C3 Maths help please [URGENT] Watch

LL2018
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Where do I even start with this question ? (C3 Solomon paper D)

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LL2018
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username4011812
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Not sure about the first but for the second I would do
ln x^2/e = lnx^2 - ln e
= lnx^2 - 1
= 2lnx - 1
= 2y - 1

I'm not sure if that's right tho you'll have to check x
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ladifas
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Here are some clues.

For part (i):
Let a = log2(x). Now consider what that actually means (i.e. turn it into a power).

For part (ii):
Think about your log laws. Try expanding into multiple ln(something) terms.
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JoshMach65
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y = ln(x), therefore x = e^y.

i) logbase(2)(x) = logbase(2)(e^y) = ylogbase(2)(e).

ii) ln(x^2/e) = ln(x^2) - ln(e) = 2y - 1.

Part ii gives a nicer answer.
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FluuidKarma
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(Original post by LL2018)
Question
write x in terms of y
lnx=y
therefore
e^y=x

then it's just
i)
log2(e^y)
(y)log2(e)

the answer above me is right for q2
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LL2018
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Thanks all!
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LL2018
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(Original post by FluuidKarma)
write x in terms of y
lnx=y
therefore
e^y=x

then it's just
i)
log2(e^y)
(y)log2(e)

the answer above me is right for q2
I've worked out the first two parts but how would I go about solving the last part? (part b):
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LL2018
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(Original post by JoshMach65)
y = ln(x), therefore x = e^y.

i) logbase(2)(x) = logbase(2)(e^y) = ylogbase(2)(e).

ii) ln(x^2/e) = ln(x^2) - ln(e) = 2y - 1.

Part ii gives a nicer answer.
How can I solve part b?
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FluuidKarma
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(Original post by LL2018)
I've worked out the first two parts but how would I go about solving the last part? (part b):
just sub in what you found in part i and ii into the new equation and it should just be basic algebra from there
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LL2018
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(Original post by ladifas)
Here are some clues.

For part (i):
Let a = log2(x). Now consider what that actually means (i.e. turn it into a power).

For part (ii):
Think about your log laws. Try expanding into multiple ln(something) terms.
Could you help me with part b please?Name:  c3 question 2.PNG
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LL2018
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(Original post by FluuidKarma)
just sub in what you found in part i and ii into the new equation and it should just be basic algebra from there
This is the solomon answer which I have no idea how they derived: Name:  c3 2.PNG
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M4cc4n4
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(Original post by LL2018)
This is the solomon answer which I have no idea how they derived: Name:  c3 2.PNG
Views: 28
Size:  11.0 KB
Solomon used C2 Log rules to get the answer for part a (changing the base)

using their method you will and up with
Name:  1529342698240211420577338549223.jpg
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