natural log,
say ln of x.
It's got an x intercept at one, zero.
Notice it doesn't have
any horizontal asymptote
or y intercept,
but it does have a vertical intercept.
That's in general.
We're gonna be dealing
with ln of x minus four.
Well, we know from our
transformation of functions,
this will shift the
graph four to the right.
In general, ln of x, this graph here
has a domain of all reals.
Excuse me, not a domain of all reals,
a domain of x greater than zero
and a range of all reals.
Put this where you can see it.
And a domain is x greater than zero.
That's in general.
Now, this has been
shifted four to the right,
which is going to shift our
domain four to the right.
The other way to think about it,
and this will be good because
we can generalize this.
You need what your taking
the natural log of to
be greater than zero.
So, x minus four must
be greater than zero.
And so,
except this is not a domain question.
Excuse me.
I'm doin' it again.
Range.
Shifting it to the right four is not going
to change the range.
The problem with this problem,
or what can be tricky,
is that, if you graph this,
yo don't really see it.
In general, the graph of
a natural log or a log
has a range of all real numbers.
So, shifting it to the right
is not gonna change that.
Now, which of the following is
an x-intercept of f of x
equals ln of x minus four?
Well, this x-intercept has
been shifted four to the right.
So, it's gonna be at the point five zero.
The other way to see that is
to set your equation
equal to zero and solve.
Ln of x minus four.
Ln is base e, so rewriting
this as an exponential,
e to the zero equals x minus four.
Anything to the zero power is one.
Don't forget that.
I hear zero a lot.
Anything to the zero power is one
equals x minus four.
Add four to each side.
You get x equals five.
So there's a couple different
ways of looking at that one.
The y-intercept of ln of x minus four.
That's the output when the input is zero.
So, if we do that.
Ln of zero minus four
equals ln of negative four,
which is undefined.
Well, because we're
shiftin' this to the right,
it's not gonna hit the y-axis.
If we shifted this to the left,
if we're ln of x plus four,
we would've had a y-intercept.
So, no y-intercept.
Which of the following is
a vertical asymptote
of ln of x minus four?
Well, two ways to look at it.
Once again, this had a
vertical intercept at zero.
Shifting that four to the right
would be the line x equals
four, 'cause it's a vertical.
The other way to think about it is
your vertical intercept will be.
Excuse me, your vertical asymptote will be
at whatever x value makes what
your taking the log of zero.
Where x minus four equals zero will be
our vertical asymptote,
so it's at x equals four.
So, there's a couple different ways
to solve a few of these problems.
