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www.mathcentre.ac.uk/.../Exponential%20and%20logarithm%20functions.mp4

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    In this video, we'll be looking
    at exponential functions and
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    logarithm functions, and I'd
    like to start off by thinking
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    about functions of the form F of
    X equals A to the power of X,
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    where a is representing real
    positive numbers. I'm going to
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    split this up into three cases.
    First of all, the case when a
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    equals 1 hour, then going to
    look at the case when A is more
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    than one, and finally I'll look
    at the case where.
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    Is between zero and one. So
    first of all.
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    When a equals 1, this will give
    us the function F of X equals 1
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    to the power of X and we can see
    that once the power of anything
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    is actually one. So this is the
    linear function F of X equals 1.
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    So that's quite straightforward,
    and Secondly, I'd like to look
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    at the case where a is more than
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    one. And to demonstrate what
    happens in this case, I'd like
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    to consider a specific example.
    In this case, I'll choose A to
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    be equal to two, which gives us
    the function F of X equals 2 to
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    the power of X.
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    Now a good place to start with
    these kind of functions is to
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    look at for some different
    values of the argument.
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    So starts off by looking at F of
    0, which is actually equal to 2
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    to the power 0.
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    And we know that anything to the
    power 0 equals 1.
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    Next, we'll look at F of one.
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    Which is 2 to the power of 1.
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    And two to the power of one is
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    2. And we can look at F of
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    two. Which is 2 squared,
    which is 4. So quite
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    straightforward, and finally F
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    of three. Which is 2 to the
    power of three which actually
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    gives us 8.
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    Also want to consider some
    negative arguments as well, so
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    if we look at F of minus one.
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    This is 2 to the power of
    minus one. And remember when
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    we have a negative power,
    that means that we have to
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    invert our number so we
    actually end up with one
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    half.
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    If you look at F of minus two
    I guess is 2 to the power of
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    minus two. Once again, this
    negative power makes we've got
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    one over 2 squared and it's 2
    squared is 4's actually gives us
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    1/4. And final arguments are
    consider is F of minus three.
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    Which is 2 to the power of
    minus three, which gives us
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    one over 2 cubed and two
    cubes 8. So we get one 8th.
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    I'm going to take these
    results now and put them into
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    a table and we can use that
    table to help us plot a graph
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    of the function.
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    So our table, the
    values of X&F of X.
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    We just come from minus 3
    - 2 - 1 zero, 1
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    two and three and the value
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    hardware 1/8.
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    1/4
    1/2 one, 2,
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    four and eight.
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    So we're going to plot these now
    so we can get a graph of the
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    function. So do RF of X
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    axis here. And X axis
    horizontally. So on the X axis
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    we need to go from minus three
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    to +3. So minus 1
    - 2 - 3 and
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    one. Two and three this
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    way. And on the vertical scale,
    the F of X axis, we need to go
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    up to 8. So
    12345678. Make
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    sure we
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    label that.
    So now let's plot the
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    points.
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    Minus 3 1/8.
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    Minus 2
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    1/4. Minus
    1 1/2.
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    Zero and one.
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    12
    214
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    And finally, three and eight.
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    And so we need to try and draw a
    smooth curve through the points.
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    And this is the graph
    of the function F of
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    X equals 2 to the
    power of X.
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    Now this is actually quite
    clearly shows the general shape
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    of graphs of the functions where
    F of X equals A to the X, and a
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    is more than one. However, what
    happens when we vary the value
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    of A? Well, by looking at a few
    sketches of a few different
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    graphs, they should become
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    clear. So I have my axes again.
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    F of X.
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    X horizontally.
    We've just spotted the graph of
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    this function. Which one through
    1 F of X axis and this was F of
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    X equals 2 to the power of X.
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    If we were to look at this.
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    Graph.
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    This might represent F of X
    equals 5 to the power of X.
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    If I was look at this graph.
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    This might represent the graph
    of the function F of X equals 10
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    to the power of X.
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    So what's actually happening
    here? Well, for bigger values of
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    a. We can see that the output
    increases more quickly as the
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    arguments increases.
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    Another couple of important
    points to notice here are first
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    of all that every single graph
    that I've sketched here comes
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    through this .0 one, and in
    fact, regardless of our value of
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    A. F of 0.
