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    This question is from the specimen paper.

    Question 6f

    I have the equation.

     -1.8 = -6e^{-0.1t}

    When I solve it, t = 12 except that's not the answer....the answer's 38 :confused: Help please?

    Thank you!
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    I get 10ln(10/3). What's your working?
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    I get 12.0379 aswell are you sure 38 is the correct answer?
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    (Original post by Drederick Tatum)
    I get 10ln(10/3). What's your working?
    My working is:

    -1.8/ -6 = e^{-0.1t}

    0.3 = e^{-0.1t}

    ln(0.3) = -0.1t

    ln(0.3)/-0.1 = t

    t = 12.039....
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    (Original post by Apple24)
    I get 12.0379 aswell are you sure 38 is the correct answer?
    The mark scheme says:

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    I get 12 too.
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    You found t (time) correctly, but the question asks for capital T, temperature. To find this, substitute your value for t into the T=... equation on the second line of the question.
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    (Original post by gottastudy)
    The mark scheme says:

    1) You're clearly going wrong somewhere as the question's presumably asking you to work out T, where as you're working out t.

    2) It's always best to give the full question.
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    Ahh I've got it.

    From reading the original question in the paper you have to sub the rate at what you differentiated which is 12 back into the original equation to find T and you will get 38.
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    (Original post by ttoby)
    You found t (time) correctly, but the question asks for capital T, temperature. To find this, substitute your value for t into the T=... equation on the second line of the question.
    (Original post by Swayum)
    1) You're clearly going wrong somewhere as the question's presumably asking you to work out T, where as you're working out t.

    2) It's always best to give the full question.
    Whoops! Thank you! I've substituted t into the original and now have 38. :yes: Thank you all!
 
 
 
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