JVR12
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I have the equation y = a *10^{bt}, where a and b are constant to be determined.

How do I get the gradient and intercept from the log equation of this? Or am I just seeing it wrong?
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JVR12
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Also, log(y) is obviously the y axis and t is on the x axis.
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RDKGames
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(Original post by JVR12)
I have the equation y = a *10^{bt}, where a and b are constant to be determined.

How do I get the gradient and intercept from the log equation of this? Or am I just seeing it wrong?
You can't get any specific ones without any more info.

Otherwise, take logs of both sides and apply log rules.
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JVR12
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(Original post by RDKGames)
You can't get any specific ones without any more info.

Otherwise, take logs of both sides and apply log rules.
Ok so the question is:

The numbers of flu viruses detected in the first few weeks of the 2012 - 2013 flu epidemic in the UK were as follows

week 1 = 7
week 2 = 10
week 3 = 24
week 4 = 32
week 5 = 40
week 6 = 38
week 7 = 63
week 8 = 96
week 9 = 234
week 10 = 480

These data may be modelled by an equation of the form y = a * 10^{bt}, where y is the number of flu viruses detected in week t of the epidemic, and a and b are constants to be determined.

(i) Explain why this model leads to a straight line graph of log(y) against t. State the gradient and intercept of this graph in terms of a and b.

Later we need to plot the graph and find said values of a and b. Am I being incredibly stupid again?
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RDKGames
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(Original post by JVR12)
Ok so the question is:

The numbers of flu viruses detected in the first few weeks of the 2012 - 2013 flu epidemic in the UK were as follows

week 1 = 7
week 2 = 10
week 3 = 24
week 4 = 32
week 5 = 40
week 6 = 38
week 7 = 63
week 8 = 96
week 9 = 234
week 10 = 480

These data may be modelled by an equation of the form y = a * 10^{bt}, where y is the number of flu viruses detected in week t of the epidemic, and a and b are constants to be determined.

(i) Explain why this model leads to a straight line graph of log(y) against t. State the gradient and intercept of this graph in terms of a and b.

Later we need to plot the graph and find said values of a and b. Am I being incredibly stupid again?
No, just take logs and put it in the form Y=mt+c where Y=\log(y)
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JVR12
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(Original post by RDKGames)
No, just take logs and put it in the form Y=mt+c where Y=\log(y)
I've done that, and all I'm getting is log(y) = btlog(a), and I'm not sure what the gradient would be, as t is the x axis and both b and log(a) are to be determined?
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RDKGames
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(Original post by JVR12)
I've done that, and all I'm getting is log(y) = btlog(a), and I'm not sure what the gradient would be, as t is the x axis and both b and log(a) are to be determined?
That is clearly not correct because if I take everything as the exponent of 10 I get y=a^{bt}

Try again and show your steps.
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JVR12
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(Original post by RDKGames)
That is clearly not correct because if I take everything as the exponent of 10 I get y=a^{bt}

Try again and show your steps.
with y=a^{bt} i then get log_a y = bt, which would mean a = 10 and the gradient is just b?

Sorry this is taking long to get through my thick skull, this is last years C2 paper and my teacher will have my neck if I don't answer this question (im terrible with log graphs)
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RDKGames
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(Original post by JVR12)
with y=a^{bt} i then get log_a y = bt, which would mean a = 10 and the gradient is just b?

Sorry this is taking long to get through my thick skull, this is last years C2 paper and my teacher will have my neck if I don't answer this question (im terrible with log graphs)
That's not what I-- Never mind. (at least you got the gradient right)

You're over-complicating it.

y=a\cdot 10^{bt} \Leftrightarrow \log(y)=\log(a\cdot 10^{bt}) \Leftrightarrow Y=\log(a)+bt\log(10) so the gradient and Y-intercepts are.....?
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JVR12
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(Original post by RDKGames)
That's not what I-- Never mind. (at least you got the gradient right)

You're over-complicating it.

y=a\cdot 10^{bt} \Leftrightarrow \log(y)=\log(a\cdot 10^{bt}) \Leftrightarrow Y=\log(a)+bt\log(10) so the gradient and Y-intercepts are.....?
OH MY GOD I'M INCREDIBLY THICK!! I guess this is why you don't lose your formula sheet that has all the log laws written out

Thanks a lot for the help, I thought I was fairly decent with log laws (i always get full marks on the individual questions, just not graphs)....

All of this because I logged the entire thing individually....
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JVR12
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(Original post by RDKGames)
That's not what I-- Never mind. (at least you got the gradient right)

You're over-complicating it.

y=a\cdot 10^{bt} \Leftrightarrow \log(y)=\log(a\cdot 10^{bt}) \Leftrightarrow Y=\log(a)+bt\log(10) so the gradient and Y-intercepts are.....?
I realised my mistake before you replied, but can't thank you enough for your help (i'm incredibly thick)
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RDKGames
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(Original post by JVR12)
I realised my mistake before you replied, but can't thank you enough for your help (i'm incredibly thick)
No worries
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