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    Will equal 1 for every single
    value of a.
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    Secondly, we notice that F of
    X is always more than 0.
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    In other words, are output for
    this function is always
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    positive. As a very
    important feature of these
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    kind of functions.
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    The last case I would like to
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    consider. Is the case where a is
    between zero and one case where
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    a is between zero and one.
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    To demonstrate this case, I
    would like it to look at a
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    specific example. In this case I
    will choose a equals 1/2, so
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    this means I'm looking at the
    function F of X.
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    Equals 1/2 to
    the power of X.
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    Now with the last example, a
    good place to start is by
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    looking at some different values
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    for the arguments. So let's
    first of all consider F of 0.
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    This will give us 1/2 to the
    power of 0.
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    And as we said before, anything
    to the power of 0 is one.
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    Secondly, we look at F of one.
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    Which is 1/2 to the power of 1.
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    Which is equal to 1/2.
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    Half of 2.
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    Equals 1/2 squared.
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    So on the top that just gives
    US1 squared, which is one.
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    On the bottom 2 squared, which
    is 4. So we end up with one
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    quarter. An F of three.
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    Is equal to 1/2 cubed.
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    So on the top we get one cubed
    which just gives US1 and under
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    bottom 2 cubed which gives us 8.
    So we end up with one 8th.
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    And as before, we also need
    to consider some negative
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    arguments.
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    So half of minus one.
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    Gives us 1/2 to the power of
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    minus one. And remember the
    minus sign on the power actually
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    inverts are fraction, so we end
    up with two over 1 to the power
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    of 1, which is just two.
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    F of
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    minus 2. Gives
    us 1/2 to the power of minus
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    2. Which is 2 over 1 squared.
    You can see on the top we get 2
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    squared which is 4 and on the
    bottom we just get one. So
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    that's actually equal to 4.
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    And finally, F of minus three.
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    Is one half to the power of
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    -3? Which means we get two over
    one and we deal with the minus
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    sign cubed. So 2 cubed in the
    top which is 8 and again just
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    the one on the bottom. So that
    just gives us 8.
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    So once again, we're going to
    take these results and put them
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    into a table so we can plot a
    graph. A graph of the function.
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    So X&F of X
    again for our table.
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    And we have values of the
    argument ranging from minus
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    three, all the way up to
    three again.
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    And this time the
    values where 8421.
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    1/2 one quarter.
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    And one 8th.
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    So let's plot this now
    on a graph.
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    So vertically we get F of X and
    a horizontal axis. We've got X
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    and we're going from minus three
    to three again.
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    So minus 1 - 2 -
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    3. 123
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    here. And
    then we're going up
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    to 8 on the
    vertical axis, 12345678.
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    So let's
    plot the
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    points. First point
    is minus three 8.
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    Which is around about here.
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    Minus two and four.
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    But here. Minus one and two.
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    But the. Zero and one.
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    One 1/2.
    2 one quarter.
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    And finally 3 1/8.
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    As with our previous example, we
    need to try and draw a smooth
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    curve through the points.
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    So.
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    And this represents the function
    F of X equals 1/2 to
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    the power of X.
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    And actually this demonstrates
    the general shape for functions
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    of the form F of X equals A to
    the X when A is between zero and
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    one. But what happens when we
    vary a within those boundaries?
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    Well, sketching a few
    graphs of this function
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    will help us to see.
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    Just do some axes again.
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    So we've got F of X.
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    And X. We've just seen.
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    This curve, which was.
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    F of X equals 1/2 to
    the power of X.
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    We might have seen.
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    This would have represented.
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    F of X equals.
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    1/5 to the power of X or we may
    even have seen.
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    Something like this?
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    Which would have represented
    maybe F of X equals 110th to the
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    power of X.
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    So we can see that for bigger
    values of a this is. We come
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    down here. The output decreases
    more slowly as the arguments
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    increases. And a few important
    points to notice here. First of
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    all, as with the previous
    example, F of 0 equals 1
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    regardless of the value of A and
    in fact, as long as a is
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    positive and real, this will
    always be the case.
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    Second thing to notice, as with
    the previous example, is that
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    our output F of X is always
    positive, so output is always
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    more than 0.
