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2006 paper 1 question 10 - HSN forum

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2006 paper 1 question 10


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#1 Amazoph

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Posted 14 May 2007 - 09:01 PM

Clueless as what to do here...

Two variables, x and y, are connected by the law  y = a ^ x . The graph of log to the base 4 y against x is a straight line passing through the origin and the point A(6,3). Find the value of a.

#2 BuckminsterFullerene

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Posted 14 May 2007 - 09:05 PM

Hello Amazoph

Firstly
For all questions as such, you begin with what they've given you ie y=a^x. They have also told you that their graph is of log (base) 4 and x.

Hence, you should, take the log(base) 4 of both sides

Ie. Log(4)y=Log(4)a^x
Log (4)y = x . log(4)a

You should now recognise that log(4)a is the gradient because this is similar to y=mx+c
hence;

You know the gradient of the line is 0.5
so

Log(4)a=0.5

Hence

4^0.5 = a
a = 2

Glad to help!
Chemical Engineering - Edinburgh

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#3 antollo

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Posted 17 April 2008 - 09:51 PM

QUOTE(BuckminsterFullerene @ May 14 2007, 10:05 PM) View Post
Hello Amazoph

Firstly
For all questions as such, you begin with what they've given you ie y=a^x. They have also told you that their graph is of log (base) 4 and x.

Hence, you should, take the log(base) 4 of both sides

Ie. Log(4)y=Log(4)a^x
Log (4)y = x . log(4)a

You should now recognise that log(4)a is the gradient because this is similar to y=mx+c
hence;

You know the gradient of the line is 0.5
so

Log(4)a=0.5

Hence

4^0.5 = a
a = 2

Glad to help!


hi,
i follow how u get the gradient to equal log(4)a,
but im confused how you got the gradient of the line to be 0.5 =/

it's probably something really simple, but i cant seem to figure it out


#4 Steve

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Posted 21 April 2008 - 10:44 PM

It's just because of the information you are given. You know that the line passes through (0,0) and (6,3), so the gradient is 1/2.
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