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y=mx+c and Logs help! Please Watch

1. ok so can anyone tell me how you could re-arange this equation (T^2=(4pie^2*A*B)/C*D)) into something that looks like y-mx+c using Logs Where T^2=y
and B=x.
Any help would be appreciated
2. Anyone?
3. (Original post by ryan051991)
Anyone?
This looks too similar to an AQA Physics ISA I did on simple harmonic motion... And it doesn't involve logs by the way.
4. (Original post by Freerider101)
This looks too similar to an AQA Physics ISA I did on simple harmonic motion... And it doesn't involve logs by the way.
]

yeh dude ive done it i was just so confused about it coz i thought you would have to do logs to get a (+c) because it did not go through origin
5. (Original post by ryan051991)
ok so can anyone tell me how you could re-arange this equation (T^2=(4pie^2*A*B)/C*D)) into something that looks like y-mx+c using Logs Where T^2=y
and B=x.
Any help would be appreciated
From that equation you simply want to write

c=0
6. Ah dude i am totally lost think i am just never going to get it LOL
7. (Original post by ryan051991)
]

yeh dude ive done it i was just so confused about it coz i thought you would have to do logs to get a (+c) because it did not go through origin
There is a good reason for that.
8. (Original post by Freerider101)
This looks too similar to an AQA Physics ISA I did on simple harmonic motion... And it doesn't involve logs by the way.
Yeah you can Log both sides as it would give you m x and c (where aplicable) so would show relationship

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