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    So now we've looked at what
    happens for all the different
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    values of a when A is positive
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    and real. No one
    interesting thing you
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    might have noticed is
    this. We've got some
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    symmetry going on here. If
    I actually just put some
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    axes down again here.
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    So F of X find X you'll
    remember.
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    This curve here.
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    Passing through the point one
    was F of X equals 2 to the power
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    of X. And also
    this curve here.
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    Was F of X equals 1/2
    to the power of X?
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    And we can see this link it
    'cause one of these graphs is a
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    reflection of the other in the F
    of X axis. And in fact this
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    could have been 1/5 to the X and
    this could have been 5 to the X
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    'cause generally speaking.
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    F of X equal to.
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    AX.
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    Is a
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    reflection. Of F of
    X equals 1 over
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    8 to the X
    and that is in
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    the F of X
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    axis. And one over 8 to the
    X can also be written as A to
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    the minus X.
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    So that's an interesting point
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    to note. Now you might recall
    from Chapter 2.3 that the
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    exponential number E.
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    Is approximately
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    2.718. Which means it falls
    into the first category, where
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    a is more than one. If we're
    going to consider the function
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    F of X equals E to the X, which
    is the exponential function.
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    So if I do some axes again.
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    F of X vertically.
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    An ex horizontally.
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    This graph here.
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    Might represent F of X equals E
    to the X and this is what the
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    exponential function actually
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    looks like. I might also like to
    consider the function F of X
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    equals E to the minus X.
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    As we've just discovered.
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    That is a reflection in the F of
    X axis, so this will be F of X
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    equals E to the minus X. And
    remember this important .1 on
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    the F of X axis.
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    So look that exponential
    functions and we've looked the
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    functions of the form A to the
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    X. At my work to consider
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    logarithm functions. And
    logarithm functions take the
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    form F of X equals the log
    of X. So particular base. In
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    this case a.
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    And as with the previous
    example, I'd like to split my
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    analysis of this into three
    parts. First of all, looking at
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    when a is one.
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    Second of all, looking at when
    there is more than one, and
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    finally when a is a number
    between zero and one, and as
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    with the previous example, a is
    only going to be positive real
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    numbers. So first of all, what
    happens when a equals 1?
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    Well, this means we'll get a
    function F of X equals the log
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    of X to base one.
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    Remember, this is equivalent.
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    To say that this number one.
  • 19:17 - 19:20
    To some power.
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    F of X is equal to X, just
    like earlier on, this is
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    equivalent. Generally. 2A
    to the power of F of X
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    equals X.
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    So when we look at this, we can
    see that we can only get
  • 19:40 - 19:44
    solutions when we consider the
    arguments X equals 1.
  • 19:44 - 19:49
    And in fact, if we look at the
    arguments X equals 1, there is
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    an infinite number of answers be
    cause one to any power will give
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    us one. So actually this is not
    a valid function because we've
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    got many outputs for just one
    single input. So that's what
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    happens for a equals 1 would
    happen for a is more than one.
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    Let's look at the case when I is
    more than one.
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    So this means we're looking at
    the function F of X equals and
  • 20:15 - 20:20
    in this case I will choose a
    equals 2 again to demonstrate
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    what's happening. So F of X
    equals the log of X to base 2.
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    And remember, this is
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    equivalent. The same 2 to the
    power of F of X.
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    Equals X. So as with all
    functions, when we get to this
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    kind of situation, we want to
    start looking at some different
  • 20:43 - 20:47
    values for the argument to help
    us plot a graph of the function.
  • 20:48 - 20:51
    However, before we do that, we
    might just want to take a closer
  • 20:51 - 20:53
    look at what's going on here.
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    Because 2 to the power of F of
    X. In other words, two to some
  • 21:00 - 21:05
    power can never ever be negative
    or 0. This is a positive number.
  • 21:05 - 21:09
    Whatever we choose, and since
    that's more than zero, and
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    that's equal to X, this means
    that X must be more than 0.
  • 21:14 - 21:17
    And X represents our
    arguments, which means I'm
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    only going to look at
    positive arguments for this
  • 21:20 - 21:20
    reason.
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    So start off by looking
    at F of one.
  • 21:26 - 21:30
    Now F of one gives us.
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    The log of one to base 2.
