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FP1 - AQA Linear Laws

The question shows a straight line graph passing through the points (0,2) and (3,6). The axes are log y and log x.

The question asks to express y in the form ax^b where a and b are constants to be determined to 3 sf.


So I did,

logy = mlogx + c
log y = (4/3)logx +2
log y = logx^(4/3) + log 100
log y = log(100x^(4/3))
y = 100x^(4/3)


The 100 should be 7.39.

Please point out the errors

Thanks
Reply 1
Avatar for davros
davros
16
Original post by maggiehodgson
The question shows a straight line graph passing through the points (0,2) and (3,6). The axes are log y and log x.

The question asks to express y in the form ax^b where a and b are constants to be determined to 3 sf.


So I did,

logy = mlogx + c
log y = (4/3)logx +2
log y = logx^(4/3) + log 100
log y = log(100x^(4/3))
y = 100x^(4/3)


The 100 should be 7.39.

Please point out the errors

Thanks


You're using base-10 logs when you should be using natural (base-e) logs :smile:
Original post by davros
You're using base-10 logs when you should be using natural (base-e) logs :smile:



The diagram has the axes labelled log10 and the question says the graph of log10y against log10x.


So is the answer in the book incorrect for the question given? It looks like it, from what you say and indeed e^2 gives the answer. The book's misled me hasn't it.

Thanks.
Reply 3
Avatar for davros
davros
16
Original post by maggiehodgson
The diagram has the axes labelled log10 and the question says the graph of log10y against log10x.


So is the answer in the book incorrect for the question given? It looks like it, from what you say and indeed e^2 gives the answer. The book's misled me hasn't it.

Thanks.


Looks like it! Your (0, 2) point says c = log a = 2 so if logs are based 10 you get a = 100 as you did; if you take natural logs you get the book's value.

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