Logarithms and exponential functions

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Shas72
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The temp T degree Celsius of a hot drink , t mins after it's made, can be modelled by the equation T=25+Ke^(-nt), where k and n are constants.
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Shas72
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(Original post by Shas72)
The temp T degree Celsius of a hot drink , t mins after it's made, can be modelled by the equation T=25+Ke^(-nt), where k and n are constants.
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Shas72
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(Original post by Shas72)
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I converted to linear equation and got ln T= -nt+lnk +ln 25.
When I find the gradient after converting T to ln T, I get n=0.04. But the ans is 0.08.
I dont get it
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mqb2766
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(Original post by Shas72)
I converted to linear equation and got ln T= -nt+lnk +ln 25.
When I find the gradient after converting T to ln T, I get n=0.04. But the ans is 0.08.
I dont get it
That conversion isn't right. How did you take logs?
Last edited by mqb2766; 1 month ago
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Shas72
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(Original post by mqb2766)
That conversion isn't right. How did you take logs?
I did lnT=ln25 +lnk-nt
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mqb2766
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(Original post by Shas72)
I did lnT=ln25 +lnk-nt
But you've said
log(A+B) = logA) + log(B)
Which is wrong
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Shas72
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(Original post by mqb2766)
But you've said
log(A+B) = logA) + log(B)
Which is wrong
I did not understand. So should I move 25 to the left to make T-25=ke^-nt
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mqb2766
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(Original post by Shas72)
I did not understand. So should I move 25 to the left to make T-25=ke^-nt
Yes
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Shas72
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(Original post by mqb2766)
Yes
Thanks
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