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    North actually means remember.
  • 21:38 - 21:45
    Is that two to some power? Half
    of one is equal to 1, so two to
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    what power will give me one?
    Well, it must be 2 to the power
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    zero. Give me one because
    anything to the power zero gives
  • 21:55 - 22:01
    me one. So this half of one must
    be 0. So therefore F of one
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    equals 0. Next I look
    at F of two.
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    And this is means that we've got
    the log of two.
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    To base 2.
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    And remember that this is
    equivalent to saying that two to
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    some power. In this case F of
    two is equal to two SO2 to what
  • 22:22 - 22:24
    power will give Me 2?
  • 22:25 - 22:30
    Well, we know that 2 to the
    power of one will give Me 2, so
  • 22:30 - 22:33
    this F of two must be equal to
  • 22:33 - 22:40
    1. And also going to look
    at F of F of four.
  • 22:40 - 22:47
    Now F of four means that
    we've got the log of four
  • 22:47 - 22:51
    to base two. Remember, this is
  • 22:51 - 22:53
    equivalent. Just saying that
  • 22:53 - 22:58
    we've got. Two to some
    power. In this case, F
  • 22:58 - 23:00
    of four is equal to 4.
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    Now 2 to what power will give me
  • 23:04 - 23:10
    4? Well, it's actually going to
    be 2 squared, so it must be 2 to
  • 23:10 - 23:14
    the power of two. Will give me
    4, so therefore we can see that
  • 23:14 - 23:16
    F of four will actually give Me
  • 23:16 - 23:22
    2. I also want to consider
    some fractional arguments here
  • 23:22 - 23:24
    between zero and one.
  • 23:25 - 23:28
    So if I look at for example.
  • 23:28 - 23:31
    F of 1/2.
  • 23:32 - 23:38
    OK, this means we've got the log
    of 1/2 to the base 2.
  • 23:38 - 23:41
    And this is equivalent
    to saying.
  • 23:43 - 23:50
    The two to some power. In this
    case F of 1/2 equals 1/2.
  • 23:51 - 23:54
    Now this time it's a little bit
    more tricky to see actually
  • 23:54 - 23:55
    what's going on here.
  • 23:56 - 24:02
    But remember, we can write 1/2
    as 2 to the power of minus one.
  • 24:03 - 24:06
    Because remember about that
    minus power that we talked about
  • 24:06 - 24:10
    earlier on. So 2 to the power of
    sampling equals 2 to the power
  • 24:10 - 24:13
    of minus one. This something
    must be minus one.
  • 24:14 - 24:21
    So therefore. Half
    of 1/2 actually equals minus
  • 24:21 - 24:28
    one. And the final argument
    I want to consider is F of
  • 24:28 - 24:34
    1/4. And this gives us
    the log of 1/4.
  • 24:35 - 24:36
    So the base 2.
  • 24:36 - 24:40
    So this is equivalent to saying.
  • 24:41 - 24:48
    That two to some power. In this
    case F of 1/4 is equal to
  • 24:48 - 24:53
    1/4. And once again, it's not an
    easy step just to see exactly
  • 24:53 - 24:56
    what's going on here. Straight
    away. We want to try and rewrite
  • 24:56 - 24:58
    this right hand side.
  • 24:58 - 25:04
    Now one over 4 is the same as
    one over 2 squared. +2 squared
  • 25:04 - 25:09
    is just the same as four, and
    remember about the minus sign so
  • 25:09 - 25:16
    we can put that onto the top and
    we get 2 to the power of minus
  • 25:16 - 25:21
    two. So here 2 to the power of
    something equals 2 to the power
  • 25:21 - 25:26
    of minus two. That something
    must be minus two, so therefore
  • 25:26 - 25:28
    F of 1/4 equals minus 2.
  • 25:29 - 25:33
    So what we're going to do now?
    We're going to put these results
  • 25:33 - 25:37
    into a table so we can plot a
    graph of the function.
  • 25:38 - 25:44
    So we've got
    X&F of X.
  • 25:44 - 25:51
    Now X values we
    chose. We had one
  • 25:51 - 25:54
    quarter. We had one half.
  • 25:55 - 25:59
    1, two and four.
