WEBVTT

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Hi, I'm Fiona.

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This is Tamila 2020 paper one.

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In question 15,
we're asked to find the positive

b1e6e8ce-52da-473a-8ab7-e04240b0c38f-1
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difference between the two real values of
X for which log to the base 2 of X all to

b1e6e8ce-52da-473a-8ab7-e04240b0c38f-2
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the power of 4 + 12,
log to the base 2 of 1 / X all squared -2

b1e6e8ce-52da-473a-8ab7-e04240b0c38f-3
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to the power of 6 = 0.

6860b5f9-c51f-452b-b283-63f1157a213b-0
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And here are our options to choose from
once we get to that point in our solving.

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OK, let's jump into this question.

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And we might be tempted to think, oh,
how am I going to solve this equation for

d1a06286-6a0e-4b27-8729-581ba29a8e21-1
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X?

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But typical of Tamua,
this question is designed for us to be

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able to spot an underlying structure that
will unlock the question.

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So take a moment now to pause the video
and see if you can spot an underlying

1b3316e5-86d2-4539-96db-7326e48c589a-1
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structure in this equation that will
really blow open this question and allow

1b3316e5-86d2-4539-96db-7326e48c589a-2
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us to solve it very efficiently.

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OK,
the trick lies in this second term here,

d509a035-151d-4d96-93b3-bbf561623ee5-1
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because log to the base 2 of 1 / X is the
same as log to the base 2 of X to the

d509a035-151d-4d96-93b3-bbf561623ee5-2
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power of -1.

c3cfa9f9-fcf1-47fa-a490-47a75561d78d-0
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And using my laws of the logs,
I can take this power over here.

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And that is the same as minus log to the
base 2 of X.

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But of course, in this term,
this expression is being squared.

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That means its value will be the same
whether we square it's negative or

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whether we square it's positive.

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So now we are able to spot that the
underlying structure of this equation is

c1b90f78-92d4-46ab-905f-916960bb66ac-1
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in fact a quadratic in log to the base 2
of XO squared.

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I'm going to write that out on the board
to save time.

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So here's the equation that we're going
to solve in a form where the underlying

66a5d230-5f23-4f5c-808b-1db69d2a4bb0-1
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structure of this quadratic is very
obvious.

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So now what I need to do is to factorise
this quadratic.

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2 to the power of 6 is 64 and so when I'm
factorising this I'm looking for two

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factors that add together to give 12 so
that will be +16 -, 4.

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This factor here because my quadratic is
in log the base 2 of XL squared it will

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give non real solutions and so I'm going
to focus on this factor here which will

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give a solution of log to the base 2 of X
= ± sqrt 4 which is 2.

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So I've got log to the base 2 of X = a
positive 2 and log to the base 2 of X = -

f3e6911a-7759-4394-b610-0c08bfc5f284-1
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2.

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This equation is going to give a solution
of X = 4 and this equation is going to

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give a solution of X = 1/4.

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I need the positive difference between
these two real values of X which satisfy

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this equation.

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And the positive difference I'm going to
find by taking 4 -, 1/4,

cd997054-b2fc-4070-afe7-9e2007388b1f-1
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four is 16 quarters.

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Take away one of those quarters and I get
15 quarters, which is option C.

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So that's the answer to this question.

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Now let's take a moment to reflect.

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This question involved a hidden quadratic
in order to unlock the question and 1st

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to get into the flow of things.

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If you've never come across something
like that before,

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then now you can add that to your
mathematical toolkit as something to be

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looking out for because it can come up in
these kinds of exams.

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When we're looking at problem solving in
this way,

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you might have been tempted to substitute
log to the base 2 of XO squared for

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something like Y and then solve your
equation for Y.

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And I would avoid doing something like
that because then you might not be able

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to as easily spot that this factor
doesn't give you real solutions.

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And also you might forget to convert back
to find solutions for X to get this

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positive difference.

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For my final reflection on this question,
I want to focus on this part of the board

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because some people find it hard to
deduce from log to the base 2 of X = 2

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that X takes a value of four.

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So if we think of a typical logarithmic
expression,

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we might have something like log to a
base, which I'll call B of a number,

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which I'll call north.

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And the value of this expression is the
power on the base to get my number N And

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so I'm going to set that equal to a value
and I'll call it P because logs are all

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about the power.

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I often think of the early 90s song that
many people are familiar with that goes,

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I got the power when I'm doing logarithm,
when I'm dealing with logs and

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logarithmic expressions,
because logs are all about the power.

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Sometimes I even imagine this little box
on my base B because the value of this

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expression, I'll say it again,
is the power that goes on my base to get

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my number N and that gives me my P.

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So looking back down here,
hopefully now you can see that this

e4d10c1d-e9ae-44f1-ab9a-34650d481e7a-1
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equation,
log to the base 2 of X = 2 gives a value

e4d10c1d-e9ae-44f1-ab9a-34650d481e7a-2
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of X = 4 because I get the four from 2 to
the power of 2.

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The other result we had was log to the
base 2 of X = - 2 and I can see now that

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X should be 2 to the power of -2 which is
1 / 2 ^2 and that gives me my value of

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1/4 which I found earlier.