  • 25:59 - 26:05
    And the corresponding outputs
    Here were minus 2 - 1
  • 26:05 - 26:08
    zero one and two.
  • 26:08 - 26:12
    So let's look at plotting the
    graph of this function now, so
  • 26:12 - 26:14
    as before, we want some axes
  • 26:14 - 26:18
    here. So F of
  • 26:18 - 26:21
    X vertically. And X
  • 26:21 - 26:28
    horizontally. Horizontally we
    need to go from
  • 26:28 - 26:32
    one 3:45, so just
  • 26:32 - 26:38
    go 1234. Includes everything we
    need and vertically we need to
  • 26:38 - 26:40
    go from minus 2 + 2.
  • 26:41 - 26:43
    So minus 1 - 2.
  • 26:43 - 26:48
    I'm one and two, so let's
    put the points.
  • 26:49 - 26:57
    1/4 and minus 2. First of
    all, 1/4 minus two 1/2 -
  • 26:57 - 27:04
    1. So it's 1/2
    - 1 one and 0.
  • 27:05 - 27:06
    21
  • 27:07 - 27:11
    And finally, four and two.
  • 27:12 - 27:16
    So now we want to try and
    draw a smooth curve through
  • 27:16 - 27:19
    the point so we can see the
    graph of the function.
  • 27:20 - 27:28
    Excellent, so this
    is F of
  • 27:28 - 27:36
    X equals the
    log of X.
  • 27:36 - 27:40
    These two. And actually, this
    represents the general shape for
  • 27:40 - 27:45
    functions of the form F of X
    equals the log of X to base a
  • 27:45 - 27:49
    when A is more than one. But
    what happens as we very well,
  • 27:49 - 27:53
    let's have a look at a few
    sketches of some graphs of some
  • 27:53 - 27:57
    different functions and that
    should help us to see what's
  • 27:57 - 28:03
    going on. So if I have
    F of X vertically.
  • 28:03 - 28:07
    And X horizontally.
  • 28:08 - 28:10
    We've just seen.
  • 28:11 - 28:15
    F of
  • 28:15 - 28:21
    X equals. Log
    of X to base 2.
  • 28:22 - 28:24
    And if I was to draw this.
  • 28:27 - 28:29
    This might represent.
  • 28:30 - 28:36
    F of X equals the log of X
    to the base E. Remember E being
  • 28:36 - 28:40
    the number 2.718, the
    exponential number and actually
  • 28:40 - 28:44
    this is called the natural log
    and is sometimes written.
  • 28:45 - 28:46
    LNX
  • 28:47 - 28:50
    And finally, it might have.
  • 28:53 - 28:57
    F of X equals.
  • 28:57 - 29:02
    Log of AXA Base 5 maybe so we
    can see that what's happening
  • 29:02 - 29:05
    here for bigger values of a.
  • 29:06 - 29:09
    The output is increasing more
  • 29:09 - 29:12
    slowly. As the arguments
  • 29:12 - 29:17
    increases. Now a few important
    points to notice here, the first
  • 29:17 - 29:22
    one. It's a notice this point
    here, this one on the X axis.
  • 29:23 - 29:26
    Regardless of our value of A.
  • 29:26 - 29:33
    F of one will always be 0.
    That's true for all values of a
  • 29:33 - 29:37
    here. Second thing to notice is
    something we touched upon
  • 29:37 - 29:40
    earlier on is just point about
    the arguments always being
  • 29:40 - 29:43
    positive and we can see this
    graphically. Here we see we've
  • 29:43 - 29:46
    got no points to the left of the
    F of X axis.
  • 29:47 - 29:51
    And so X is always more than
  • 29:51 - 29:56
    0. The final case
    I want to look at.
  • 29:57 - 30:03
    Is the case worth a is between
    zero and one. And to demonstrate
  • 30:03 - 30:08
    this case I will look at the
    specific example where a equals
  • 30:08 - 30:14
    1/2. And if I equals 1/2, the
    function we're going to be
  • 30:14 - 30:20
    looking at is F of X equals the
    log of X to the base 1/2.
  • 30:21 - 30:27
    So remember we can rewrite this.
    This is equivalent to saying
  • 30:27 - 30:34
    that one half to some power. In
    this case F of X is equal
  • 30:34 - 30:39
    to X. Answer the previous
    example. We need to think
  • 30:39 - 30:42
    carefully about which arguments
    we're going to consider now.
  • 30:42 - 30:48
    Because 1/2 to any power will
    always give me a positive
  • 30:48 - 30:53
    number. In other words, 1/2. So
    the F of X.
  • 30:53 - 30:55
    Is always more than 0.
  • 30:56 - 31:00
    And since this is equal to X,
    this means that X is more than
  • 31:00 - 31:03
    0. And so many going to consider
  • 31:03 - 31:08
    positive arguments. So first of
    all, I can set up F of one.
  • 31:09 - 31:15
    Half of 1 means that we've
    got the log of one to the
  • 31:15 - 31:17
    base of 1/2.
  • 31:18 - 31:20
    Remember, this is equivalent.
  • 31:21 - 31:28
    Just saying.
    1/2 to some power. In this case,
  • 31:28 - 31:32
    F of one is equal to 1.
  • 31:34 - 31:38
    So how does this work? 1/2 to
    some power equals 1. Remember
  • 31:38 - 31:42
    anything to the power of 0
    equals 1, so F of one.
  • 31:43 - 31:46
    Must be 0.
  • 31:46 - 31:52
    Secondly,
    F of two.
  • 31:53 - 31:55
    This gives us the log of two.
  • 31:56 - 31:58
    So the base 1/2.
  • 31:58 - 32:00
    Remember, this is equivalent to
  • 32:00 - 32:07
    saying. That we've got 1/2 to
    some power. In this case F of
  • 32:07 - 32:10
    two. Is equal to 2.
  • 32:11 - 32:15
    So how do we find out what this
    powers got to be?
  • 32:16 - 32:20
    Well, we want to look at
    rewriting this number 2, and
  • 32:20 - 32:24
    this is where the minus negative
    powers come in useful.
  • 32:24 - 32:30
    So we can actually write this as
    1/2 to the power of minus one.
  • 32:30 - 32:36
    If 1/2 to the power of F of two
    equals 1/2 to the power of minus
  • 32:36 - 32:41
    one, then these powers must be
    the same, so F of two equals
  • 32:41 - 32:47
    minus one. After two
    equals minus one.
  • 32:49 - 32:56
    Now look at
    F of four.
  • 32:56 - 33:01
    Half of four gives us the log of
    four, so the base 1/2.
  • 33:02 - 33:07
    Remember, this is equivalent to
    saying that one half to some
  • 33:07 - 33:11
    power. In this case F of four.
  • 33:12 - 33:14
    Must be equal to 4.
  • 33:14 - 33:18
    And once again, we need to think
    about rewriting this right hand
  • 33:18 - 33:23
    side to get a half so we can see
    what the power is.
  • 33:23 - 33:26
    Now for we know we can write us
  • 33:26 - 33:32
    2 squared. And then using our
    negative powers we can rewrite
  • 33:32 - 33:34
    this as one half.
  • 33:34 - 33:40
    So the minus two, so if 1/2 to
    the power of F of four equals
  • 33:40 - 33:45
    1/2 to the power of minus two,
    then these powers must be equal.
  • 33:45 - 33:47
    So F of four equals minus 2.
  • 33:48 - 33:54
    And also we want to
    consider some fractional
  • 33:54 - 34:00
    arguments. So let's look at
    F of 1/2.
  • 34:00 - 34:03
    Half of 1/2.
  • 34:03 - 34:08
    Gives us the log of 1/2
    to the base 1/2.
  • 34:09 - 34:13
    So lots of haves and this is
    equivalent to saying we've got
  • 34:13 - 34:19
    1/2 to the power of F of 1/2.
    That's going to be equal to 1/2.
  • 34:19 - 34:23
    This first sight might seem a
    little bit complicated, but it's
  • 34:23 - 34:28
    not at all because 1/2 to some
    power to give me 1/2. Well
  • 34:28 - 34:33
    that's just half to the power of
    1, so this F of 1/2 is actually
  • 34:33 - 34:34
    equal to 1.
  • 34:35 - 34:42
    And finally, like
    to consider F
  • 34:42 - 34:49
    of 1/4. Which is
    equal to the log of 1/4 to
  • 34:49 - 34:51
    the base of 1/2.
  • 34:52 - 34:58
    I remember this is equivalent to
    writing 1/2 to some power. In
  • 34:58 - 35:05
    this case F of 1/4 is equal
    to 1/4, and again, what we need
  • 35:05 - 35:10
    to do is just think about
    rewriting the right hand side
  • 35:10 - 35:16
    and actually this is the same as
    one over 2 squared.
  • 35:16 - 35:19
    Which is the same as one half.
  • 35:20 - 35:26
    Squad So 1/2
    to some power equals 1/2
  • 35:26 - 35:32
    squared. The powers must be
    equal, so F of 1/4 must equal 2.
  • 35:32 - 35:35
    After 1/4 equals 2.
  • 35:36 - 35:41
    So as usual, put these results
    into a table so we can plot a
  • 35:41 - 35:43
    graph of the function.
  • 35:44 - 35:48
    It's just the table over here
    X&F of X.
  • 35:51 - 35:54
    Arguments were one
  • 35:54 - 36:00
    quarter 1/2. 1,
    two and four.
  • 36:01 - 36:07
    And the corresponding outputs.
    There were two one 0 -
  • 36:07 - 36:14
    1 and minus two. So let's
    plot these points. So first of
  • 36:14 - 36:16
    all some axes.
  • 36:16 - 36:17
    F of X.
  • 36:18 - 36:25
    X. Vertically, we need
    to go from minus 2 + 2, so no
  • 36:25 - 36:27
    problems minus 1 - 2.
  • 36:27 - 36:30
    One and two.
  • 36:30 - 36:33
    And horizontally we need to go
    all the way up to four.
  • 36:33 - 36:37
    So 1 two.
  • 36:37 - 36:44
    314 Let's
    plot the points 1/4 and two.
  • 36:45 - 36:49
    1/2 and
    one.
  • 36:51 - 36:53
    One and 0.
  • 36:54 - 36:57
    Two negative one.
  • 36:58 - 37:01
    Four and negative
  • 37:01 - 37:07
    2. House before going to try
    and draw a smooth curve
  • 37:07 - 37:08
    through these points.
  • 37:16 - 37:23
    OK, excellent and this is F of X
    equals the log of X to base 1/2.
  • 37:23 - 37:27
    Actually this demonstrates the
    general shape for functions of
  • 37:27 - 37:35
    the form F of X equals log of
    X to the base were a is equal
  • 37:35 - 37:37
    to the number between zero and
  • 37:37 - 37:42
    one. But what happens as a
    varies within those boundaries?
  • 37:42 - 37:47
    Well, by looking at the sketch
    of a few functions like that, we
  • 37:47 - 37:51
    should be able to see what's
    going on. So just do my axes.
  • 37:53 - 37:54
    F of X.
  • 37:55 - 38:00
    And X. And
    we've just seen.
  • 38:04 - 38:11
    F of X equals log
    of X. It's a base
  • 38:11 - 38:15
    1/2. Well, we might have had.
  • 38:16 - 38:17
    Something that looked like.
  • 38:18 - 38:24
    This. This might have been
    F of X equals.
  • 38:24 - 38:28
    The log of X so base one
    over E. Remember either
  • 38:28 - 38:31
    exponential number or we
    might even have hard.
  • 38:35 - 38:39
    Something which looked like
    this. This might have been F of
  • 38:39 - 38:42
    X equals. Log of X.
  • 38:43 - 38:48
    Base 1/5
    So what's happening for
  • 38:48 - 38:52
    different values of a well, we
    can see that for the bigger
  • 38:52 - 38:57
    values of a. The output
    decreases more quickly as
  • 38:57 - 38:59
    the arguments increases.
  • 39:01 - 39:05
    As a couple of other important
    points to notice here as well,
  • 39:05 - 39:11
    firstly, is this .1 again on the
    X axis, and in fact we notice
  • 39:11 - 39:16
    that F of one equals 0
    regardless of our value of A and
  • 39:16 - 39:18
    that's true for any value.
  • 39:19 - 39:23
    Secondly, is once again this
    thing about the positive
  • 39:23 - 39:27
    arguments we can see
    graphically. Once again, the X
  • 39:27 - 39:30
    has to be more than 0.
  • 39:31 - 39:35
    So now we know what happens
    for all the different values
  • 39:35 - 39:38
    of a when we considering
    logarithm functions.
  • 39:39 - 39:43
    Once again, you may have
    notice some symmetry.
  • 39:44 - 39:50
    This time the symmetry was
    centered on the X axis.
  • 39:51 - 39:55
    If I actually draw two of
    the curves here.
  • 40:01 - 40:07
    This one might have represented
    F of X equals the log of X to
  • 40:07 - 40:14
    base 2. This will might
    represent F of X equals log of X
  • 40:14 - 40:19
    to the base 1/2, and then forget
    very important .1.
  • 40:19 - 40:21
    And we can see here that
  • 40:21 - 40:27
    actually. The base two function
    is a reflection in the X axis of
  • 40:27 - 40:31
    the function, which has a base
    1/2, and in fact that could have
  • 40:31 - 40:37
    been five and one, 5th or E and
    one over E as the base,
  • 40:37 - 40:44
    generally speaking. F of X
    equals the log of X to
  • 40:44 - 40:47
    base A. Is
  • 40:47 - 40:50
    a reflection.
  • 40:50 - 40:57
    Of.
    F of X equals.
  • 40:58 - 41:04
    Log of X to base one over A That
    is in the X axis.
  • 41:04 - 41:10
    Accent, so now we've looked at
    reflections in the X axis in the
  • 41:10 - 41:15
    F of X axis. We've looked at
    exponential functions, and we've
  • 41:15 - 41:20
    looked at logarithm functions.
    The final thing I'd like to look
  • 41:20 - 41:26
    at in this video is whether
    there is a link between these
  • 41:26 - 41:31
    two functions. Firstly, the
    function F of X equals Y to the
  • 41:31 - 41:33
    X. Remember the exponential
  • 41:33 - 41:37
    function. And the second
    function F of X equals the
  • 41:37 - 41:40
    natural log of X, which we
    mentioned briefly earlier on
  • 41:40 - 41:44
    now. Remember, this means the
    log of X to base A.
  • 41:45 - 41:49
    Well, good place to start would
    be to look at the graphs of the
  • 41:49 - 41:54
    functions. So I'll do that now.
    We've got F of X.
  • 41:56 - 42:01
    And X. And remember, F of X
    equals E to the X.
  • 42:05 - 42:10
    Run along like this. Is this
    important? .1 F of X axis.
  • 42:10 - 42:14
    Stuff of X equals E to the X.
  • 42:15 - 42:17
    And if of X equals
    a natural log of X.
  • 42:18 - 42:20
    Came from here.
  • 42:21 - 42:25
    I'm run along something like
    this once again going through
  • 42:25 - 42:29
    that important .1 on the X axis.
  • 42:29 - 42:31
    Sex equals and natural log of X.
  • 42:31 - 42:37
    Now instead of link, I think
    helpful line to draw in here is
  • 42:37 - 42:39
    this dotted line.
  • 42:41 - 42:42
    This dotted line.
  • 42:43 - 42:49
    Represents the graph of the
    linear function F of X equals X.
  • 42:49 - 42:55
    I want to put that in. We can
    see almost immediately that the
  • 42:55 - 42:58
    exponential function is a
    reflection of the natural
  • 42:58 - 43:03
    logarithm function in the line F
    of X equals X. What does that
  • 43:03 - 43:07
    mean? Well, this is equivalent
    to saying that the axes have
  • 43:07 - 43:12
    been swapped around, so to move
    from this function to this
  • 43:12 - 43:14
    function, all my ex file use.
  • 43:14 - 43:20
    Have gone to become F of X
    values and all my F of X
  • 43:20 - 43:24
    values have gone to become
    X values. In other words,
  • 43:24 - 43:27
    the inputs and outputs
    have been swapped around.
  • 43:28 - 43:30
    So In summary.
  • 43:31 - 43:38
    This means that the function F
    of X equals E to the X.
  • 43:38 - 43:41
    Is the inverse?
  • 43:42 - 43:48
    Of the function F
    of X equals a
  • 43:48 - 43:52
    natural log of X.
Title:
www.mathcentre.ac.uk/.../Exponential%20and%20logarithm%20functions.mp4
Video Language:
English

English subtitles

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